NCERT Solutions for class 12 chapter 1 solutions

Answer:

Total mass of the solution = Mass of \( \text{C}_6\text{H}_6 \) + Mass of \( \text{CCl}_4 \)

= \( 22\text{ g} + 122\text{ g} = 144\text{ g} \)

Mass percentage of \( \text{C}_6\text{H}_6 \):

\[ \text{Mass \% of } \text{C}_6\text{H}_6 = \frac{\text{Mass of } \text{C}_6\text{H}_6}{\text{Total mass of solution}} \times 100 \] \[ = \frac{22}{144} \times 100 = 15.28\% \]

Mass percentage of \( \text{CCl}_4 \):

\[ \text{Mass \% of } \text{CCl}_4 = \frac{\text{Mass of } \text{CCl}_4}{\text{Total mass of solution}} \times 100 \] \[ = \frac{122}{144} \times 100 = 84.72\% \]

Alternatively, Mass percentage of \( \text{CCl}_4 = (100 – 15.28)\% = 84.72\% \)

Answer:

Let the total mass of the solution be 100 g. Then, mass of benzene (\( \text{C}_6\text{H}_6 \)) = 30 g and mass of carbon tetrachloride (\( \text{CCl}_4 \)) = \( 100 – 30 = 70\text{ g} \).

Molar mass of benzene (\( \text{C}_6\text{H}_6 \)) = \( (6 \times 12 + 6 \times 1) = 78\text{ g mol}^{-1} \)

Number of moles of \( \text{C}_6\text{H}_6 = \frac{30}{78}\text{ mol} = 0.3846\text{ mol} \)

Molar mass of \( \text{CCl}_4 = (1 \times 12 + 4 \times 35.5) = 154\text{ g mol}^{-1} \)

Number of moles of \( \text{CCl}_4 = \frac{70}{154}\text{ mol} = 0.4545\text{ mol} \)

Mole fraction of \( \text{C}_6\text{H}_6 \):

\[ x_{\text{C}_6\text{H}_6} = \frac{n_{\text{C}_6\text{H}_6}}{n_{\text{C}_6\text{H}_6} + n_{\text{CCl}_4}} \] \[ = \frac{0.3846}{0.3846 + 0.4545} = 0.458 \]

Answer:

Molarity is given by: \( \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} \)

(a) Molar mass of \( \text{Co(NO}_3)_2 \cdot 6\text{H}_2\text{O} = 59 + 2(14 + 3 \times 16) + 6 \times 18 = 291\text{ g mol}^{-1} \)

Moles of \( \text{Co(NO}_3)_2 \cdot 6\text{H}_2\text{O} = \frac{30}{291}\text{ mol} = 0.103\text{ mol} \)

Molarity = \( \frac{0.103\text{ mol}}{4.3\text{ L}} = 0.023\text{ M} \)

(b) Number of moles in 1000 mL of 0.5 M \( \text{H}_2\text{SO}_4 = 0.5\text{ mol} \)

Number of moles in 30 mL of 0.5 M \( \text{H}_2\text{SO}_4 = \frac{0.5 \times 30}{1000} = 0.015\text{ mol} \)

New Volume = 500 mL = 0.5 L

Molarity = \( \frac{0.015\text{ mol}}{0.5\text{ L}} = 0.03\text{ M} \)

Answer:

Molar mass of urea (\( \text{NH}_2\text{CONH}_2 \)) = \( 2(1 \times 14 + 2 \times 1) + 1 \times 12 + 1 \times 16 = 60\text{ g mol}^{-1} \)

0.25 molal aqueous solution means that 0.25 mol of urea is present in 1000 g of water.

Mass of urea in 1000 g water = \( 0.25 \times 60 = 15\text{ g} \)

Total mass of solution = \( 1000 + 15 = 1015\text{ g} = 1.015\text{ kg} \)

Mass of urea required for 1.015 kg solution = 15 g

Mass of urea required for 2.5 kg solution:

\[ = \frac{15 \times 2.5}{1.015} = 36.946\text{ g} \approx 37\text{ g} \]

Answer:

(a) Molar mass of KI = \( 39 + 127 = 166\text{ g mol}^{-1} \)

20% (w/w) KI means 20 g of KI is in 100 g of solution. Mass of water = \( 100 – 20 = 80\text{ g} = 0.08\text{ kg} \).

Moles of KI = \( \frac{20}{166} = 0.12\text{ mol} \)

Molality (\( m \)) = \( \frac{0.12\text{ mol}}{0.08\text{ kg}} = 1.5\text{ m} \)

(b) Density of solution = \( 1.202\text{ g mL}^{-1} \)

Volume of 100 g solution = \( \frac{100\text{ g}}{1.202\text{ g mL}^{-1}} = 83.19\text{ mL} = 83.19 \times 10^{-3}\text{ L} \)

Molarity (\( M \)) = \( \frac{0.12\text{ mol}}{83.19 \times 10^{-3}\text{ L}} = 1.44\text{ M} \)

(c) Moles of water = \( \frac{80}{18} = 4.44\text{ mol} \)

Mole fraction of KI (\( x_{KI} \)) = \( \frac{0.12}{0.12 + 4.44} = 0.0263 \)

Answer:

\( K_H = 1.67 \times 10^8\text{ Pa} \)

\( p = 2.5\text{ atm} = 2.5 \times 1.01325 \times 10^5\text{ Pa} = 2.533125 \times 10^5\text{ Pa} \)

According to Henry’s law: \( p = K_H x \)

\[ x = \frac{p}{K_H} = \frac{2.533125 \times 10^5}{1.67 \times 10^8} = 0.00152 \]

For 500 mL of water, \( n_{H_2O} \approx \frac{500}{18} = 27.78\text{ mol} \)

\[ x = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}} \] \[ n_{CO_2} = x \times n_{H_2O} = 0.00152 \times 27.78 = 0.042\text{ mol} \]

Mass of \( \text{CO}_2 = 0.042 \times 44 = 1.848\text{ g} \)

Homogeneous mixtures of two or more than two components are known as solutions. There are three main types:

• Gaseous solution: Solvent is a gas (e.g., mixture of oxygen and nitrogen).
• Liquid solution: Solvent is a liquid (e.g., ethanol in water).
• Solid solution: Solvent is a solid (e.g., alloys like brass).

(i) Mole fraction: Ratio of moles of a component to the total moles of all components.

(ii) Molality (m): Moles of solute per kilogram of solvent.

(iii) Molarity (M): Moles of solute per litre of solution.

(iv) Mass percentage: Mass of solute in grams present in 100 g of solution.

Let mass of \( \text{Na}_2\text{CO}_3 = x\text{ g} \). Then \( \text{NaHCO}_3 = (1-x)\text{ g} \).

Moles of \( \text{Na}_2\text{CO}_3 = \frac{x}{106} \); Moles of \( \text{NaHCO}_3 = \frac{1-x}{84} \).

Since equimolar: \( \frac{x}{106} = \frac{1-x}{84} \Rightarrow x = 0.5579\text{ g} \).

Moles of each = 0.0053 mol.

Total moles of HCl required = \( 2 \times 0.0053 \) (for carbonate) + \( 1 \times 0.0053 \) (for bicarbonate) = 0.0159 mol.

Volume of 0.1 M HCl = \( \frac{0.0159 \times 1000}{0.1} = 159\text{ mL} \).

If vapour pressure is higher than predicted, it is positive deviation (\( \Delta_{sol}H = \text{positive} \)). If lower, it is negative deviation (\( \Delta_{sol}H = \text{negative} \)).