Solutions Class 12 Chemistry NCERT Solutions

Q1Exercises

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Solution

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

There are nine types of solutions, based on the physical state of the solute and the solvent. These are:

Gaseous Solutions:
- Gas in Gas: A mixture of two or more gases. Example: A mixture of oxygen and nitrogen gases (air).
- Liquid in Gas: Liquid particles dispersed in a gaseous medium. Example: Chloroform mixed with nitrogen gas, or water vapour in air.
- Solid in Gas: Solid particles dispersed in a gaseous medium. Example: Camphor in nitrogen gas, or dust particles in air.
Liquid Solutions:
- Gas in Liquid: A gas dissolved in a liquid. Example: Oxygen dissolved in water, or carbon dioxide in soda water.
- Liquid in Liquid: A liquid dissolved in another liquid. Example: Ethanol dissolved in water.
- Solid in Liquid: A solid dissolved in a liquid. Example: Glucose or salt dissolved in water.
Solid Solutions:
- Gas in Solid: Gas particles trapped in a solid matrix. Example: Solution of hydrogen in palladium.
- Liquid in Solid: A liquid dispersed in a solid. Example: Amalgam of mercury with sodium.
- Solid in Solid: A solid dissolved in another solid. Example: Copper dissolved in gold (alloys like brass).

Q2Exercises

Give an example of a solid solution in which the solute is a gas.

Solution

An example of a solid solution in which the solute is a gas is the solution of hydrogen gas in palladium metal. In this solution, hydrogen gas (solute) is adsorbed into the solid palladium (solvent).

Q3Exercises

Define the following terms:

(i)

Mole fraction

(ii)

Molality

(iii)

Molarity

(iv)

Mass percentage.

Solution

(i)

Mole fraction ( $x$ ): The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution. For a component A in a solution of components A and B:

x_A = \frac{\text{Moles of A}}{\text{Moles of A} + \text{Moles of B}} = \frac{n_A}{n_A + n_B}

(ii)

Molality ( $m$ ): Molality is defined as the number of moles of the solute dissolved in 1 kilogram (kg) of the solvent.

m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Its unit is mol/kg or molal (m).

(iii)

Molarity ( $M$ ): Molarity is defined as the number of moles of the solute dissolved in 1 litre (L) of the solution.

M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}

Its unit is mol/L or molar (M).

(iv)

Mass percentage (w/w): The mass percentage of a component in a solution is the mass of that component per 100 grams of the solution.

\text{Mass \%} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100

Q4Exercises

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?

Solution

Given: Mass percentage of nitric acid (

HNO_3

) = 68% by mass. Density of the solution (

d

) = 1.504 g

mL^{-1}

.

To Find: Molarity of the nitric acid solution.

Calculation: 68% by mass means that 100 g of the solution contains 68 g of

HNO_3

. Mass of

HNO_3

= 68 g Mass of solution = 100 g

First, calculate the molar mass of

HNO_3

: Molar mass of

HNO_3

= 1 (H) + 14 (N) + 3

\times

16 (O) = 1 + 14 + 48 = 63 g

mol^{-1}

.

Next, calculate the number of moles of

HNO_3

in 68 g: Moles of

HNO_3

=

\frac{\text{Mass}}{\text{Molar mass}} = \frac{68 \text{ g}}{63 \text{ g mol}^{-1}} = 1.079 \text{ mol}

.

Now, calculate the volume of 100 g of the solution using its density: Volume of solution =

\frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{100 \text{ g}}{1.504 \text{ g mL}^{-1}} = 66.49 \text{ mL}

.

Convert the volume to litres: Volume in L =

66.49 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.06649 \text{ L}

.

Finally, calculate the molarity: Molarity (M) =

\frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{1.079 \text{ mol}}{0.06649 \text{ L}} = 16.23 \text{ M}

.

Final Answer: The molarity of the concentrated nitric acid sample is 16.23 M.

Q5Exercises

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL⁻¹, then what shall be the molarity of the solution?

Solution

Given: Glucose solution is 10% w/w. Density of solution (

d

) = 1.2 g

mL^{-1}

.

To Find:

Molality ( $m$ )
Mole fraction of glucose ( $x_{glucose}$ ) and water ( $x_{water}$ )
Molarity ( $M$ )

Calculation: 10% w/w means 10 g of glucose is present in 100 g of solution. Mass of glucose (

w_2

) = 10 g. Mass of solution = 100 g. Mass of water (

w_1

) = Mass of solution - Mass of glucose = 100 g - 10 g = 90 g.

Molar mass of glucose (

C_6H_{12}O_6

) = 6

\times

12 + 12

\times

1 + 6

\times

16 = 180 g

mol^{-1}

. Molar mass of water (

H_2O

) = 2

\times

1 + 16 = 18 g

mol^{-1}

.

Moles of glucose (

n_2

) =

\frac{10 \text{ g}}{180 \text{ g mol}^{-1}} = 0.0555 \text{ mol}

. Moles of water (

n_1

) =

\frac{90 \text{ g}}{18 \text{ g mol}^{-1}} = 5 \text{ mol}

.

1. Molality ( $m$ ): Mass of solvent (water) = 90 g = 0.090 kg. Molality =

\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0555 \text{ mol}}{0.090 \text{ kg}} = 0.617 \text{ m}

.

2. Mole fraction ( $x$ ): Total moles =

n_1 + n_2 = 5 + 0.0555 = 5.0555 \text{ mol}

.

x_{glucose} = \frac{n_2}{n_1 + n_2} = \frac{0.0555}{5.0555} = 0.011

.

x_{water} = \frac{n_1}{n_1 + n_2} = \frac{5}{5.0555} = 0.989

. (Check:

0.011 + 0.989 = 1.000

)

3. Molarity ( $M$ ): Mass of solution = 100 g. Density of solution = 1.2 g

mL^{-1}

. Volume of solution =

\frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.2 \text{ g mL}^{-1}} = 83.33 \text{ mL}

. Volume in L =

83.33 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.08333 \text{ L}

. Molarity =

\frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.0555 \text{ mol}}{0.08333 \text{ L}} = 0.666 \text{ M}

.

Final Answer:

Molality of the solution is 0.617 m.
Mole fraction of glucose is 0.011 and mole fraction of water is 0.989.
Molarity of the solution is 0.666 M.

Q6Exercises

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Solution

Given: Total mass of mixture (

Na_2CO_3

and

NaHCO_3

) = 1 g. The mixture contains equimolar amounts of both compounds. Molarity of HCl solution = 0.1 M.

To Find: Volume of 0.1 M HCl required.

Calculation: Let the number of moles of

Na_2CO_3

be

x

mol. Since the mixture is equimolar, the number of moles of

NaHCO_3

is also

x

mol.

Molar mass of

Na_2CO_3

= 2

\times

23 + 12 + 3

\times

16 = 46 + 12 + 48 = 106 g

mol^{-1}

. Molar mass of

NaHCO_3

= 23 + 1 + 12 + 3

\times

16 = 84 g

mol^{-1}

.

Mass of

Na_2CO_3

in the mixture =

x \times 106

g. Mass of

NaHCO_3

in the mixture =

x \times 84

g.

Total mass of the mixture = 1 g.

106x + 84x = 1

190x = 1

x = \frac{1}{190} = 0.005263 \text{ mol}

. So, moles of

Na_2CO_3

= 0.005263 mol and moles of

NaHCO_3

= 0.005263 mol.

The reactions with HCl are:

Na_2CO_3(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)

NaHCO_3(s) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) + CO_2(g)

From the stoichiometry: 1 mole of

Na_2CO_3

reacts with 2 moles of HCl. Moles of HCl for

Na_2CO_3

=

2 \times 0.005263 = 0.010526 \text{ mol}

.

1 mole of

NaHCO_3

reacts with 1 mole of HCl. Moles of HCl for

NaHCO_3

=

1 \times 0.005263 = 0.005263 \text{ mol}

.

Total moles of HCl required =

0.010526 + 0.005263 = 0.015789 \text{ mol}

.

Now, calculate the volume of 0.1 M HCl solution: Molarity =

\frac{\text{Moles}}{\text{Volume in L}}

Volume in L =

\frac{\text{Moles}}{\text{Molarity}} = \frac{0.015789 \text{ mol}}{0.1 \text{ mol L}^{-1}} = 0.15789 \text{ L}

.

Volume in mL =

0.15789 \text{ L} \times 1000 \text{ mL L}^{-1} = 157.89 \text{ mL}

.

Final Answer: 157.89 mL of 0.1 M HCl is required.

Q7Exercises

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Solution

Given: Solution 1: 300 g of a 25% solution by mass. Solution 2: 400 g of a 40% solution by mass.

To Find: The mass percentage of the resulting solution.

Calculation: First, calculate the mass of solute in each solution.

Mass of solute in Solution 1 = 25% of 300 g =

\frac{25}{100} \times 300 \text{ g} = 75 \text{ g}

.

Mass of solute in Solution 2 = 40% of 400 g =

\frac{40}{100} \times 400 \text{ g} = 160 \text{ g}

.

Now, find the total mass of solute and total mass of the solution after mixing.

Total mass of solute = Mass of solute from Solution 1 + Mass of solute from Solution 2 =

75 \text{ g} + 160 \text{ g} = 235 \text{ g}

.

Total mass of the resulting solution = Mass of Solution 1 + Mass of Solution 2 =

300 \text{ g} + 400 \text{ g} = 700 \text{ g}

.

Finally, calculate the mass percentage of the resulting solution.

Mass percentage =

\frac{\text{Total mass of solute}}{\text{Total mass of solution}} \times 100

=

\frac{235 \text{ g}}{700 \text{ g}} \times 100 = 33.57\%

.

