SolutionsClass 12 Chemistry Notes

Introduction to Solutions

In our daily lives, we mostly interact with mixtures rather than pure substances. From the air we breathe to the water we drink, mixtures are everywhere. A special type of mixture, which is uniform throughout, is called a solution.

A solution is a homogeneous mixture of two or more components. This means its composition and properties are the same throughout the entire mixture.

• The component present in the largest quantity is called the solvent. The solvent usually determines the physical state (solid, liquid, or gas) of the solution.
• The other components, present in smaller quantities, are called solutes.

In this chapter, we will focus on binary solutions, which consist of only two components: one solute and one solvent.

Types of Solutions

Solutions can be classified based on the physical state of the solute and the solvent. There are three main types: Gaseous, Liquid, and Solid solutions.

Expressing Concentration of Solutions

The concentration of a solution describes the amount of solute dissolved in a given amount of solvent or solution. We can describe it qualitatively (e.g., dilute or concentrated), but a quantitative description is more precise and useful. Here are the common ways to express concentration quantitatively.

Mass Percentage (w/w)

This is the mass of a component per 100 grams of the solution. It's often used in industrial applications.

Mass percentage of a component is calculated as: $\text{Mass \% of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100$

Volume Percentage (V/V)

This is the volume of a component per 100 mL of the solution. It's common for solutions where both solute and solvent are liquids.

Volume percentage of a component is calculated as: $\text{Volume \% of a component} = \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100$

Mass by Volume Percentage (w/V)

Commonly used in medicine and pharmacy, this is the mass of solute dissolved in 100 mL of the solution.

Parts Per Million (ppm)

This unit is used when a solute is present in very small, or trace, quantities. It represents the number of parts of a component per million ($10^6$) parts of the solution.

Parts per million is calculated as: $\text{ppm} = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components of the solution}} \times 10^6$

Mole Fraction (x)

The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components in the solution.

For a binary solution with components A and B, the mole fractions are: $x_A = \frac{n_A}{n_A + n_B} \quad \text{and} \quad x_B = \frac{n_B}{n_A + n_B}$ where $n_A$ and $n_B$ are the number of moles of A and B, respectively.

Given

• Mass percentage of ethylene glycol = 20%
• This means 20 g of ethylene glycol is in 100 g of solution.
• Mass of ethylene glycol ($w_{glycol}$) = 20 g
• Mass of water ($w_{water}$) = 100 g - 20 g = 80 g

To Find

• Mole fraction of ethylene glycol ($x_{glycol}$)
• Mole fraction of water ($x_{water}$)

Formula

• Molar mass of $C_2H_6O_2 = (12 \times 2) + (1 \times 6) + (16 \times 2) = 62 \text{ g mol}^{-1}$
• Molar mass of $H_2O = (1 \times 2) + 16 = 18 \text{ g mol}^{-1}$
• Moles = Mass / Molar mass
• $x_{glycol} = \frac{\text{moles of } C_2H_6O_2}{\text{moles of } C_2H_6O_2 + \text{moles of } H_2O}$

Solution

First, calculate the number of moles of each component. $\text{Moles of } C_2H_6O_2 = \frac{20 \text{ g}}{62 \text{ g mol}^{-1}} = 0.322 \text{ mol}$ $\text{Moles of water} = \frac{80 \text{ g}}{18 \text{ g mol}^{-1}} = 4.444 \text{ mol}$

Now, calculate the mole fraction of ethylene glycol. $x_{glycol} = \frac{0.322 \text{ mol}}{0.322 \text{ mol} + 4.444 \text{ mol}} = \frac{0.322}{4.766} = 0.068$

The mole fraction of water can be found similarly or by using the property that the sum of mole fractions is 1. $x_{water} = 1 - x_{glycol} = 1 - 0.068 = 0.932$

Final Answer The mole fraction of ethylene glycol is 0.068, and the mole fraction of water is 0.932.

Molarity (M)

Molarity is defined as the number of moles of solute dissolved in one litre of solution. Its unit is moles per litre ($\text{mol L}^{-1}$) or simply M.

