Some Basic Concepts Of ChemistryClass 11 Chemistry NCERT Solutions

Given: Chemical formulas (i) H₂O, (ii) CO₂, (iii) CH₄

To Find: Molar mass of each compound.

Solution: We will use the standard atomic masses of the elements:

(i) Molar mass of H₂O: $\text{Molar mass of H}_2\text{O} = (2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})$ $= (2 \times 1.008) + (1 \times 16.00)$ $= 2.016 + 16.00$ $= 18.016 \text{ g/mol}$

(ii) Molar mass of CO₂: $\text{Molar mass of CO}_2 = (1 \times \text{Atomic mass of C}) + (2 \times \text{Atomic mass of O})$ $= (1 \times 12.011) + (2 \times 16.00)$ $= 12.011 + 32.00$ $= 44.011 \text{ g/mol}$

(iii) Molar mass of CH₄: $\text{Molar mass of CH}_4 = (1 \times \text{Atomic mass of C}) + (4 \times \text{Atomic mass of H})$ $= (1 \times 12.011) + (4 \times 1.008)$ $= 12.011 + 4.032$ $= 16.043 \text{ g/mol}$

Given: Three different scenarios for the combustion of carbon.

To Find: The amount of carbon dioxide (CO₂) produced in each case.

Balanced Chemical Equation: The combustion of carbon in oxygen is represented by: $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ From the equation, 1 mole of Carbon reacts with 1 mole of Oxygen (O₂) to produce 1 mole of Carbon Dioxide (CO₂). Molar mass of O₂ = $2 \times 16 = 32$ g/mol Molar mass of CO₂ = $12.01 + (2 \times 16) = 44.01$ g/mol

(i) 1 mole of carbon is burnt in air. Air contains an excess of oxygen. Therefore, carbon is the limiting reactant. According to the stoichiometry, 1 mole of C will produce 1 mole of CO₂. Amount of CO₂ produced = 1 mole. Mass of CO₂ produced = $1 \text{ mol} \times 44.01 \text{ g/mol} = 44.01 \text{ g}$.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen. First, find the moles of each reactant: Moles of C = 1 mol Moles of O₂ = $\frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5$ mol According to the equation, 1 mole of C requires 1 mole of O₂. We have 1 mole of C but only 0.5 moles of O₂. Thus, O₂ is the limiting reagent. The amount of CO₂ produced will be determined by the amount of O₂. Since 1 mole of O₂ produces 1 mole of CO₂, 0.5 moles of O₂ will produce 0.5 moles of CO₂. Mass of CO₂ produced = $0.5 \text{ mol} \times 44.01 \text{ g/mol} = 22.005 \text{ g}$.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen. First, find the moles of each reactant: Moles of C = 2 mol Moles of O₂ = $\frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5$ mol According to the equation, 1 mole of C requires 1 mole of O₂. To react with 2 moles of C, we would need 2 moles of O₂. We only have 0.5 moles of O₂. Thus, O₂ is the limiting reagent. The amount of CO₂ produced will be determined by the amount of O₂. 0.5 moles of O₂ will produce 0.5 moles of CO₂. Mass of CO₂ produced = $0.5 \text{ mol} \times 44.01 \text{ g/mol} = 22.005 \text{ g}$.

Given: Amount of ethane (C₂H₆) = 3 moles

To solve this, we match each prefix with its corresponding power of 10 multiple.

The correct matches are:

Given: Contamination level of chloroform (CHCl₃) = 15 ppm (by mass).

(i) Express this in per cent by mass. 'ppm' stands for parts per million. 15 ppm by mass means 15 parts of chloroform are present in $10^6$ parts of the solution by mass. $\text{Mass per cent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$ $\text{Mass per cent} = \frac{15}{10^6} \times 100 = 15 \times 10^{-6} \times 100 = 15 \times 10^{-4} \%$ $= 0.0015 \%$

(ii) Determine the molality of chloroform in the water sample. Molality (m) is defined as moles of solute per kg of solvent. $\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$ From 15 ppm, let's assume we have $10^6$ g of the water sample (solution). Mass of solution = $10^6$ g Mass of solute (CHCl₃) = 15 g Mass of solvent (water) = Mass of solution - Mass of solute Mass of solvent = $10^6$ g - 15 g ≈ $10^6$ g (since the amount of solute is very small) Mass of solvent in kg = $10^6 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 1000$ kg