Final Answer: The mass percentage of the resulting solution is 33.57%.

Q8Exercises

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?

Solution

Given: Mass of ethylene glycol (

C_2H_6O_2

) = 222.6 g (solute). Mass of water = 200 g (solvent). Density of the solution = 1.072 g

mL^{-1}

.

To Find:

Molality ( $m$ )
Molarity ( $M$ )

Calculation: Molar mass of ethylene glycol (

C_2H_6O_2

) = 2

\times

12 + 6

\times

1 + 2

\times

16 = 24 + 6 + 32 = 62 g

mol^{-1}

.

First, calculate the moles of ethylene glycol: Moles of ethylene glycol =

\frac{\text{Mass}}{\text{Molar mass}} = \frac{222.6 \text{ g}}{62 \text{ g mol}^{-1}} = 3.59 \text{ mol}

.

1. Molality ( $m$ ): Mass of solvent (water) = 200 g = 0.200 kg. Molality =

\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{3.59 \text{ mol}}{0.200 \text{ kg}} = 17.95 \text{ m}

.

2. Molarity ( $M$ ): Total mass of the solution = Mass of solute + Mass of solvent =

222.6 \text{ g} + 200 \text{ g} = 422.6 \text{ g}

.

Volume of the solution =

\frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{422.6 \text{ g}}{1.072 \text{ g mL}^{-1}} = 394.22 \text{ mL}

.

Convert the volume to litres: Volume in L =

394.22 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.39422 \text{ L}

.

Molarity =

\frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{3.59 \text{ mol}}{0.39422 \text{ L}} = 9.11 \text{ M}

.

Final Answer:

The molality of the solution is 17.95 m.
The molarity of the solution is 9.11 M.

Q9Exercises

A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i)

express this in percent by mass

(ii)

determine the molality of chloroform in the water sample.

Solution

Given: Concentration of chloroform (

CHCl_3

) = 15 ppm (by mass).

To Find:

(i)

Concentration in percent by mass.

(ii)

Molality of chloroform.

Calculation: 15 ppm (by mass) means 15 parts of chloroform are present in

10^6

parts of the solution by mass. This means 15 g of

CHCl_3

is present in

10^6

g of the solution.

(i) Percent by mass: Mass percent =

\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100

=

\frac{15}{10^6} \times 100 = 15 \times 10^{-4} \% = 0.0015\%

.

(ii) Molality: Mass of solute (

CHCl_3

) = 15 g. Mass of solution =

10^6

g. Mass of solvent (water) = Mass of solution - Mass of solute =

10^6 \text{ g} - 15 \text{ g} \approx 10^6 \text{ g}

(since the mass of solute is negligible). Mass of solvent in kg =

\frac{10^6 \text{ g}}{1000 \text{ g kg}^{-1}} = 1000 \text{ kg}

.

Molar mass of chloroform (

CHCl_3

) = 12 + 1 + 3

\times

35.5 = 119.5 g

mol^{-1}

.

Moles of chloroform =

\frac{\text{Mass}}{\text{Molar mass}} = \frac{15 \text{ g}}{119.5 \text{ g mol}^{-1}} = 0.1255 \text{ mol}

.

Molality (

m

) =

\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.1255 \text{ mol}}{1000 \text{ kg}} = 1.255 \times 10^{-4} \text{ m}

.

Final Answer:

(i)

The concentration in percent by mass is

1.5 \times 10^{-3}\ \%

.

(ii)

The molality of chloroform in the water sample is

1.255 \times 10^{-4}

m.

Q10Exercises

What role does the molecular interaction play in a solution of alcohol and water?

Solution

Molecular interactions play a crucial role in determining the properties of a solution of alcohol (e.g., ethanol) and water. This type of solution exhibits a positive deviation from Raoult's law.

Interactions in Pure Components: In pure water, there are strong intermolecular hydrogen bonds between water molecules (water-water interactions). Similarly, in pure ethanol, there are hydrogen bonds between ethanol molecules (ethanol-ethanol interactions).
Interactions in the Solution: When ethanol and water are mixed, the ethanol molecules occupy spaces between the water molecules. This disrupts the original hydrogen bonding network in both pure liquids. New hydrogen bonds are formed between ethanol and water molecules (ethanol-water interactions).
Relative Strength of Interactions: The newly formed ethanol-water interactions are weaker than the original water-water and ethanol-ethanol interactions. The solute-solvent (A-B) interactions are weaker than the solute-solute (A-A) and solvent-solvent (B-B) interactions.
Consequences:
- Increased Vapour Pressure: Because the intermolecular forces of attraction are weaker in the solution, the molecules of both components can escape into the vapour phase more easily. This leads to a total vapour pressure of the solution that is higher than predicted by Raoult's law (positive deviation).
- Volume Increase on Mixing ( $\Delta_{mix}V > 0$ ): The weaker forces cause the molecules to be held less tightly, resulting in a slight increase in the total volume upon mixing.
- Endothermic Mixing ( $\Delta_{mix}H > 0$ ): Energy is required to break the stronger H-bonds in the pure components, and less energy is released when the new, weaker bonds are formed. Therefore, the overall process of mixing is endothermic.

Q11Exercises

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Solution

The dissolution of a gas in a liquid is generally an exothermic process. This means that when a gas dissolves, heat is released. The process can be represented by the following equilibrium:

Gas (solute) + Liquid (solvent)

\rightleftharpoons

Saturated Solution + Heat (

\Delta_{sol}H < 0

)

According to Le Chatelier's principle, if a change of condition is applied to a system in equilibrium, the system will shift in a direction that counteracts the change. In this case, raising the temperature means adding heat to the system.

To counteract the addition of heat, the equilibrium will shift in the endothermic direction, which is the reverse direction (to the left). This means that the dissolved gas will come out of the solution and return to the gaseous phase. As a result, the solubility of the gas in the liquid decreases.

This is why, for example, a cold carbonated drink fizzes more when it warms up, as the dissolved carbon dioxide gas becomes less soluble and escapes from the liquid.

Q12Exercises

State Henry's law and mention some important applications.

Solution

Henry's Law: Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.

Mathematically, it is expressed as:

p = K_H \cdot x

where:

$p$ is the partial pressure of the gas in the vapour phase.
$x$ is the mole fraction of the gas in the solution (a measure of its solubility).
$K_H$ is the Henry's law constant, which is specific to the gas and solvent at a given temperature.

Important Applications of Henry's Law:

Carbonated Beverages: To increase the solubility of carbon dioxide ( $CO_2$ ) in soft drinks and soda water, the bottles are sealed under high pressure. When the bottle is opened, the pressure above the liquid decreases, causing the solubility of $CO_2$ to decrease, and the gas bubbles out.
Scuba Diving: Scuba divers breathe compressed air at high pressure underwater. This increases the concentration of gases, particularly nitrogen, dissolved in their blood. If a diver ascends too quickly, the external pressure decreases rapidly. This causes the dissolved nitrogen to come out of the blood as bubbles, leading to a painful and dangerous condition called "the bends" (decompression sickness). To avoid this, divers use tanks filled with air diluted with helium (which has lower solubility in blood) and ascend slowly.
High Altitude Sickness (Anoxia): At high altitudes, the partial pressure of oxygen in the atmosphere is lower than at sea level. According to Henry's law, this leads to a lower concentration of dissolved oxygen in the blood and tissues of people living at high altitudes or climbers. This lack of oxygen can cause weakness, fatigue, and difficulty in thinking, a condition known as anoxia.
Functioning of Lungs: The exchange of gases ( $O_2$ and $CO_2$ ) in the lungs and tissues is governed by differences in partial pressures, as explained by Henry's law. Oxygen from the air in the lungs (high partial pressure) dissolves into the blood (low partial pressure), while carbon dioxide from the blood (high partial pressure) is released into the lungs (low partial pressure).

Q13Exercises

The partial pressure of ethane over a solution containing $6.56 \times 10^{-3}$ g of ethane is 1 bar. If the solution contains $5.00 \times 10^{-2}$ g of ethane, then what shall be the partial pressure of the gas?

Solution

Given: Case 1: Mass of ethane (

m_1

) =

6.56 \times 10^{-3}

g, Partial pressure (

p_1

) = 1 bar. Case 2: Mass of ethane (

m_2

) =

5.00 \times 10^{-2}

g.

To Find: Partial pressure of the gas in Case 2 (

p_2

).

Formula: According to Henry's law, the mass of a gas dissolved in a given volume of a liquid is directly proportional to the partial pressure of the gas.

m = k \cdot p

where

m

is the mass of the dissolved gas,

p

is the partial pressure, and

k

is a proportionality constant.

Therefore, we can write the relationship for the two cases as:

\frac{m_1}{p_1} = \frac{m_2}{p_2}

Calculation: Rearranging the formula to solve for

p_2

:

p_2 = p_1 \times \frac{m_2}{m_1}

Substituting the given values:

p_2 = 1 \text{ bar} \times \frac{5.00 \times 10^{-2} \text{ g}}{6.56 \times 10^{-3} \text{ g}}

p_2 = 1 \times \frac{5.00}{0.656}

p_2 = 7.622 \text{ bar}

Final Answer: The partial pressure of the gas will be 7.62 bar.

Q14Exercises

What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{mix}H$ related to positive and negative deviations from Raoult's law?

Solution

Raoult's law describes the behaviour of ideal solutions. When a solution does not obey Raoult's law, it is called a non-ideal solution and exhibits either positive or negative deviation.

Positive Deviation from Raoult's Law: A solution shows positive deviation when the observed total vapour pressure is higher than the value predicted by Raoult's law. This occurs when the intermolecular attractive forces between the solute and solvent molecules (A-B interactions) are weaker than the forces between the molecules of the pure components (A-A and B-B interactions).