$\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}$

Given

• Mass of NaOH = 5 g
• Volume of solution = 450 mL = 0.450 L

To Find

Molarity (M) of the solution.

Formula

• Molar mass of NaOH = 23 + 16 + 1 = $40 \text{ g mol}^{-1}$
• Moles = Mass / Molar mass
• Molarity = Moles of solute / Volume of solution in L

Solution

First, find the moles of NaOH. $\text{Moles of NaOH} = \frac{5 \text{ g}}{40 \text{ g mol}^{-1}} = 0.125 \text{ mol}$

Now, calculate the molarity. $\text{Molarity} = \frac{0.125 \text{ mol}}{0.450 \text{ L}} = 0.278 \text{ mol L}^{-1}$

Final Answer The molarity of the solution is $0.278 \text{ M}$.

Molality (m)

Molality is defined as the number of moles of solute per kilogram of the solvent. Its unit is moles per kilogram ($\text{mol kg}^{-1}$) or simply m.

$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Given

• Mass of ethanoic acid ($w_2$) = 2.5 g
• Mass of benzene ($w_1$) = 75 g = 0.075 kg

To Find

Molality (m) of the solution.

Formula

• Molar mass of $CH_3COOH$ (or $C_2H_4O_2$) = $(12 \times 2) + (1 \times 4) + (16 \times 2) = 60 \text{ g mol}^{-1}$
• Moles = Mass / Molar mass
• Molality = Moles of solute / Mass of solvent in kg

Solution

First, find the moles of ethanoic acid. $\text{Moles of } C_2H_4O_2 = \frac{2.5 \text{ g}}{60 \text{ g mol}^{-1}} = 0.0417 \text{ mol}$

Now, calculate the molality. $\text{Molality} = \frac{0.0417 \text{ mol}}{0.075 \text{ kg}} = 0.556 \text{ mol kg}^{-1}$

Final Answer The molality of ethanoic acid in benzene is $0.556 \text{ m}$.

Solubility

Solubility is the maximum amount of a substance (solute) that can be dissolved in a specified amount of solvent at a specific temperature. It depends on the nature of the solute and solvent, as well as temperature and pressure.

Solubility of a Solid in a Liquid

The principle that governs the solubility of a solid in a liquid is "like dissolves like."

• Polar solutes (like sodium chloride and sugar) tend to dissolve in polar solvents (like water).
• Non-polar solutes (like naphthalene) tend to dissolve in non-polar solvents (like benzene). This happens because the intermolecular interactions between the solute and solvent molecules are similar to the interactions within the pure components.

When a solid solute is added to a solvent, two processes occur:

1. Dissolution: Solute particles dissolve and their concentration in the solution increases.
2. Crystallisation: Some dissolved solute particles collide with the solid solute and separate out of the solution.

A state of dynamic equilibrium is reached when the rate of dissolution equals the rate of crystallisation. $\text{Solute} + \text{Solvent} \rightleftharpoons \text{Solution}$ A solution in this state is called a saturated solution, meaning no more solute can be dissolved at that temperature and pressure. An unsaturated solution is one in which more solute can be dissolved.

Effect of Temperature

Temperature significantly affects the solubility of solids. According to Le Chatelier's Principle:

• If the dissolution process is endothermic ($\Delta_{sol}H > 0$, heat is absorbed), solubility increases with a rise in temperature.
• If the dissolution process is exothermic ($\Delta_{sol}H < 0$, heat is released), solubility decreases with a rise in temperature.

Effect of Pressure

Pressure has no significant effect on the solubility of solids in liquids because both solids and liquids are highly incompressible.

Solubility of a Gas in a Liquid

The solubility of gases in liquids is greatly affected by both pressure and temperature.

Effect of Pressure and Henry's Law

The solubility of a gas in a liquid increases as the pressure of the gas above the liquid increases. This relationship is described by 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.

