Solutions

Given:
- Mass of benzene = 22 g
- Mass of carbon tetrachloride = 122 g
- Total mass of solution = 22 + 122 = 144 g

Formula:
$$\text{Mass percentage of a component} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$

Mass percentage of benzene:
$$= \frac{22}{144} \times 100 = 15.28\%$$

Mass percentage of CCl₄:
$$= \frac{122}{144} \times 100 = 84.72\%$$

Answer: Mass percentage of benzene = 15.28% and of CCl₄ = 84.72%

Given:
- 30% by mass benzene means 30 g of benzene in 100 g of solution.
- Mass of CCl₄ = 100 − 30 = 70 g

Molar masses:
- Benzene (C₆H₆): $M = 6 \times 12 + 6 \times 1 = 78\,\text{g mol}^{-1}$
- CCl₄: $M = 12 + 4 \times 35.5 = 154\,\text{g mol}^{-1}$

Moles:
$$n_{\text{benzene}} = \frac{30}{78} = 0.3846\,\text{mol}$$
$$n_{\text{CCl}_4} = \frac{70}{154} = 0.4545\,\text{mol}$$

Mole fraction of benzene:
$$x_{\text{benzene}} = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{CCl}_4}} = \frac{0.3846}{0.3846 + 0.4545} = \frac{0.3846}{0.8391} = 0.459$$

Mole fraction of CCl₄:
$$x_{\text{CCl}_4} = 1 - 0.459 = 0.541$$

Answer: Mole fraction of benzene = 0.459 and of CCl₄ = 0.541

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

(a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L:

Molar mass of Co(NO₃)₂·6H₂O:
$$= 58.9 + 2(14 + 48) + 6(18) = 58.9 + 124 + 108 = 290.9 \approx 291\,\text{g mol}^{-1}$$

Moles of Co(NO₃)₂·6H₂O:
$$= \frac{30}{291} = 0.103\,\text{mol}$$

Molarity:
$$= \frac{0.103}{4.3} = 0.024\,\text{mol L}^{-1}$$

(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL:

Using dilution formula: $M_1V_1 = M_2V_2$
$$0.5 \times 30 = M_2 \times 500$$
$$M_2 = \frac{0.5 \times 30}{500} = \frac{15}{500} = 0.03\,\text{mol L}^{-1}$$

Answer: (a) 0.024 M (b) 0.03 M

Given:
- Molality (m) = 0.25 mol kg⁻¹
- Total mass of solution = 2.5 kg
- Molar mass of urea (NH₂CONH₂) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol⁻¹

Let mass of urea = $w$ g, then mass of water = $(2500 - w)$ g = $(2500 - w)/1000$ kg

Using molality formula:
$$m = \frac{\text{moles of urea}}{\text{mass of water in kg}}$$
$$0.25 = \frac{w/60}{(2500 - w)/1000}$$
$$0.25 = \frac{1000w}{60(2500 - w)}$$
$$0.25 \times 60 \times (2500 - w) = 1000w$$
$$15(2500 - w) = 1000w$$
$$37500 - 15w = 1000w$$
$$37500 = 1015w$$
$$w = \frac{37500}{1015} = 36.95\,\text{g} \approx 36.946\,\text{g}$$

Answer: Mass of urea required = 36.946 g

Given:
- 20% (w/w) KI solution means 20 g KI in 100 g solution
- Mass of water = 100 − 20 = 80 g = 0.080 kg
- Density = 1.202 g mL⁻¹
- Molar mass of KI = 39 + 127 = 166 g mol⁻¹
- Molar mass of H₂O = 18 g mol⁻¹

Moles of KI:
$$n_{\text{KI}} = \frac{20}{166} = 0.1205\,\text{mol}$$

Moles of water:
$$n_{\text{H}_2\text{O}} = \frac{80}{18} = 4.444\,\text{mol}$$

(a) Molality:
$$m = \frac{n_{\text{KI}}}{\text{mass of water in kg}} = \frac{0.1205}{0.080} = 1.506 \approx 1.5\,\text{mol kg}^{-1}$$

(b) Molarity:

Volume of 100 g solution:
$$V = \frac{\text{mass}}{\text{density}} = \frac{100}{1.202} = 83.19\,\text{mL} = 0.08319\,\text{L}$$

$$M = \frac{0.1205}{0.08319} = 1.448 \approx 1.45\,\text{mol L}^{-1}$$

(c) Mole fraction of KI:
$$x_{\text{KI}} = \frac{n_{\text{KI}}}{n_{\text{KI}} + n_{\text{H}_2\text{O}}} = \frac{0.1205}{0.1205 + 4.444} = \frac{0.1205}{4.5645} = 0.0263$$

Answer: (a) Molality = 1.5 mol kg⁻¹, (b) Molarity = 1.45 mol L⁻¹, (c) Mole fraction of KI = 0.0263