Relationship with $\Delta_{mix}H$ : Because the A-B interactions are weaker, energy is required to overcome the stronger A-A and B-B interactions. The energy released upon forming the new, weaker A-B bonds is less than the energy absorbed to break the old bonds. Therefore, the overall process of mixing is endothermic. For positive deviation, $\Delta_{mix}H > 0$ (positive).
Example: A mixture of ethanol and acetone.

Negative Deviation from Raoult's Law: A solution shows negative deviation when the observed total vapour pressure is lower than the value predicted by Raoult's law. This occurs when the intermolecular attractive forces between the solute and solvent molecules (A-B interactions) are stronger than the forces between the molecules of the pure components (A-A and B-B interactions).

Relationship with $\Delta_{mix}H$ : Because the new A-B interactions are stronger, more energy is released upon forming these bonds than is absorbed to break the weaker A-A and B-B bonds. Therefore, the overall process of mixing is exothermic. For negative deviation, $\Delta_{mix}H < 0$ (negative).
Example: A mixture of chloroform and acetone, where hydrogen bonds form between the two components.

Summary:

Deviation Type	Vapour Pressure	Intermolecular Forces	$\Delta_{mix}H$ Sign
Positive	Higher than predicted	A-B < A-A, B-B	Positive (> 0)
Negative	Lower than predicted	A-B > A-A, B-B	Negative (< 0)

Q15Exercises

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution

Given: Solution: 2% non-volatile solute in water. Pressure of the solution (

p_1

) = 1.004 bar. The pressure is measured at the normal boiling point of the solvent (water).

To Find: Molar mass of the solute (

M_2

).

Formula: At the normal boiling point of water (100°C), the vapour pressure of pure water (

p_1^0

) is equal to the standard atmospheric pressure, which is 1 atm or 1.013 bar.

According to Raoult's law for relative lowering of vapour pressure:

\frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2}

where

x_2

is the mole fraction of the solute,

n_1

is moles of solvent, and

n_2

is moles of solute.

Calculation: A 2% solution by mass means 2 g of solute is dissolved in (100 - 2) = 98 g of water. Mass of solute (

w_2

) = 2 g. Mass of solvent (water,

w_1

) = 98 g.

Molar mass of water (

M_1

) = 18 g

mol^{-1}

. Let the molar mass of the solute be

M_2

.

Moles of water (

n_1

) =

\frac{w_1}{M_1} = \frac{98 \text{ g}}{18 \text{ g mol}^{-1}} = 5.444 \text{ mol}

. Moles of solute (

n_2

) =

\frac{w_2}{M_2} = \frac{2}{M_2} \text{ mol}

.

Now, substitute the values into Raoult's law equation:

\frac{1.013 - 1.004}{1.013} = \frac{n_2}{n_1 + n_2}

\frac{0.009}{1.013} = \frac{\frac{2}{M_2}}{5.444 + \frac{2}{M_2}}

0.00888 = \frac{2/M_2}{5.444 + 2/M_2}

For a dilute solution, we can approximate

n_1 + n_2 \approx n_1

:

0.00888 \approx \frac{n_2}{n_1} = \frac{2/M_2}{5.444}

0.00888 = \frac{2}{5.444 \times M_2}

M_2 = \frac{2}{5.444 \times 0.00888} = \frac{2}{0.04834}

M_2 = 41.37 \text{ g mol}^{-1}

Final Answer: The molar mass of the solute is approximately 41.37 g

mol^{-1}

.

Q16Exercises

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Solution

Given: For an ideal solution of heptane and octane at 373 K: Vapour pressure of pure heptane (

p_{heptane}^0

) = 105.2 kPa. Vapour pressure of pure octane (

p_{octane}^0

) = 46.8 kPa. Mass of heptane = 26.0 g. Mass of octane = 35 g.

To Find: The total vapour pressure of the mixture (

p_{total}

).

Formula: According to Raoult's law for an ideal solution:

p_{total} = p_{heptane} + p_{octane} = x_{heptane} \cdot p_{heptane}^0 + x_{octane} \cdot p_{octane}^0

where

x

represents the mole fraction of each component in the liquid phase.

Calculation: First, calculate the molar masses: Molar mass of heptane (

C_7H_{16}

) = 7

\times

12 + 16

\times

1 = 84 + 16 = 100 g

mol^{-1}

. Molar mass of octane (

C_8H_{18}

) = 8

\times

12 + 18

\times

1 = 96 + 18 = 114 g

mol^{-1}

.

Next, calculate the number of moles of each component: Moles of heptane (

n_{heptane}

) =

\frac{26.0 \text{ g}}{100 \text{ g mol}^{-1}} = 0.260 \text{ mol}

. Moles of octane (

n_{octane}

) =

\frac{35 \text{ g}}{114 \text{ g mol}^{-1}} = 0.307 \text{ mol}

.

Total moles =

0.260 + 0.307 = 0.567 \text{ mol}

.

Now, calculate the mole fractions:

x_{heptane} = \frac{n_{heptane}}{\text{Total moles}} = \frac{0.260}{0.567} = 0.458

.

x_{octane} = \frac{n_{octane}}{\text{Total moles}} = \frac{0.307}{0.567} = 0.542

. (Check:

0.458 + 0.542 = 1.000

)

Finally, calculate the total vapour pressure:

p_{total} = (0.458 \times 105.2 \text{ kPa}) + (0.542 \times 46.8 \text{ kPa})

p_{total} = 48.18 \text{ kPa} + 25.37 \text{ kPa}

p_{total} = 73.55 \text{ kPa}

.

Final Answer: The vapour pressure of the mixture is 73.55 kPa.

Q17Exercises

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Solution

Given: Vapour pressure of pure water (

p_1^0

) = 12.3 kPa. Concentration of the solution = 1 molal (1 m).

To Find: The vapour pressure of the solution (

p_1

).

Formula: According to Raoult's law, the vapour pressure of a solution containing a non-volatile solute is given by:

p_1 = x_1 \cdot p_1^0

where

x_1

is the mole fraction of the solvent (water).

Calculation: A 1 molal solution means 1 mole of solute is dissolved in 1 kg (1000 g) of the solvent (water).

Moles of solute (

n_2

) = 1 mol. Mass of water (

w_1

) = 1000 g. Molar mass of water (

M_1

) = 18 g

mol^{-1}

.

Moles of water (

n_1

) =

\frac{w_1}{M_1} = \frac{1000 \text{ g}}{18 \text{ g mol}^{-1}} = 55.55 \text{ mol}

.

Total moles in the solution =

n_1 + n_2 = 55.55 + 1 = 56.55 \text{ mol}

.

Mole fraction of the solvent (water,

x_1

) =

\frac{n_1}{n_1 + n_2} = \frac{55.55}{56.55} = 0.9823

.

Now, calculate the vapour pressure of the solution:

p_1 = 0.9823 \times 12.3 \text{ kPa}

p_1 = 12.084 \text{ kPa}

.

Final Answer: The vapour pressure of the 1 molal solution is 12.084 kPa.

Q18Exercises

Calculate the mass of a non-volatile solute (molar mass 40 g mol⁻¹) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution

Given: Mass of solvent (octane,

w_1

) = 114 g. Molar mass of solute (

M_2

) = 40 g

mol^{-1}

. The vapour pressure of the solution is 80% of the pure solvent's vapour pressure. So,

p_1 = 0.80 \times p_1^0

.

To Find: Mass of the non-volatile solute (

w_2

).

Formula: According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute:

\frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2}

Calculation: First, calculate the relative lowering of vapour pressure:

\frac{p_1^0 - 0.80 p_1^0}{p_1^0} = \frac{0.20 p_1^0}{p_1^0} = 0.20

So,

x_2 = 0.20

.

Since

x_1 + x_2 = 1

, the mole fraction of the solvent (octane) is

x_1 = 1 - 0.20 = 0.80

.

Now, let's find the number of moles of the solvent (octane,

C_8H_{18}

): Molar mass of octane (

M_1

) = 8

\times

12 + 18

\times

1 = 96 + 18 = 114 g

mol^{-1}

. Moles of octane (

n_1

) =

\frac{w_1}{M_1} = \frac{114 \text{ g}}{114 \text{ g mol}^{-1}} = 1 \text{ mol}

.

Let the moles of solute be

n_2

. We have:

x_1 = \frac{n_1}{n_1 + n_2} = 0.80

\frac{1}{1 + n_2} = 0.80

1 = 0.80 \times (1 + n_2)

1 = 0.80 + 0.80 n_2

0.20 = 0.80 n_2

n_2 = \frac{0.20}{0.80} = 0.25 \text{ mol}

Finally, calculate the mass of the solute (

w_2

): Mass = Moles

\times

Molar mass

w_2 = n_2 \times M_2 = 0.25 \text{ mol} \times 40 \text{ g mol}^{-1} = 10 \text{ g}

.

Final Answer: 10 g of the solute should be dissolved.

Q19Exercises

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(i)

molar mass of the solute

(ii)

vapour pressure of water at 298 K.

Solution

Given: Let the molar mass of the solute be

M_2

and the vapour pressure of pure water be

p_1^0

. Mass of solute (

w_2

) = 30 g. Molar mass of water (

M_1

) = 18 g

mol^{-1}

.

Case 1: Mass of water (

w_1

) = 90 g. Vapour pressure of solution (

p_{sol,1}

) = 2.8 kPa.

Case 2: 18 g of water is added. New mass of water (

w'_1

) = 90 + 18 = 108 g. New vapour pressure of solution (

p_{sol,2}

) = 2.9 kPa.