The most common form of Henry's Law is expressed using mole fraction: $p = K_H x$ where:

• $p$ is the partial pressure of the gas in the vapour phase.
• $x$ is the mole fraction of the gas in the solution.
• $K_H$ is the Henry's law constant.

The value of $K_H$ is different for different gases and temperatures. [!note] A higher value of $K_H$ at a given pressure means a lower solubility of the gas.

Applications of Henry's Law

• Carbonated Beverages: To increase the solubility of $CO_2$ in soft drinks and soda water, the bottles are sealed under high pressure.
• Scuba Diving: At high pressures underwater, more atmospheric gases (especially nitrogen) dissolve in a diver's blood. If the diver ascends too quickly, the pressure decreases, and the dissolved nitrogen forms bubbles in the blood, causing a painful and dangerous condition called "the bends". To prevent this, divers use tanks filled with air diluted with helium, which has very low solubility in blood.
• High Altitudes: At high altitudes, the partial pressure of oxygen is lower. This leads to lower concentrations of dissolved oxygen in the blood, causing climbers to feel weak and unable to think clearly, a condition known as anoxia.

Effect of Temperature

The solubility of gases in liquids decreases with a rise in temperature. The dissolution of a gas is an exothermic process (heat is released). According to Le Chatelier's Principle, increasing the temperature will shift the equilibrium to favor the reactants (gas + liquid), thus reducing solubility.

Given

• Partial pressure of $N_2$, $p = 0.987$ bar
• Henry's law constant, $K_H = 76.48 \text{ kbar} = 76480$ bar
• Volume of water = 1 litre

To Find

Number of millimoles of $N_2$ dissolved in 1 litre of water.

Formula

$p = K_H x$

Solution

First, calculate the mole fraction of nitrogen ($x_{Nitrogen}$) in the solution. $x_{Nitrogen} = \frac{p}{K_H} = \frac{0.987 \text{ bar}}{76480 \text{ bar}} = 1.29 \times 10^{-5}$

Next, we need the number of moles of water in 1 litre. Since the density of water is approximately 1 kg/L, 1 litre of water has a mass of 1000 g. $\text{Moles of water} (n_{water}) = \frac{1000 \text{ g}}{18 \text{ g mol}^{-1}} = 55.5 \text{ mol}$

The mole fraction of nitrogen is also given by: $x_{Nitrogen} = \frac{n_{N_2}}{n_{N_2} + n_{water}}$ Since the amount of dissolved nitrogen is very small, we can approximate $n_{N_2} + n_{water} \approx n_{water}$. $x_{Nitrogen} \approx \frac{n_{N_2}}{55.5}$

Now, solve for the moles of $N_2$ ($n_{N_2}$). $n_{N_2} = x_{Nitrogen} \times 55.5 = (1.29 \times 10^{-5}) \times 55.5 \text{ mol} = 7.16 \times 10^{-4} \text{ mol}$

Finally, convert moles to millimoles. $7.16 \times 10^{-4} \text{ mol} \times \frac{1000 \text{ mmol}}{1 \text{ mol}} = 0.716 \text{ mmol}$

Final Answer 0.716 mmol of $N_2$ gas would dissolve in 1 litre of water.

Vapour Pressure of Liquid Solutions

Vapour Pressure of Liquid-Liquid Solutions

Let's consider a binary solution of two volatile liquids (components 1 and 2). In a closed container, both components will evaporate, and an equilibrium will be established between the liquid and vapour phases. The relationship between the partial vapour pressures of the components and their mole fractions in the solution is given by Raoult's Law.

Raoult's Law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction in the solution.

For component 1: $p_1 = p_1^0 x_1$

For component 2: $p_2 = p_2^0 x_2$

where:

• $p_1$ and $p_2$ are the partial vapour pressures of components 1 and 2.
• $x_1$ and $x_2$ are their mole fractions in the liquid solution.
• $p_1^0$ and $p_2^0$ are the vapour pressures of the pure components 1 and 2 at the same temperature.