Now, calculate the moles of chloroform (CHCl₃). Molar mass of CHCl₃ = $12.01 + 1.008 + (3 \times 35.45) = 12.01 + 1.008 + 106.35 = 119.368$ g/mol $\text{Moles of CHCl}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{15 \text{ g}}{119.368 \text{ g/mol}} = 0.1256 \text{ mol}$

Now, calculate molality: $\text{Molality (m)} = \frac{0.1256 \text{ mol}}{1000 \text{ kg}} = 1.256 \times 10^{-4} \text{ m}$

Answer: Scientific notation expresses a number in the form $N \times 10^n$, where N is a number between 1.0 and 9.999..., and n is an integer.

Answer: We apply the rules for determining significant figures.

Answer: We round each number to retain only three significant figures.

(a) Law of Chemical Combination To analyze the data, let's fix the mass of one reactant (dinitrogen) and observe the mass of the other (dioxygen).

For experiments (i) and (ii), the mass of dinitrogen is fixed at 14 g. The masses of dioxygen that combine with 14 g of dinitrogen are 16 g and 32 g. The ratio of these masses is $16:32$, which simplifies to 1:2.

For experiments (iii) and (iv), let's fix the mass of dinitrogen to 14 g. To do this, we halve the given masses. In (iii), 28 g of dinitrogen reacts with 32 g of dioxygen. So, 14 g of dinitrogen reacts with 16 g of dioxygen. In (iv), 28 g of dinitrogen reacts with 80 g of dioxygen. So, 14 g of dinitrogen reacts with 40 g of dioxygen. Now, let's look at the masses of dioxygen that combine with a fixed mass (14 g) of dinitrogen across all experiments: 16 g, 32 g, 16 g, 40 g. The different masses are 16 g, 32 g and 40 g. The ratio of these masses is $16:32:40$, which simplifies to 2:4:5.

Since the masses of one element (dioxygen) that combine with a fixed mass of another element (dinitrogen) are in a simple whole-number ratio, the data obeys the Law of Multiple Proportions.

Statement of the Law: If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Given: Mass of N₂ = $2.00 \times 10^3$ g = 2000 g Mass of H₂ = $1.00 \times 10^3$ g = 1000 g Balanced Equation: N₂(g) + 3H₂(g) → 2NH₃(g)

Solution: Step 1: Convert mass of reactants to moles. Molar mass of N₂ = 2 × 14.01 = 28.02 g/mol Molar mass of H₂ = 2 × 1.008 = 2.016 g/mol

$\text{Moles of N}_2 = \frac{2000 \text{ g}}{28.02 \text{ g/mol}} = 71.38 \text{ mol}$ $\text{Moles of H}_2 = \frac{1000 \text{ g}}{2.016 \text{ g/mol}} = 496.03 \text{ mol}$

Step 2: Identify the limiting reagent. From the balanced equation, the stoichiometric ratio is 1 mole of N₂ reacts with 3 moles of H₂. Let's find out how many moles of H₂ are required to react with 71.38 mol of N₂: $\text{Moles of H}_2 \text{ required} = 71.38 \text{ mol N}_2 \times \frac{3 \text{ mol H}_2}{1 \text{ mol N}_2} = 214.14 \text{ mol H}_2$ We have 496.03 mol of H₂, which is more than the 214.14 mol required. Therefore, H₂ is in excess and N₂ is the limiting reagent.

(i) Calculate the mass of ammonia produced. The amount of product formed is determined by the limiting reagent (N₂). From the equation, 1 mole of N₂ produces 2 moles of NH₃. $\text{Moles of NH}_3 \text{ produced} = 71.38 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 142.76 \text{ mol NH}_3$ Molar mass of NH₃ = 14.01 + (3 × 1.008) = 17.034 g/mol $\text{Mass of NH}_3 \text{ produced} = 142.76 \text{ mol} \times 17.034 \text{ g/mol} = 2432.1 \text{ g}$ $= 2.432 \times 10^3 \text{ g}$

(ii) Will any of the two reactants remain unreacted? Yes, the reactant in excess (dihydrogen, H₂) will remain unreacted.