To Find:

(i)

Molar mass of the solute (

M_2

).

(ii)

Vapour pressure of pure water (

p_1^0

).

Formula: According to Raoult's law:

p_{sol} = x_1 \cdot p_1^0

, where

x_1

is the mole fraction of the solvent.

x_1 = \frac{n_1}{n_1 + n_2}

Calculation: Moles of solute (

n_2

) =

\frac{30}{M_2}

.

For Case 1: Moles of water (

n_1

) =

\frac{90}{18} = 5 \text{ mol}

.

2.8 = \left( \frac{5}{5 + \frac{30}{M_2}} \right) \cdot p_1^0 \quad \quad \text{---(Equation 1)}

For Case 2: New moles of water (

n'_1

) =

\frac{108}{18} = 6 \text{ mol}

.

2.9 = \left( \frac{6}{6 + \frac{30}{M_2}} \right) \cdot p_1^0 \quad \quad \text{---(Equation 2)}

Now, we divide Equation 2 by Equation 1:

\frac{2.9}{2.8} = \frac{\left( \frac{6}{6 + \frac{30}{M_2}} \right) \cdot p_1^0}{\left( \frac{5}{5 + \frac{30}{M_2}} \right) \cdot p_1^0}

\frac{2.9}{2.8} = \frac{6}{5} \times \frac{5 + \frac{30}{M_2}}{6 + \frac{30}{M_2}}

\frac{2.9}{2.8} \times \frac{5}{6} = \frac{5 + \frac{30}{M_2}}{6 + \frac{30}{M_2}}

0.863 = \frac{5M_2 + 30}{6M_2 + 30}

0.863 (6M_2 + 30) = 5M_2 + 30

5.178 M_2 + 25.89 = 5M_2 + 30

0.178 M_2 = 4.11

M_2 = \frac{4.11}{0.178} = 23.09 \text{ g mol}^{-1}

(i) Molar mass of the solute ( $M_2$ ) is 23.09 g $mol^{-1}$ .

Now, substitute the value of

M_2

into Equation 1 to find

p_1^0

.

2.8 = \left( \frac{5}{5 + \frac{30}{23.09}} \right) \cdot p_1^0

2.8 = \left( \frac{5}{5 + 1.3} \right) \cdot p_1^0

2.8 = \left( \frac{5}{6.3} \right) \cdot p_1^0

p_1^0 = \frac{2.8 \times 6.3}{5} = 3.528 \text{ kPa}

(ii) Vapour pressure of water at 298 K ( $p_1^0$ ) is 3.528 kPa.

Final Answer:

(i)

The molar mass of the solute is 23.09 g

mol^{-1}

.

(ii)

The vapour pressure of water at 298 K is 3.528 kPa.

Q20Exercises

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Solution

Given: Freezing point of pure water (

T_f^0

) = 273.15 K.

For Cane Sugar Solution: Concentration = 5% by mass. Freezing point of solution (

T_f

) = 271 K. Molar mass of cane sugar (

C_{12}H_{22}O_{11}

) = 342 g

mol^{-1}

.

For Glucose Solution: Concentration = 5% by mass. Molar mass of glucose (

C_6H_{12}O_6

) = 180 g

mol^{-1}

.

To Find: The freezing point of the 5% glucose solution.

Formula: Depression in freezing point,

\Delta T_f = K_f \cdot m

, where

K_f

is the molal freezing point depression constant and

m

is the molality.

Calculation: Step 1: Calculate $K_f$ for water using the cane sugar data. For the 5% cane sugar solution: Mass of sugar = 5 g. Mass of water = 100 g - 5 g = 95 g = 0.095 kg.

Moles of sugar =

\frac{5 \text{ g}}{342 \text{ g mol}^{-1}} = 0.0146 \text{ mol}

. Molality of sugar solution (

m_{sugar}

) =

\frac{0.0146 \text{ mol}}{0.095 \text{ kg}} = 0.1537 \text{ m}

.

Depression in freezing point for sugar solution (

\Delta T_{f,sugar}

) =

T_f^0 - T_f = 273.15 - 271 = 2.15 \text{ K}

.

Now, find

K_f

:

K_f = \frac{\Delta T_{f,sugar}}{m_{sugar}} = \frac{2.15 \text{ K}}{0.1537 \text{ m}} = 13.99 \text{ K kg mol}^{-1}

.

Step 2: Calculate the freezing point of the glucose solution. For the 5% glucose solution: Mass of glucose = 5 g. Mass of water = 95 g = 0.095 kg.

Moles of glucose =

\frac{5 \text{ g}}{180 \text{ g mol}^{-1}} = 0.0278 \text{ mol}

. Molality of glucose solution (

m_{glucose}

) =

\frac{0.0278 \text{ mol}}{0.095 \text{ kg}} = 0.2926 \text{ m}

.

Depression in freezing point for glucose solution (

\Delta T_{f,glucose}

) =

K_f \cdot m_{glucose}

=

13.99 \text{ K kg mol}^{-1} \times 0.2926 \text{ m} = 4.09 \text{ K}

.

New freezing point (

T'_{f}

) =

T_f^0 - \Delta T_{f,glucose}

=

273.15 \text{ K} - 4.09 \text{ K} = 269.06 \text{ K}

.

Final Answer: The freezing point of the 5% glucose solution is 269.06 K.

Q21Exercises

Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol⁻¹. Calculate atomic masses of A and B.

Solution

Given: Mass of solvent (benzene,

w_1

) = 20 g = 0.020 kg. Molar depression constant for benzene (

K_f

) = 5.1 K kg

mol^{-1}

.

For compound AB₂: Mass of solute (

w_2

) = 1.0 g. Depression in freezing point (

\Delta T_f

) = 2.3 K.

For compound AB₄: Mass of solute (

w'_2

) = 1.0 g. Depression in freezing point (

\Delta T'_f

) = 1.3 K.

To Find: Atomic masses of A and B.

Formula: Molar mass of solute (

M_2

) can be calculated using the formula for depression in freezing point:

M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}

(where

w_1

is in grams)

Calculation: Step 1: Calculate the molar mass of AB₂ ( $M_{AB_2}$ ).

M_{AB_2} = \frac{5.1 \text{ K kg mol}^{-1} \times 1.0 \text{ g} \times 1000 \text{ g kg}^{-1}}{2.3 \text{ K} \times 20 \text{ g}}

M_{AB_2} = \frac{5100}{46} = 110.87 \text{ g mol}^{-1}

Step 2: Calculate the molar mass of AB₄ ( $M_{AB_4}$ ).

M_{AB_4} = \frac{5.1 \text{ K kg mol}^{-1} \times 1.0 \text{ g} \times 1000 \text{ g kg}^{-1}}{1.3 \text{ K} \times 20 \text{ g}}

M_{AB_4} = \frac{5100}{26} = 196.15 \text{ g mol}^{-1}

Step 3: Set up equations to find atomic masses. Let the atomic mass of element A be

m_A

and the atomic mass of element B be

m_B

.

From the molar masses calculated:

m_A + 2m_B = 110.87 \quad \quad \text{---(Equation 1)}

m_A + 4m_B = 196.15 \quad \quad \text{---(Equation 2)}

Step 4: Solve the simultaneous equations. Subtract Equation 1 from Equation 2:

(m_A + 4m_B) - (m_A + 2m_B) = 196.15 - 110.87

2m_B = 85.28

m_B = \frac{85.28}{2} = 42.64 \text{ g mol}^{-1}

.

Now, substitute the value of

m_B

into Equation 1:

m_A + 2(42.64) = 110.87

m_A + 85.28 = 110.87

m_A = 110.87 - 85.28 = 25.59 \text{ g mol}^{-1}

.

Final Answer: The atomic mass of A is 25.59 g

mol^{-1}

. The atomic mass of B is 42.64 g

mol^{-1}

.

Q22Exercises

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Solution

Given: Temperature (T) = 300 K (constant).

Case 1: Mass of glucose = 36 g. Volume of solution = 1 L. Osmotic pressure (

\Pi_1

) = 4.98 bar.

Case 2: Osmotic pressure (

\Pi_2

) = 1.52 bar.

To Find: The concentration of the solution in Case 2 (

C_2

).

Formula: The osmotic pressure (

\Pi

) is given by the formula:

\Pi = CRT

where C is the molar concentration, R is the gas constant, and T is the temperature.

Since R and T are constant, the osmotic pressure is directly proportional to the concentration:

\Pi \propto C

Therefore, we can write the relationship:

\frac{\Pi_1}{C_1} = \frac{\Pi_2}{C_2}

Calculation: Step 1: Calculate the concentration of the solution in Case 1 ( $C_1$ ). Molar mass of glucose (

C_6H_{12}O_6

) = 180 g

mol^{-1}

. Moles of glucose =

\frac{36 \text{ g}}{180 \text{ g mol}^{-1}} = 0.2 \text{ mol}

. Concentration (

C_1

) =

\frac{\text{Moles}}{\text{Volume in L}} = \frac{0.2 \text{ mol}}{1 \text{ L}} = 0.2 \text{ M}

.

Step 2: Use the proportionality to find the concentration in Case 2 ( $C_2$ ).

\frac{4.98 \text{ bar}}{0.2 \text{ M}} = \frac{1.52 \text{ bar}}{C_2}

C_2 = 0.2 \text{ M} \times \frac{1.52 \text{ bar}}{4.98 \text{ bar}}

C_2 = 0.2 \times 0.3052

C_2 = 0.061 \text{ M}

Final Answer: The concentration of the solution would be 0.061 M.

Q23Exercises

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i)

n-hexane and n-octane

(ii)

I₂ and CCl₄

(iii)

NaClO₄ and water

(iv)

methanol and acetone

(v)

acetonitrile (CH₃CN) and acetone (C₃H₆O).