According to Dalton's law of partial pressures, the total vapour pressure ($p_{total}$) above the solution is the sum of the partial pressures: $p_{total} = p_1 + p_2 = p_1^0 x_1 + p_2^0 x_2$ Since $x_1 = 1 - x_2$, we can write: $p_{total} = p_1^0(1 - x_2) + p_2^0 x_2 = p_1^0 + (p_2^0 - p_1^0)x_2$

The composition of the vapour phase can also be determined. If $y_1$ and $y_2$ are the mole fractions of components 1 and 2 in the vapour phase, then: $p_1 = y_1 p_{total} \quad \text{and} \quad p_2 = y_2 p_{total}$

Given

• Vapour pressure of pure $CHCl_3$, $p_{CHCl_3}^0$ = 200 mm Hg
• Vapour pressure of pure $CH_2Cl_2$, $p_{CH_2Cl_2}^0$ = 415 mm Hg
• Mass of $CHCl_3$ = 25.5 g
• Mass of $CH_2Cl_2$ = 40 g

To Find

(i) Total vapour pressure of the solution, $p_{total}$ (ii) Mole fractions in the vapour phase, $y_{CHCl_3}$ and $y_{CH_2Cl_2}$

Formula

• Molar mass of $CHCl_3 = 12 + 1 + (35.5 \times 3) = 119.5 \text{ g mol}^{-1}$
• Molar mass of $CH_2Cl_2 = 12 + 2 + (35.5 \times 2) = 85 \text{ g mol}^{-1}$
• $p_{total} = p_{CHCl_3}^0 x_{CHCl_3} + p_{CH_2Cl_2}^0 x_{CH_2Cl_2}$
• $p_i = y_i p_{total}$

Solution

(i) Calculate the total vapour pressure

First, find the moles and mole fractions of each component in the liquid phase. $\text{Moles of } CHCl_3 = \frac{25.5 \text{ g}}{119.5 \text{ g mol}^{-1}} = 0.213 \text{ mol}$ $\text{Moles of } CH_2Cl_2 = \frac{40 \text{ g}}{85 \text{ g mol}^{-1}} = 0.470 \text{ mol}$ Total moles = $0.213 + 0.470 = 0.683$ mol.

$x_{CHCl_3} = \frac{0.213}{0.683} = 0.312$ $x_{CH_2Cl_2} = \frac{0.470}{0.683} = 0.688$

Now, calculate the total vapour pressure. $p_{total} = (200 \text{ mm Hg} \times 0.312) + (415 \text{ mm Hg} \times 0.688)$ $p_{total} = 62.4 + 285.52 = 347.9 \text{ mm Hg}$

Answer for part (i) = $347.9 \text{ mm Hg}$

(ii) Calculate the mole fractions in the vapour phase

First, calculate the partial pressure of each component. $p_{CHCl_3} = p_{CHCl_3}^0 x_{CHCl_3} = 200 \times 0.312 = 62.4 \text{ mm Hg}$ $p_{CH_2Cl_2} = p_{CH_2Cl_2}^0 x_{CH_2Cl_2} = 415 \times 0.688 = 285.5 \text{ mm Hg}$

Now, find the mole fractions in the vapour phase. $y_{CHCl_3} = \frac{p_{CHCl_3}}{p_{total}} = \frac{62.4}{347.9} = 0.18$ $y_{CH_2Cl_2} = \frac{p_{CH_2Cl_2}}{p_{total}} = \frac{285.5}{347.9} = 0.82$

Answer for part (ii) = $y_{CHCl_3} = 0.18$, $y_{CH_2Cl_2} = 0.82$

Raoult's Law as a special case of Henry's Law

If we compare Raoult's Law ($p_i = p_i^0 x_i$) and Henry's Law ($p = K_H x$), we see that the partial pressure of the volatile component is proportional to its mole fraction in both cases. The only difference is the proportionality constant ($p_i^0$ vs $K_H$). Therefore, Raoult's law can be seen as a special case of Henry's law where $K_H$ becomes equal to $p_i^0$.