(iii) If yes, which one and what would be its mass? Reactant remaining is dihydrogen (H₂).

$\text{Moles of H}_2 \text{ remaining} = \text{Initial moles of H}_2 - \text{Reacted moles of H}_2$ $= 496.03 \text{ mol} - 214.14 \text{ mol} = 281.89 \text{ mol}$

$\text{Mass of H}_2 \text{ remaining} = \text{Moles remaining} \times \text{Molar mass of H}_2$ $= 281.89 \text{ mol} \times 2.016 \text{ g/mol} = 568.29 \text{ g}$

To Convert: The given quantities into their basic SI units. The basic unit for length is the metre (m). The basic unit for mass is the kilogram (kg).

Conversion Factors: 1 picometre (pm) = $10^{-12}$ metres (m) 1 milligram (mg) = $10^{-3}$ grams (g) 1 kilogram (kg) = $10^3$ grams (g), so 1 mg = $10^{-6}$ kg

(i) 28.7 pm $28.7 \text{ pm} = 28.7 \times 10^{-12} \text{ m}$ In standard scientific notation: $= 2.87 \times 10^{-11} \text{ m}$

(ii) 15.15 pm $15.15 \text{ pm} = 15.15 \times 10^{-12} \text{ m}$ In standard scientific notation: $= 1.515 \times 10^{-11} \text{ m}$

(iii) 25365 mg $25365 \text{ mg} = 25365 \times 10^{-6} \text{ kg}$ In standard scientific notation: $= 2.5365 \times 10^4 \times 10^{-6} \text{ kg} = 2.5365 \times 10^{-2} \text{ kg}$

Given: Mass of CO₂ produced = 3.38 g Mass of H₂O produced = 0.690 g Volume of gas at STP = 10.0 L Mass of 10.0 L of gas = 11.6 g

(i) Empirical Formula Step 1: Find the mass of Carbon (C) and Hydrogen (H) in the sample. Molar mass of CO₂ = 44.01 g/mol; Atomic mass of C = 12.01 g/mol $\text{Mass of C} = \text{Mass of CO}_2 \times \frac{\text{Atomic mass of C}}{\text{Molar mass of CO}_2} = 3.38 \text{ g} \times \frac{12.01}{44.01} = 0.922 \text{ g}$ Molar mass of H₂O = 18.02 g/mol; Atomic mass of H = 1.008 g/mol $\text{Mass of H} = \text{Mass of H}_2\text{O} \times \frac{2 \times \text{Atomic mass of H}}{\text{Molar mass of H}_2\text{O}} = 0.690 \text{ g} \times \frac{2 \times 1.008}{18.02} = 0.077 \text{ g}$

Step 2: Convert mass to moles. $\text{Moles of C} = \frac{0.922 \text{ g}}{12.01 \text{ g/mol}} = 0.0768 \text{ mol}$ $\text{Moles of H} = \frac{0.077 \text{ g}}{1.008 \text{ g/mol}} = 0.0764 \text{ mol}$

Step 3: Find the simplest whole-number ratio. Divide by the smallest value (0.0764): $\text{Ratio for C} = \frac{0.0768}{0.0764} \approx 1$ $\text{Ratio for H} = \frac{0.0764}{0.0764} = 1$ The ratio of C:H is 1:1. The empirical formula is CH.

(ii) Molar Mass of the Gas At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 L. We are given that 10.0 L of the gas weighs 11.6 g. $\text{Mass of 22.4 L of gas} = \frac{11.6 \text{ g}}{10.0 \text{ L}} \times 22.4 \text{ L} = 26.0 \text{ g}$ So, the molar mass of the gas is approximately 26.0 g/mol.

(iii) Molecular Formula Step 1: Find the empirical formula mass. Empirical formula mass of CH = 12.01 + 1.008 = 13.018 g/mol

Step 2: Find the ratio 'n'. $n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{26.0 \text{ g/mol}}{13.018 \text{ g/mol}} \approx 2$

Step 3: Determine the molecular formula. $\text{Molecular Formula} = n \times (\text{Empirical Formula}) = 2 \times (\text{CH}) = \text{C}_2\text{H}_2$ The gas is acetylene (ethyne).