Solution

(i)

n-hexane and n-octane: Both are non-polar molecules. The most important type of intermolecular interaction is the London dispersion forces (or van der Waals forces).

(ii)

I₂ and CCl₄: Both are non-polar molecules. The most important type of intermolecular interaction is the London dispersion forces.

(iii)

NaClO₄ and water: NaClO₄ is an ionic compound that dissociates into

Na^+

and

ClO_4^-

ions in water. Water is a polar molecule (a dipole). The most important interaction is ion-dipole interaction between the ions and the polar water molecules.

(iv)

Methanol and acetone: Methanol (

CH_3OH

) is a polar molecule capable of hydrogen bonding. Acetone ((

CH_3)_2CO

) is a polar molecule but cannot form hydrogen bonds with itself. The most important interactions are dipole-dipole interactions and hydrogen bonding (between the -OH group of methanol and the oxygen atom of acetone).

(v)

Acetonitrile (CH₃CN) and acetone (C₃H₆O): Both are polar molecules (due to the C≡N and C=O groups, respectively). The most important type of intermolecular interaction is dipole-dipole interaction.

Q24Exercises

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.

Solution

Principle: The principle of solubility is "like dissolves like". This means that non-polar solutes dissolve best in non-polar solvents, and polar solutes dissolve best in polar solvents.

Solvent: n-Octane (

C_8H_{18}

) is a non-polar hydrocarbon.

Solutes:

KCl: An ionic compound, which is extremely polar.
CH₃OH (Methanol): A highly polar molecule capable of hydrogen bonding.
CH₃CN (Acetonitrile): A polar molecule.
Cyclohexane ( $C_6H_{12}$ ): A non-polar hydrocarbon.

Explanation: Since n-octane is a non-polar solvent, the solubility of the given solutes will depend on how non-polar they are. The more non-polar the solute, the higher its solubility in n-octane.

KCl: As an ionic compound, it has very strong ion-ion interactions. It is insoluble in non-polar solvents like n-octane because the weak solute-solvent interactions cannot overcome the strong lattice energy of the ionic crystal. It will be the least soluble.
CH₃OH: Methanol is very polar and forms strong hydrogen bonds. It is immiscible with non-polar n-octane. Its solubility will be very low, but slightly higher than KCl.
CH₃CN: Acetonitrile is polar but less polar than methanol and does not have hydrogen bonding as strong as methanol. It will be more soluble than methanol but less soluble than the non-polar cyclohexane.
Cyclohexane: Cyclohexane is a non-polar molecule, very similar in nature to n-octane. Due to the similar non-polar characteristics, they are completely miscible. It will be the most soluble.

Order of Increasing Solubility: The order of increasing solubility in n-octane is:

KCl < CH_3OH < CH_3CN < \text{Cyclohexane}

Q25Exercises

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i)

phenol

(ii)

toluene

(iii)

formic acid

(iv)

ethylene glycol

(v)

chloroform

(vi)

pentanol.

Solution

Principle: Water is a highly polar solvent that forms strong hydrogen bonds. Solubility in water depends on the ability of a compound to form hydrogen bonds with water and the size of its non-polar (hydrophobic) part.

(i)

Phenol ( $C_6H_5OH$ ): It has a polar -OH group that can form hydrogen bonds with water, but it also has a large non-polar benzene ring. Therefore, phenol is partially soluble in water.

(ii)

Toluene ( $C_6H_5CH_3$ ): It is a non-polar hydrocarbon. It cannot form hydrogen bonds with water. Therefore, toluene is insoluble in water.

(iii)

Formic acid (HCOOH): It is a small molecule with a polar carboxyl group that can readily form hydrogen bonds with water. Therefore, formic acid is highly soluble in water.

(iv)

Ethylene glycol ( $HOCH_2CH_2OH$ ): It has two polar -OH groups and a small hydrocarbon part. It can form extensive hydrogen bonds with water. Therefore, ethylene glycol is highly soluble in water.

(v)

Chloroform ( $CHCl_3$ ): It is a slightly polar molecule but cannot form strong hydrogen bonds with water. Its interactions with water are weak. Therefore, chloroform is insoluble (or very slightly soluble) in water.

(vi)

Pentanol ( $C_5H_{11}OH$ ): It has a polar -OH group for hydrogen bonding, but the non-polar alkyl chain (pentyl group) is quite large. The hydrophobic nature of the alkyl chain dominates. Therefore, pentanol is partially soluble (or sparingly soluble) in water.

Q26Exercises

If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.

Solution

Given: Density of lake water solution = 1.25 g

mL^{-1}

. Mass of

Na^+

ions (solute) = 92 g. Mass of water (solvent) = 1 kg = 1000 g.

To Find: Molarity of

Na^+

ions.

Calculation: Step 1: Calculate the total mass of the solution. Total mass of solution = Mass of solvent + Mass of solute = 1000 g + 92 g = 1092 g.

Step 2: Calculate the volume of the solution. Volume of solution =

\frac{\text{Mass of solution}}{\text{Density of solution}}

=

\frac{1092 \text{ g}}{1.25 \text{ g mL}^{-1}} = 873.6 \text{ mL}

.

Convert the volume to litres: Volume in L =

\frac{873.6}{1000} = 0.8736 \text{ L}

.

Step 3: Calculate the moles of the solute ( $Na^+$ ions). Molar mass of Na = 23 g

mol^{-1}

. Moles of

Na^+

=

\frac{\text{Mass of Na}^+}{\text{Molar mass of Na}}

=

\frac{92 \text{ g}}{23 \text{ g mol}^{-1}} = 4 \text{ mol}

.

Step 4: Calculate the molarity. Molarity (M) =

\frac{\text{Moles of solute}}{\text{Volume of solution in L}}

=

\frac{4 \text{ mol}}{0.8736 \text{ L}} = 4.579 \text{ M}

.

Final Answer: The molarity of

Na^+

ions in the lake is 4.58 M.

Q27Exercises

If the solubility product of CuS is $6 \times 10^{-16}$ , calculate the maximum molarity of CuS in aqueous solution.

Solution

Given: The solubility product constant (

K_{sp}

) of CuS =

6 \times 10^{-16}

.

To Find: The maximum molarity of CuS in the solution.

Explanation: The maximum molarity of CuS is its molar solubility (S). The dissolution equilibrium for CuS in water is:

CuS(s) \rightleftharpoons Cu^{2+}(aq) + S^{2-}(aq)

The solubility product expression is:

K_{sp} = [Cu^{2+}][S^{2-}]

Let S be the molar solubility of CuS. At equilibrium, the concentration of the ions will be:

[Cu^{2+}] = S

mol/L

[S^{2-}] = S

mol/L

Calculation: Substitute these concentrations into the

K_{sp}

expression:

K_{sp} = (S)(S) = S^2

S^2 = 6 \times 10^{-16}

S = \sqrt{6 \times 10^{-16}}

S = \sqrt{6} \times \sqrt{10^{-16}}

S = 2.45 \times 10^{-8} \text{ M}

Final Answer: The maximum molarity of CuS in aqueous solution is

2.45 \times 10^{-8}

M.

Q28Exercises

Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.

Solution

Given: Mass of aspirin (

C_9H_8O_4

, solute) = 6.5 g. Mass of acetonitrile (

CH_3CN

, solvent) = 450 g.

To Find: The mass percentage of aspirin in the solution.

Formula: Mass percentage =

\frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100

Calculation: Step 1: Calculate the total mass of the solution. Total mass of solution = Mass of solute + Mass of solvent = 6.5 g + 450 g = 456.5 g.

Step 2: Calculate the mass percentage of aspirin. Mass % of aspirin =

\frac{\text{Mass of aspirin}}{\text{Total mass of solution}} \times 100

=

\frac{6.5 \text{ g}}{456.5 \text{ g}} \times 100 = 1.424\%

.

Final Answer: The mass percentage of aspirin in acetonitrile is 1.424%.

Q29Exercises

Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5 \times 10^{-3}$ m aqueous solution required for the above dose.

Solution

Given: Dose of nalorphene (mass of solute,

w_2

) = 1.5 mg =

1.5 \times 10^{-3}

g. Molality of the aqueous solution (

m

) =

1.5 \times 10^{-3}

m.

To Find: Total mass of the solution required.

Calculation: Step 1: Calculate the molar mass of nalorphene ( $C_{19}H_{21}NO_3$ ). Molar mass (

M_2

) = (19

\times

12) + (21

\times

1) + 14 + (3

\times

16) = 228 + 21 + 14 + 48 = 311 g

mol^{-1}

.

Step 2: Calculate the moles of nalorphene in the dose. Moles of solute (

n_2

) =

\frac{\text{Mass}}{\text{Molar mass}} = \frac{1.5 \times 10^{-3} \text{ g}}{311 \text{ g mol}^{-1}} = 4.823 \times 10^{-6} \text{ mol}

.

Step 3: Use the molality to find the mass of the solvent (water). Molality (

m

) =

\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Mass of solvent in kg (

w_1

) =

\frac{\text{Moles of solute}}{\text{Molality}} = \frac{4.823 \times 10^{-6} \text{ mol}}{1.5 \times 10^{-3} \text{ mol kg}^{-1}} = 3.215 \times 10^{-3} \text{ kg}

.

Mass of solvent in g =

3.215 \times 10^{-3} \text{ kg} \times 1000 \text{ g kg}^{-1} = 3.215 \text{ g}

.