Vapour Pressure of Solutions of Solids in Liquids

When a non-volatile solute (like sugar or salt) is added to a volatile solvent (like water), the vapour pressure of the solution is lower than that of the pure solvent.

Why does this happen? In a pure liquid, the entire surface is occupied by volatile solvent molecules. When a non-volatile solute is added, some of the surface area is now occupied by solute particles. This reduces the number of solvent molecules at the surface that can escape into the vapour phase. Consequently, the vapour pressure is reduced.

For such a solution, the vapour pressure is contributed only by the solvent. Raoult's law can be stated as: $p_1 = p_1^0 x_1$ where $p_1$ is the vapour pressure of the solvent over the solution, $p_1^0$ is the vapour pressure of the pure solvent, and $x_1$ is the mole fraction of the solvent.

Ideal and Non-ideal Solutions

Ideal Solutions

An ideal solution is a solution that obeys Raoult's law over the entire range of concentration. For an ideal solution, two other important conditions are met:

• Enthalpy of mixing is zero: $\Delta_{mix}H = 0$. No heat is absorbed or evolved when the components are mixed.
• Volume of mixing is zero: $\Delta_{mix}V = 0$. The total volume of the solution is exactly the sum of the volumes of the individual components.

This behaviour occurs when the intermolecular attractive forces between different components (A-B) are nearly equal to the forces between identical components (A-A and B-B). [!example] Solutions of n-hexane and n-heptane, or benzene and toluene, behave nearly ideally.

Non-ideal Solutions

A non-ideal solution is one that does not obey Raoult's law. The vapour pressure of such solutions is either higher or lower than predicted by Raoult's law. This leads to two types of deviations.

Positive Deviation from Raoult's Law

• Cause: The intermolecular attractive forces between solute and solvent molecules (A-B) are weaker than the forces between solute-solute (A-A) and solvent-solvent (B-B) molecules.
• Result: Molecules find it easier to escape from the solution than from the pure liquids. This leads to an increase in vapour pressure, so the observed vapour pressure is higher than predicted.
• Characteristics: $\Delta_{mix}H > 0$ (endothermic mixing), $\Delta_{mix}V > 0$ (volume expansion).
• Example: A mixture of ethanol and acetone. In pure ethanol, molecules are held by hydrogen bonds. Acetone molecules get in between ethanol molecules and break some of these bonds, weakening the overall interactions.

Negative Deviation from Raoult's Law

• Cause: The intermolecular attractive forces between solute and solvent molecules (A-B) are stronger than the forces between A-A and B-B molecules.
• Result: Molecules have a lower tendency to escape from the solution. This leads to a decrease in vapour pressure, so the observed vapour pressure is lower than predicted.
• Characteristics: $\Delta_{mix}H < 0$ (exothermic mixing), $\Delta_{mix}V < 0$ (volume contraction).
• Example: A mixture of chloroform and acetone. A hydrogen bond forms between the hydrogen atom of chloroform and the oxygen atom of acetone, creating a stronger A-B interaction.

Azeotropes

Some liquids, on mixing, form azeotropes. These are binary mixtures that have the same composition in both the liquid and vapour phase and boil at a constant temperature. Because of this, the components of an azeotrope cannot be separated by fractional distillation.

• Minimum boiling azeotropes: Formed by solutions showing a large positive deviation from Raoult's law. The boiling point of the azeotrope is lower than that of either pure component. (e.g., ethanol-water mixture at ~95% ethanol).
• Maximum boiling azeotropes: Formed by solutions showing a large negative deviation from Raoult's law. The boiling point of the azeotrope is higher than that of either pure component. (e.g., nitric acid-water mixture at ~68% nitric acid).

Colligative Properties and Determination of Molar Mass

Colligative properties are properties of solutions that depend only on the number of solute particles relative to the total number of particles present, and not on the nature or identity of the solute. They are all related to the lowering of the solvent's vapour pressure.

The four main colligative properties are:

1. Relative lowering of vapour pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure

Relative Lowering of Vapour Pressure

We've seen that adding a non-volatile solute lowers the vapour pressure of the solvent. The lowering of vapour pressure ($\Delta p_1$) is $p_1^0 - p_1$.

The relative lowering of vapour pressure is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent. Raoult showed that this is equal to the mole fraction of the solute ($x_2$). $\frac{\Delta p_1}{p_1^0} = \frac{p_1^0 - p_1}{p_1^0} = x_2$ Since $x_2 = \frac{n_2}{n_1 + n_2}$, we have: $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1 + n_2}$ For dilute solutions, $n_2$ is much smaller than $n_1$, so we can approximate $n_1 + n_2 \approx n_1$. $\frac{p_1^0 - p_1}{p_1^0} \approx \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$ This equation allows us to calculate the molar mass of the solute ($M_2$) if all other quantities are known.

Given

• Vapour pressure of pure benzene, $p_1^0 = 0.850$ bar
• Vapour pressure of solution, $p = 0.845$ bar
• Mass of solute, $w_2 = 0.5$ g
• Mass of solvent (benzene), $w_1 = 39.0$ g
• Molar mass of solvent (benzene), $M_1 = 78 \text{ g mol}^{-1}$

To Find

Molar mass of the solute, $M_2$.

Formula

$\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1}$

Solution

Substitute the given values into the formula. $\frac{0.850 - 0.845}{0.850} = \frac{0.5 \text{ g} \times 78 \text{ g mol}^{-1}}{M_2 \times 39 \text{ g}}$ $\frac{0.005}{0.850} = \frac{39}{M_2 \times 39}$ $0.00588 = \frac{1}{M_2}$ $M_2 = \frac{1}{0.00588} = 170 \text{ g mol}^{-1}$

Final Answer The molar mass of the solid substance is $170 \text{ g mol}^{-1}$.

Elevation of Boiling Point

The boiling point of a liquid is the temperature at which its vapour pressure equals the atmospheric pressure. Since adding a non-volatile solute lowers the vapour pressure, a higher temperature is needed to make the solution's vapour pressure reach atmospheric pressure. Therefore, the boiling point of a solution is always higher than that of the pure solvent.

The elevation of boiling point ($\Delta T_b$) is the difference between the boiling point of the solution ($T_b$) and the boiling point of the pure solvent ($T_b^0$). $\Delta T_b = T_b - T_b^0$ For dilute solutions, $\Delta T_b$ is directly proportional to the molality (m) of the solution. $\Delta T_b = K_b m$

• $K_b$ is the Boiling Point Elevation Constant or Ebullioscopic Constant, which is a characteristic of the solvent.

We can use this to find the molar mass of the solute ($M_2$): $\Delta T_b = K_b \left( \frac{1000 \times w_2}{M_2 \times w_1} \right) \implies M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}$

Given

• Mass of glucose ($w_2$) = 18 g
• Mass of water ($w_1$) = 1 kg
• $K_b$ for water = $0.52 \text{ K kg mol}^{-1}$
• Boiling point of pure water ($T_b^0$) = 373.15 K (at 1.013 bar)

To Find

The boiling point of the solution ($T_b$).

Formula

• Molar mass of glucose ($C_6H_{12}O_6$) = $(12 \times 6) + (1 \times 12) + (16 \times 6) = 180 \text{ g mol}^{-1}$
• Molality (m) = Moles of solute / Mass of solvent in kg
• $\Delta T_b = K_b m$
• $T_b = T_b^0 + \Delta T_b$

Solution

First, calculate the molality of the glucose solution. $\text{Moles of glucose} = \frac{18 \text{ g}}{180 \text{ g mol}^{-1}} = 0.1 \text{ mol}$ $m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol kg}^{-1}$

Next, calculate the elevation in boiling point. $\Delta T_b = K_b m = (0.52 \text{ K kg mol}^{-1}) \times (0.1 \text{ mol kg}^{-1}) = 0.052 \text{ K}$

Finally, find the new boiling point. $T_b = 373.15 \text{ K} + 0.052 \text{ K} = 373.202 \text{ K}$

Final Answer The solution will boil at $373.202 \text{ K}$.