Step 4: Calculate the total mass of the solution. Total mass of solution = Mass of solute (

w_2

) + Mass of solvent (

w_1

) =

(1.5 \times 10^{-3} \text{ g}) + 3.215 \text{ g} = 0.0015 \text{ g} + 3.215 \text{ g} = 3.2165 \text{ g}

.

Final Answer: The mass of the aqueous solution required is 3.2165 g.

Q30Exercises

Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Solution

Given: Volume of solution (V) = 250 mL = 0.250 L. Molarity of solution (M) = 0.15 M (or 0.15 mol

L^{-1}

). Solute: Benzoic acid (

C_6H_5COOH

).

To Find: Mass of benzoic acid required.

Calculation: Step 1: Calculate the molar mass of benzoic acid ( $C_6H_5COOH$ ). Molar mass = (7

\times

12) + (6

\times

1) + (2

\times

16) = 84 + 6 + 32 = 122 g

mol^{-1}

.

Step 2: Calculate the moles of benzoic acid needed. Molarity =

\frac{\text{Moles of solute}}{\text{Volume of solution in L}}

Moles of solute = Molarity

\times

Volume of solution in L =

0.15 \text{ mol L}^{-1} \times 0.250 \text{ L} = 0.0375 \text{ mol}

.

Step 3: Calculate the mass of benzoic acid. Mass = Moles

\times

Molar mass =

0.0375 \text{ mol} \times 122 \text{ g mol}^{-1} = 4.575 \text{ g}

.

Final Answer: 4.575 g of benzoic acid is required.

Q31Exercises

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Solution

Principle: The depression in freezing point (

\Delta T_f

) is a colligative property, given by the formula:

\Delta T_f = i \cdot K_f \cdot m

where:

$i$ is the van't Hoff factor (number of particles in solution per formula unit).
$K_f$ is the cryoscopic constant of the solvent (water).
$m$ is the molality of the solution.

For the same molality (

m

) and the same solvent (

K_f

), the depression in freezing point (

\Delta T_f

) is directly proportional to the van't Hoff factor (

i

).

\Delta T_f \propto i

Explanation: Acetic acid (

CH_3COOH

), trichloroacetic acid (

CCl_3COOH

), and trifluoroacetic acid (

CF_3COOH

) are all weak acids that dissociate in water to some extent:

HA \rightleftharpoons H^+ + A^-

The van't Hoff factor for a dissociating solute is given by

i = 1 + \alpha(n-1)

, where

\alpha

is the degree of dissociation and

n

is the number of ions produced (here, n=2). So,

i = 1 + \alpha

. A larger degree of dissociation (

\alpha

) results in a larger van't Hoff factor (

i

).

The degree of dissociation depends on the acid strength. A stronger acid will dissociate more, leading to a higher concentration of ions and a larger value of

i

.

The acid strength of these three acids is determined by the inductive effect of the atoms attached to the carboxyl group.

Acetic acid ( $CH_3COOH$ ): The methyl group ( $-CH_3$ ) is electron-donating, which destabilizes the acetate anion and makes it the weakest acid of the three.
Trichloroacetic acid ( $CCl_3COOH$ ): Chlorine is an electronegative atom. The three chlorine atoms have a strong electron-withdrawing inductive effect (–I effect), which pulls electron density away from the carboxyl group, stabilizes the resulting carboxylate anion, and increases the acid strength.
Trifluoroacetic acid ( $CF_3COOH$ ): Fluorine is more electronegative than chlorine. The three fluorine atoms exert an even stronger electron-withdrawing inductive effect than the chlorine atoms, making trifluoroacetic acid the strongest acid of the three.

Therefore, the order of acid strength is: Trifluoroacetic acid > Trichloroacetic acid > Acetic acid

This means the degree of dissociation (

\alpha

) and consequently the van't Hoff factor (

i

) also follow the same order: $i_{CF_3COOH} > i_{CCl_3COOH} > i_{CH_3COOH}$

Since

\Delta T_f

is proportional to

i

, the depression in freezing point will increase in the same order. $\Delta T_{f(CF_3COOH)} > \Delta T_{f(CCl_3COOH)} > \Delta T_{f(CH_3COOH)}$

Q32Exercises

Calculate the depression in the freezing point of water when 10 g of $CH_3CH_2CHClCOOH$ is added to 250 g of water. $K_a = 1.4 \times 10^{-3}$ , $K_f = 1.86 \text{ K kg mol}^{-1}$ .

Solution

Given: Mass of solute (

CH_3CH_2CHClCOOH

, 2-chlorobutanoic acid) = 10 g. Mass of solvent (water) = 250 g = 0.250 kg. Dissociation constant (

K_a

) =

1.4 \times 10^{-3}

. Cryoscopic constant for water (

K_f

) = 1.86 K kg

mol^{-1}

.

To Find: Depression in the freezing point (

\Delta T_f

).

Formula:

\Delta T_f = i \cdot K_f \cdot m

Calculation: Step 1: Calculate the molality (m) of the solution. Molar mass of

C_4H_7O_2Cl

= (4

\times

12) + (7

\times

1) + (2

\times

16) + 35.5 = 48 + 7 + 32 + 35.5 = 122.5 g

mol^{-1}

. Moles of solute =

\frac{10 \text{ g}}{122.5 \text{ g mol}^{-1}} = 0.0816 \text{ mol}

. Molality (

m

) =

\frac{0.0816 \text{ mol}}{0.250 \text{ kg}} = 0.3264 \text{ m}

.

Step 2: Calculate the degree of dissociation ( $\alpha$ ). The acid dissociates as:

HA \rightleftharpoons H^+ + A^-

Initial conc:

m \quad 0 \quad 0

Equil. conc:

m(1-\alpha) \quad m\alpha \quad m\alpha

K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(m\alpha)(m\alpha)}{m(1-\alpha)} = \frac{m\alpha^2}{1-\alpha}

.

1.4 \times 10^{-3} = \frac{0.3264 \alpha^2}{1-\alpha}

.

Assuming

\alpha

is small,

1-\alpha \approx 1

.

1.4 \times 10^{-3} \approx 0.3264 \alpha^2

\alpha^2 \approx \frac{1.4 \times 10^{-3}}{0.3264} = 4.289 \times 10^{-3}

\alpha \approx \sqrt{4.289 \times 10^{-3}} = 0.0655

. (The value of

\alpha

is small enough for the approximation to be valid).

Step 3: Calculate the van't Hoff factor (i). For dissociation into 2 ions,

i = 1 + \alpha

.

i = 1 + 0.0655 = 1.0655

.

Step 4: Calculate the depression in freezing point ( $\Delta T_f$ ).

\Delta T_f = i \cdot K_f \cdot m

\Delta T_f = 1.0655 \times 1.86 \text{ K kg mol}^{-1} \times 0.3264 \text{ mol kg}^{-1}

\Delta T_f = 0.646 \text{ K}

.

Final Answer: The depression in the freezing point of water is 0.646 K.

Q33Exercises

19.5 g of $CH_2FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Solution

Given: Mass of solute (

CH_2FCOOH

, fluoroacetic acid) = 19.5 g. Mass of solvent (water) = 500 g = 0.5 kg. Observed depression in freezing point (

\Delta T_{f,obs}

) = 1.0 °C = 1.0 K. For water,

K_f

= 1.86 K kg

mol^{-1}

.

To Find:

van't Hoff factor ( $i$ ).
Dissociation constant ( $K_a$ ).

Calculation: Step 1: Calculate the theoretical (calculated) depression in freezing point assuming no dissociation. Molar mass of

CH_2FCOOH

= (2

\times

12) + (3

\times

1) + 19 + (2

\times

16) = 24 + 3 + 19 + 32 = 78 g

mol^{-1}

. Moles of solute =

\frac{19.5 \text{ g}}{78 \text{ g mol}^{-1}} = 0.25 \text{ mol}

. Molality (

m

) =

\frac{0.25 \text{ mol}}{0.5 \text{ kg}} = 0.5 \text{ m}

.

Calculated

\Delta T_f

(

\Delta T_{f,calc}

) =

K_f \cdot m = 1.86 \text{ K kg mol}^{-1} \times 0.5 \text{ m} = 0.93 \text{ K}

.

Step 2: Calculate the van't Hoff factor (i).

i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} = \frac{\Delta T_{f,obs}}{\Delta T_{f,calc}}

i = \frac{1.0 \text{ K}}{0.93 \text{ K}} = 1.075

Step 3: Calculate the degree of dissociation ( $\alpha$ ). The acid dissociates into 2 ions (

HA \rightleftharpoons H^+ + A^-

), so

n=2

.

i = 1 + \alpha(n-1) = 1 + \alpha(2-1) = 1 + \alpha

.

1.075 = 1 + \alpha

\alpha = 0.075

.

Step 4: Calculate the dissociation constant ( $K_a$ ).

K_a = \frac{m\alpha^2}{1-\alpha}

, where

m

is the molality (0.5 m).

K_a = \frac{0.5 \times (0.075)^2}{1 - 0.075}

K_a = \frac{0.5 \times 0.005625}{0.925}

K_a = \frac{0.0028125}{0.925} = 3.04 \times 10^{-3}

Final Answer:

The van't Hoff factor ( $i$ ) is 1.075.
The dissociation constant ( $K_a$ ) of fluoroacetic acid is $3.04 \times 10^{-3}$ .

Q34Exercises

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Solution

Given: Vapour pressure of pure water (

p_1^0

) = 17.535 mm Hg. Mass of glucose (solute,

w_2

) = 25 g. Mass of water (solvent,

w_1

) = 450 g.

To Find: The vapour pressure of the solution (

p_1

).

Formula: According to Raoult's law:

p_1 = x_1 \cdot p_1^0

where

x_1

is the mole fraction of the solvent (water).