Depression of Freezing Point

The freezing point is the temperature at which a substance's liquid and solid phases are in equilibrium, meaning they have the same vapour pressure. Because a solution has a lower vapour pressure than the pure solvent at any given temperature, it must be cooled to a lower temperature for its vapour pressure to equal that of the pure solid solvent. Therefore, the freezing point of a solution is always lower than that of the pure solvent.

The depression of freezing point ($\Delta T_f$) is the difference between the freezing point of the pure solvent ($T_f^0$) and the freezing point of the solution ($T_f$). $\Delta T_f = T_f^0 - T_f$ For dilute solutions, $\Delta T_f$ is directly proportional to the molality (m). $\Delta T_f = K_f m$

• $K_f$ is the Freezing Point Depression Constant or Cryoscopic Constant, which is a characteristic of the solvent.

We can use this to find the molar mass of the solute ($M_2$): $M_2 = \frac{1000 \times w_2 \times K_f}{\Delta T_f \times w_1}$

Given

• Mass of solute ($w_2$) = 1.00 g
• Mass of benzene ($w_1$) = 50 g
• Depression in freezing point, $\Delta T_f = 0.40$ K
• $K_f$ for benzene = $5.12 \text{ K kg mol}^{-1}$

To Find

Molar mass of the solute, $M_2$.

Formula

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

Solution

Substitute the given values into the formula. $M_2 = \frac{(5.12 \text{ K kg mol}^{-1}) \times (1.00 \text{ g}) \times 1000}{(0.40 \text{ K}) \times (50 \text{ g})}$ $M_2 = \frac{5120}{20} = 256 \text{ g mol}^{-1}$

Final Answer The molar mass of the solute is $256 \text{ g mol}^{-1}$.

Osmosis and Osmotic Pressure

Osmosis is the spontaneous flow of solvent molecules from a pure solvent (or a dilute solution) to a more concentrated solution through a semipermeable membrane (SPM). An SPM is a membrane that allows small solvent molecules to pass through but blocks larger solute molecules.

Osmotic pressure ($\Pi$) is the excess pressure that must be applied to the solution side to just stop the flow of solvent (i.e., to prevent osmosis).

Osmotic pressure is a colligative property. For dilute solutions, it is proportional to the molarity (C) of the solution at a given temperature (T). $\Pi = CRT$ where R is the gas constant. Since $C = n_2/V$ (moles of solute per volume of solution in litres), we can write: $\Pi V = n_2 R T$ This can be used to find the molar mass of the solute ($M_2$): $M_2 = \frac{w_2 R T}{\Pi V}$

• Isotonic solutions: Two solutions with the same osmotic pressure. No net osmosis occurs between them.
• Hypertonic solution: A solution with a higher osmotic pressure than another. If blood cells are placed in a hypertonic solution (e.g., > 0.9% saline), water flows out, and they shrink.
• Hypotonic solution: A solution with a lower osmotic pressure than another. If blood cells are placed in a hypotonic solution (e.g., < 0.9% saline), water flows in, and they swell.

Given

• Volume of solution, $V = 200 \text{ cm}^3 = 0.200$ L
• Mass of protein, $w_2 = 1.26$ g
• Osmotic pressure, $\Pi = 2.57 \times 10^{-3}$ bar
• Temperature, $T = 300$ K
• Gas constant, $R = 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1}$

To Find

Molar mass of the protein, $M_2$.

Formula

$M_2 = \frac{w_2 R T}{\Pi V}$

Solution

Substitute the given values into the formula. $M_2 = \frac{(1.26 \text{ g}) \times (0.083 \text{ L bar mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})}{(2.57 \times 10^{-3} \text{ bar}) \times (0.200 \text{ L})}$ $M_2 = \frac{31.374}{0.000514} = 61038.9 \text{ g mol}^{-1} \approx 61,022 \text{ g mol}^{-1}$ (Note: Minor rounding differences may occur, the source text value is 61,022)

Final Answer The molar mass of the protein is $61,022 \text{ g mol}^{-1}$.