Calculation: Step 1: Calculate the moles of glucose and water. Molar mass of glucose (

C_6H_{12}O_6

,

M_2

) = 180 g

mol^{-1}

. Molar mass of water (

H_2O

,

M_1

) = 18 g

mol^{-1}

.

Moles of glucose (

n_2

) =

\frac{w_2}{M_2} = \frac{25 \text{ g}}{180 \text{ g mol}^{-1}} = 0.139 \text{ mol}

. Moles of water (

n_1

) =

\frac{w_1}{M_1} = \frac{450 \text{ g}}{18 \text{ g mol}^{-1}} = 25 \text{ mol}

.

Step 2: Calculate the mole fraction of water ( $x_1$ ). Total moles =

n_1 + n_2 = 25 + 0.139 = 25.139 \text{ mol}

.

x_1 = \frac{n_1}{n_1 + n_2} = \frac{25}{25.139} = 0.9945

.

Step 3: Calculate the vapour pressure of the solution ( $p_1$ ).

p_1 = x_1 \cdot p_1^0 = 0.9945 \times 17.535 \text{ mm Hg}

p_1 = 17.438 \text{ mm Hg}

.

Final Answer: The vapour pressure of the glucose solution is 17.438 mm Hg.

Q35Exercises

Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution

Given: Henry's law constant (

K_H

) =

4.27 \times 10^5

mm Hg (Note: The unit suggests this constant relates pressure to mole fraction, not molality. We will assume the standard definition

p = K_H \cdot x

). Partial pressure of methane (

p_{methane}

) = 760 mm Hg.

To Find: The solubility of methane in benzene.

Formula: According to Henry's law:

p = K_H \cdot x

where

x

is the mole fraction, which represents solubility.

Calculation: Rearranging the formula to solve for the mole fraction (

x

):

x_{methane} = \frac{p_{methane}}{K_H}

x_{methane} = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}}

x_{methane} = 1.78 \times 10^{-3}

Final Answer: The solubility of methane in benzene, expressed as a mole fraction, is

1.78 \times 10^{-3}

.

Q36Exercises

100 g of liquid A (molar mass 140 g mol⁻¹) was dissolved in 1000 g of liquid B (molar mass 180 g mol⁻¹). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Solution

Given: Mass of liquid A (

w_A

) = 100 g. Molar mass of A (

M_A

) = 140 g

mol^{-1}

. Mass of liquid B (

w_B

) = 1000 g. Molar mass of B (

M_B

) = 180 g

mol^{-1}

. Vapour pressure of pure B (

p_B^0

) = 500 torr. Total vapour pressure of solution (

p_{total}

) = 475 torr.

To Find:

Vapour pressure of pure liquid A ( $p_A^0$ ).
Vapour pressure of A in the solution ( $p_A$ ).

Formula: Assuming an ideal solution, Raoult's law applies:

p_{total} = p_A + p_B = x_A \cdot p_A^0 + x_B \cdot p_B^0

Calculation: Step 1: Calculate the moles of A and B. Moles of A (

n_A

) =

\frac{100 \text{ g}}{140 \text{ g mol}^{-1}} = 0.714 \text{ mol}

. Moles of B (

n_B

) =

\frac{1000 \text{ g}}{180 \text{ g mol}^{-1}} = 5.556 \text{ mol}

.

Step 2: Calculate the mole fractions of A and B. Total moles =

n_A + n_B = 0.714 + 5.556 = 6.270 \text{ mol}

.

x_A = \frac{n_A}{\text{Total moles}} = \frac{0.714}{6.270} = 0.114

.

x_B = \frac{n_B}{\text{Total moles}} = \frac{5.556}{6.270} = 0.886

. (Check:

0.114 + 0.886 = 1.000

)

Step 3: Calculate the vapour pressure of pure A ( $p_A^0$ ). Substitute the known values into Raoult's law:

475 = (0.114 \cdot p_A^0) + (0.886 \cdot 500)

475 = 0.114 p_A^0 + 443

475 - 443 = 0.114 p_A^0

32 = 0.114 p_A^0

p_A^0 = \frac{32}{0.114} = 280.7 \text{ torr}

.

Step 4: Calculate the vapour pressure of A in the solution ( $p_A$ ).

p_A = x_A \cdot p_A^0 = 0.114 \times 280.7 \text{ torr} = 32 \text{ torr}

. Alternatively, we know

p_{total} = p_A + p_B

. We found

p_B = x_B p_B^0 = 443

torr. So,

p_A = p_{total} - p_B = 475 - 443 = 32

torr.

Final Answer:

The vapour pressure of pure liquid A is 280.7 torr.
Its vapour pressure in the solution is 32 torr.

Q37Exercises

Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot $p_{total}$ , $p_{chloroform}$ , and $p_{acetone}$ as a function of $x_{acetone}$ . The experimental data observed for different compositions of mixture is:
$100 \times x_{acetone}$ 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1
$p_{acetone}$ / mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1
$p_{chloroform}$ / mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7
Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

$100 \times x_{acetone}$	0	11.8	23.4	36.0	50.8	58.2	64.5	72.1
$p_{acetone}$ / mm Hg	0	54.9	110.1	202.4	322.7	405.9	454.1	521.1
$p_{chloroform}$ / mm Hg	632.8	548.1	469.4	359.7	257.7	193.6	161.2	120.7

Solution

Ideal Solution Calculation (Raoult's Law): Given:

p_{acetone}^0 = 741.8 \text{ mm Hg}

p_{chloroform}^0 = 632.8 \text{ mm Hg}

According to Raoult's law for an ideal solution:

p_{acetone} = x_{acetone} \cdot p_{acetone}^0 = x_{acetone} \cdot 741.8

p_{chloroform} = x_{chloroform} \cdot p_{chloroform}^0 = (1 - x_{acetone}) \cdot 632.8

p_{total} = p_{acetone} + p_{chloroform}

We can calculate the ideal values:

When $x_{acetone} = 0$ , $p_{acetone} = 0$ , $p_{chloroform} = 632.8$ , $p_{total} = 632.8$ .
When $x_{acetone} = 1$ , $p_{acetone} = 741.8$ , $p_{chloroform} = 0$ , $p_{total} = 741.8$ .

The ideal plots for

p_{acetone}

,

p_{chloroform}

, and

p_{total}

will be straight lines connecting these points.

Experimental Data Calculation: For each experimental point, we calculate the total pressure by adding the given partial pressures.

For $x_{acetone} = 0.118$ : $p_{total} = 54.9 + 548.1 = 603.0 \text{ mm Hg}$
For $x_{acetone} = 0.234$ : $p_{total} = 110.1 + 469.4 = 579.5 \text{ mm Hg}$
For $x_{acetone} = 0.360$ : $p_{total} = 202.4 + 359.7 = 562.1 \text{ mm Hg}$
... and so on for the other points.

Plotting and Analysis: To create the plot, the x-axis will be the mole fraction of acetone (

x_{acetone}

) from 0 to 1, and the y-axis will be the vapour pressure in mm Hg.

Plot the Ideal Lines (dashed lines):
- Draw a straight line for $p_{acetone}$ from (0, 0) to (1, 741.8).
- Draw a straight line for $p_{chloroform}$ from (0, 632.8) to (1, 0).
- Draw a straight line for $p_{total}$ from (0, 632.8) to (1, 741.8).
Plot the Experimental Data (solid curves):
- Plot the experimental points for $p_{acetone}$ , $p_{chloroform}$ , and the calculated $p_{total}$ against $x_{acetone}$ .
- Connect the points to form curves.

(The plot would look similar to Fig. 1.6 (b) in the textbook.)

Conclusion: Upon plotting, it is observed that the experimental curves for the partial pressures of both components and the total vapour pressure lie below the corresponding straight lines predicted by Raoult's law for an ideal solution.

This indicates that the solution exhibits a negative deviation from Raoult's law. This happens because the intermolecular forces between acetone and chloroform molecules (due to hydrogen bonding) are stronger than the forces within pure acetone or pure chloroform.

Q38Exercises

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution

Given: Mass of benzene (

w_B

) = 80 g. Mass of toluene (

w_T

) = 100 g. Vapour pressure of pure benzene (

p_B^0

) = 50.71 mm Hg. Vapour pressure of pure toluene (

p_T^0

) = 32.06 mm Hg.

To Find: Mole fraction of benzene in the vapour phase (

y_B

).

Formula: For the liquid phase (Raoult's Law):

p_{total} = x_B \cdot p_B^0 + x_T \cdot p_T^0

For the vapour phase (Dalton's Law):

y_B = \frac{p_B}{p_{total}} = \frac{x_B \cdot p_B^0}{p_{total}}

Calculation: Step 1: Calculate moles and mole fractions in the liquid phase. Molar mass of benzene (

C_6H_6

) = 78 g

mol^{-1}

. Molar mass of toluene (

C_7H_8

) = 92 g

mol^{-1}

.

Moles of benzene (

n_B

) =

\frac{80 \text{ g}}{78 \text{ g mol}^{-1}} = 1.026 \text{ mol}

. Moles of toluene (

n_T

) =

\frac{100 \text{ g}}{92 \text{ g mol}^{-1}} = 1.087 \text{ mol}

.

Total moles =

1.026 + 1.087 = 2.113 \text{ mol}

.

Mole fraction of benzene (

x_B

) =

\frac{1.026}{2.113} = 0.486

. Mole fraction of toluene (

x_T

) =

1 - 0.486 = 0.514

.

Step 2: Calculate the partial pressures and total pressure. Partial pressure of benzene (

p_B

) =

x_B \cdot p_B^0 = 0.486 \times 50.71 \text{ mm Hg} = 24.65 \text{ mm Hg}

. Partial pressure of toluene (

p_T

) =

x_T \cdot p_T^0 = 0.514 \times 32.06 \text{ mm Hg} = 16.48 \text{ mm Hg}

.

Total pressure (

p_{total}

) =

p_B + p_T = 24.65 + 16.48 = 41.13 \text{ mm Hg}

.

Step 3: Calculate the mole fraction of benzene in the vapour phase.

y_B = \frac{p_B}{p_{total}} = \frac{24.65 \text{ mm Hg}}{41.13 \text{ mm Hg}} = 0.599 \approx 0.60

.

Final Answer: The mole fraction of benzene in the vapour phase is approximately 0.60.

Q39Exercises

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are $3.30 \times 10^7$ mm and $6.51 \times 10^7$ mm respectively, calculate the composition of these gases in water.

Solution

Given: Total pressure of air = 10 atm. Volume % of oxygen (

O_2

) = 20%. Volume % of nitrogen (

N_2

) = 79%. Henry's law constant for

O_2

(

K_{H,O_2}

) =

3.30 \times 10^7

mm Hg. Henry's law constant for

N_2

(

K_{H,N_2}

) =

6.51 \times 10^7

mm Hg.

To Find: The composition (mole fraction) of

O_2

and

N_2

in water.

Formula: According to Dalton's law, partial pressure = mole fraction

\times

total pressure. Assuming ideal gas behaviour, volume % is equal to mole %. Partial pressure of a gas (

p_{gas}

) = (Volume % / 100)

\times

Total pressure.

According to Henry's law:

p_{gas} = K_H \cdot x_{gas}

, where

x_{gas}

is the mole fraction of the gas in the liquid.

Calculation: Step 1: Convert total pressure to mm Hg. 1 atm = 760 mm Hg. Total pressure =

10 \text{ atm} \times 760 \text{ mm Hg/atm} = 7600 \text{ mm Hg}

.

Step 2: Calculate the partial pressures of $O_2$ and $N_2$ in the air. Partial pressure of

O_2

(

p_{O_2}

) =

\frac{20}{100} \times 7600 \text{ mm Hg} = 1520 \text{ mm Hg}

. Partial pressure of

N_2

(

p_{N_2}

) =

\frac{79}{100} \times 7600 \text{ mm Hg} = 6004 \text{ mm Hg}

.

Step 3: Calculate the mole fraction of each gas in water using Henry's Law. For Oxygen (

O_2

):

x_{O_2} = \frac{p_{O_2}}{K_{H,O_2}} = \frac{1520 \text{ mm Hg}}{3.30 \times 10^7 \text{ mm Hg}} = 4.61 \times 10^{-5}

.

For Nitrogen (

N_2

):

x_{N_2} = \frac{p_{N_2}}{K_{H,N_2}} = \frac{6004 \text{ mm Hg}}{6.51 \times 10^7 \text{ mm Hg}} = 9.22 \times 10^{-5}

.

Final Answer: The composition of the gases in water is:

Mole fraction of oxygen ( $x_{O_2}$ ) = $4.61 \times 10^{-5}$ .
Mole fraction of nitrogen ( $x_{N_2}$ ) = $9.22 \times 10^{-5}$ .

Q40Exercises

Determine the amount of $CaCl_2$ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Solution

Given: van't Hoff factor for

CaCl_2

(

i

) = 2.47. Volume of solution (V) = 2.5 L. Osmotic pressure (

\Pi

) = 0.75 atm. Temperature (T) = 27°C = 27 + 273 = 300 K. Gas constant (R) = 0.0821 L atm

mol^{-1} K^{-1}

.

To Find: Mass of

CaCl_2

dissolved.

Formula: The formula for osmotic pressure including the van't Hoff factor is:

\Pi = i \cdot C \cdot R \cdot T = i \cdot \frac{n}{V} \cdot R \cdot T

where

n

is the number of moles of the solute.

Calculation: Step 1: Calculate the number of moles (n) of $CaCl_2$ . Rearrange the formula to solve for n:

n = \frac{\Pi \cdot V}{i \cdot R \cdot T}

n = \frac{0.75 \text{ atm} \times 2.5 \text{ L}}{2.47 \times 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1} \times 300 \text{ K}}

n = \frac{1.875}{60.83} = 0.0308 \text{ mol}

Step 2: Calculate the mass of $CaCl_2$ . Molar mass of

CaCl_2

= 40.08 (Ca) + 2

\times

35.45 (Cl) = 40.08 + 70.90 = 110.98 g

mol^{-1}

.

Mass = Moles

\times

Molar mass Mass =

0.0308 \text{ mol} \times 110.98 \text{ g mol}^{-1} = 3.42 \text{ g}

.

Final Answer: The amount of

CaCl_2

dissolved is 3.42 g.

Q41Exercises

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25°C, assuming that it is completely dissociated.

Solution

Given: Mass of

K_2SO_4

= 25 mg =

25 \times 10^{-3}

g = 0.025 g. Volume of solution (V) = 2 L. Temperature (T) = 25°C = 25 + 273 = 298 K. The solute

K_2SO_4

is completely dissociated.

To Find: The osmotic pressure (

\Pi

) of the solution.

Formula:

\Pi = i \cdot C \cdot R \cdot T = i \cdot \frac{n}{V} \cdot R \cdot T

Calculation: Step 1: Determine the van't Hoff factor (i).

K_2SO_4

dissociates completely in water as follows:

K_2SO_4 (s) \rightarrow 2K^+(aq) + SO_4^{2-}(aq)

One formula unit of

K_2SO_4

produces 2 + 1 = 3 ions. Since dissociation is complete, the van't Hoff factor

i = 3

.

Step 2: Calculate the molar mass and moles of $K_2SO_4$ . Molar mass of

K_2SO_4

= (2

\times

39.1) + 32.07 + (4

\times

16) = 78.2 + 32.07 + 64 = 174.27 g

mol^{-1}

.

Moles of

K_2SO_4

(n) =

\frac{\text{Mass}}{\text{Molar mass}} = \frac{0.025 \text{ g}}{174.27 \text{ g mol}^{-1}} = 1.4345 \times 10^{-4} \text{ mol}

.

Step 3: Calculate the osmotic pressure ( $\Pi$ ). Using R = 0.0821 L atm

mol^{-1} K^{-1}

to get pressure in atm.

\Pi = \frac{i \cdot n \cdot R \cdot T}{V}

\Pi = \frac{3 \times (1.4345 \times 10^{-4} \text{ mol}) \times (0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}) \times (298 \text{ K})}{2 \text{ L}}

\Pi = \frac{0.01053}{2} = 5.265 \times 10^{-3} \text{ atm}

The question asks for the answer in pascals. 1 atm =

1.01325 \times 10^5

Pa.

\Pi

in Pascals =

(5.265 \times 10^{-3} \text{ atm}) \times (1.01325 \times 10^5 \text{ Pa/atm})

= 533.4 \text{ Pa}

.

Final Answer: The osmotic pressure of the solution is

5.27 \times 10^{-3}

atm or 533.4 Pa.

SolutionsClass 12 Chemistry NCERT Solutions

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Solution

Give an example of a solid solution in which the solute is a gas.

Solution

Define the following terms:

Mole fraction

Molality

Molarity

Mass percentage.

Solution

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?

Solution

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL⁻¹, then what shall be the molarity of the solution?

Solution

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Solution

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Solution

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?

Solution

A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

express this in percent by mass

determine the molality of chloroform in the water sample.

Solution

What role does the molecular interaction play in a solution of alcohol and water?

Solution

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Solution

State Henry's law and mention some important applications.

Solution

The partial pressure of ethane over a solution containing 6.56×10−36.56 \times 10^{-3}6.56×10−3 g of ethane is 1 bar. If the solution contains 5.00×10−25.00 \times 10^{-2}5.00×10−2 g of ethane, then what shall be the partial pressure of the gas?

Solution

What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔmixH\Delta_{mix}HΔmix​H related to positive and negative deviations from Raoult's law?

Solution

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Solution

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Solution

Calculate the mass of a non-volatile solute (molar mass 40 g mol⁻¹) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

molar mass of the solute

vapour pressure of water at 298 K.

Solution

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Solution

Solution

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Solution

Suggest the most important type of intermolecular attractive interaction in the following pairs.

n-hexane and n-octane

I₂ and CCl₄

NaClO₄ and water

methanol and acetone

acetonitrile (CH₃CN) and acetone (C₃H₆O).

Solution

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.

Solution

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

phenol

toluene

formic acid

ethylene glycol

chloroform

pentanol.

Solution

If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.

Solution

If the solubility product of CuS is 6×10−166 \times 10^{-16}6×10−16, calculate the maximum molarity of CuS in aqueous solution.

Solution

Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.

Solution

Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5×10−31.5 \times 10^{-3}1.5×10−3 m aqueous solution required for the above dose.

Solution

Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Solution

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

The partial pressure of ethane over a solution containing $6.56 \times 10^{-3}$ g of ethane is 1 bar. If the solution contains $5.00 \times 10^{-2}$ g of ethane, then what shall be the partial pressure of the gas?

What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{mix}H$ related to positive and negative deviations from Raoult's law?

If the solubility product of CuS is $6 \times 10^{-16}$ , calculate the maximum molarity of CuS in aqueous solution.

Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of $1.5 \times 10^{-3}$ m aqueous solution required for the above dose.

Calculate the depression in the freezing point of water when 10 g of $CH_3CH_2CHClCOOH$ is added to 250 g of water. $K_a = 1.4 \times 10^{-3}$ , $K_f = 1.86 \text{ K kg mol}^{-1}$ .

19.5 g of $CH_2FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Determine the amount of $CaCl_2$ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25°C, assuming that it is completely dissociated.