Reverse Osmosis and Water Purification

If a pressure greater than the osmotic pressure is applied to the solution side of an SPM, the direction of solvent flow can be reversed. This is called reverse osmosis (RO). Pure solvent is forced out of the solution, leaving the impurities behind. This process is widely used for the desalination of seawater to produce drinking water.

Abnormal Molar Masses

When using colligative properties to measure the molar mass of a solute, we sometimes get a value that is different from the expected (normal) value. This is called an abnormal molar mass. It occurs when the solute undergoes association or dissociation in the solution.

• Dissociation: Ionic compounds (like KCl) split into ions in solution. One mole of KCl produces two moles of particles ($K^+$ and $Cl^-$). Since colligative properties depend on the number of particles, the observed effect is larger than expected. This leads to a calculated molar mass that is lower than the true value.
• Association: Some molecules (like ethanoic acid in benzene) form dimers or larger clusters. Two molecules of ethanoic acid associate to form one particle. This reduces the total number of particles in the solution. The observed colligative effect is smaller than expected, leading to a calculated molar mass that is higher than the true value.

The van't Hoff Factor (i)

To account for association or dissociation, the van't Hoff factor (i) was introduced. It is defined in three ways: $i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$ $i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$ $i = \frac{\text{Total moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}}$

• For dissociation, $i > 1$ (e.g., for NaCl, $i \approx 2$).
• For association, $i < 1$ (e.g., for ethanoic acid dimerization, $i \approx 0.5$).
• For solutes that neither associate nor dissociate, $i = 1$.

The equations for colligative properties are modified to include the van't Hoff factor:

• Relative Lowering of Vapour Pressure: $\frac{p_1^0 - p_1}{p_1^0} = i \cdot x_2$
• Elevation of Boiling Point: $\Delta T_b = i K_b m$
• Depression of Freezing Point: $\Delta T_f = i K_f m$
• Osmotic Pressure: $\Pi = i CRT$

Given

• Mass of benzoic acid ($w_2$) = 2 g
• Mass of benzene ($w_1$) = 25 g
• Depression in freezing point, $\Delta T_f = 1.62$ K
• $K_f$ for benzene = $4.9 \text{ K kg mol}^{-1}$
• Normal molar mass of benzoic acid, $C_7H_6O_2 = (12 \times 7) + (1 \times 6) + (16 \times 2) = 122 \text{ g mol}^{-1}$

To Find

Percentage association of the acid.

Formula

• Abnormal Molar Mass: $M_{2, abnormal} = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
• van't Hoff factor: $i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
• For dimerization equilibrium $2A \rightleftharpoons A_2$, if $x$ is the degree of association, then $i = 1 - \frac{x}{2}$.

Solution

First, calculate the abnormal (experimental) molar mass from the colligative property data. $M_{2, abnormal} = \frac{(4.9 \text{ K kg mol}^{-1}) \times (2 \text{ g}) \times 1000}{(1.62 \text{ K}) \times (25 \text{ g})} = \frac{9800}{40.5} = 241.98 \text{ g mol}^{-1}$

Next, calculate the van't Hoff factor, $i$. $i = \frac{122 \text{ g mol}^{-1}}{241.98 \text{ g mol}^{-1}} = 0.504$

Now, relate $i$ to the degree of association, $x$. For dimerization, the total moles at equilibrium are $(1-x) + \frac{x}{2} = 1 - \frac{x}{2}$. Therefore, $i = 1 - \frac{x}{2}$. $0.504 = 1 - \frac{x}{2}$ $\frac{x}{2} = 1 - 0.504 = 0.496$ $x = 2 \times 0.496 = 0.992$

The percentage association is $x \times 100$.

Final Answer The percentage association of benzoic acid is 99.2%.