Given:
- Solubility of H₂S = 0.195 m (molality) = 0.195 mol per kg of water
- At STP, pressure of H₂S = 0.987 atm ≈ 1 atm = 101.325 kPa

Henry's Law: $p = K_H \cdot x$

Finding mole fraction of H₂S:

Moles of H₂S = 0.195 mol (in 1 kg = 1000 g of water)

Moles of H₂O = $\frac{1000}{18} = 55.56$ mol

$$x_{\text{H}_2\text{S}} = \frac{0.195}{0.195 + 55.56} = \frac{0.195}{55.755} = 3.499 \times 10^{-3}$$

Henry's law constant:
$$K_H = \frac{p}{x} = \frac{0.987\,\text{atm}}{3.499 \times 10^{-3}} = 282.1\,\text{atm}$$

Converting to bar (1 atm = 1.013 bar):
$$K_H = 282.1 \times 1.013 = 285.8\,\text{bar} \approx 282\,\text{atm}$$

Answer: Henry's law constant for H₂S = 282 atm (≈ 285.8 bar)

Given:
- $K_H = 1.67 \times 10^8$ Pa
- Pressure of CO₂ = 2.5 atm = $2.5 \times 101325 = 2.533 \times 10^5$ Pa
- Volume of soda water = 500 mL → mass ≈ 500 g (assuming density ≈ 1 g/mL)

Henry's Law: $p_{\text{CO}_2} = K_H \cdot x_{\text{CO}_2}$

$$x_{\text{CO}_2} = \frac{p}{K_H} = \frac{2.533 \times 10^5}{1.67 \times 10^8} = 1.517 \times 10^{-3}$$

Finding moles of CO₂:

Moles of water in 500 g = $\frac{500}{18} = 27.78$ mol

Since $x_{\text{CO}_2}$ is very small:
$$x_{\text{CO}_2} = \frac{n_{\text{CO}_2}}{n_{\text{CO}_2} + n_{\text{H}_2\text{O}}} \approx \frac{n_{\text{CO}_2}}{n_{\text{H}_2\text{O}}}$$

$$n_{\text{CO}_2} = x_{\text{CO}_2} \times n_{\text{H}_2\text{O}} = 1.517 \times 10^{-3} \times 27.78 = 0.04213\,\text{mol}$$

Mass of CO₂:
$$= 0.04213 \times 44 = 1.854\,\text{g} \approx 1.85\,\text{g}$$

Answer: The quantity of CO₂ in 500 mL of soda water = 1.85 g

Given:
- $p_A^0 = 450$ mm Hg, $p_B^0 = 700$ mm Hg
- $p_{\text{total}} = 600$ mm Hg

Using Raoult's Law:
$$p_{\text{total}} = p_A^0 x_A + p_B^0 x_B = p_A^0 x_A + p_B^0(1 - x_A)$$
$$600 = 450\,x_A + 700(1 - x_A)$$
$$600 = 450\,x_A + 700 - 700\,x_A$$
$$600 - 700 = -250\,x_A$$
$$x_A = \frac{100}{250} = 0.4$$
$$x_B = 1 - 0.4 = 0.6$$

Composition of vapour phase:

Partial pressures:
$$p_A = p_A^0 \cdot x_A = 450 \times 0.4 = 180\,\text{mm Hg}$$
$$p_B = p_B^0 \cdot x_B = 700 \times 0.6 = 420\,\text{mm Hg}$$

Mole fractions in vapour phase:
$$y_A = \frac{p_A}{p_{\text{total}}} = \frac{180}{600} = 0.30$$
$$y_B = \frac{p_B}{p_{\text{total}}} = \frac{420}{600} = 0.70$$

Answer: Liquid phase: $x_A = 0.4$, $x_B = 0.6$; Vapour phase: $y_A = 0.30$, $y_B = 0.70$

Given:
- $p_{\text{H}_2\text{O}}^0 = 23.8$ mm Hg
- Mass of urea = 50 g, Molar mass of urea = 60 g mol⁻¹
- Mass of water = 850 g, Molar mass of water = 18 g mol⁻¹

Moles:
$$n_{\text{urea}} = \frac{50}{60} = 0.833\,\text{mol}$$
$$n_{\text{H}_2\text{O}} = \frac{850}{18} = 47.22\,\text{mol}$$

Mole fraction of water:
$$x_{\text{H}_2\text{O}} = \frac{47.22}{47.22 + 0.833} = \frac{47.22}{48.053} = 0.9827$$

Vapour pressure of solution (Raoult's Law):
$$p_{\text{solution}} = x_{\text{H}_2\text{O}} \times p_{\text{H}_2\text{O}}^0 = 0.9827 \times 23.8 = 23.4\,\text{mm Hg}$$

Relative lowering of vapour pressure:
$$\frac{\Delta p}{p^0} = \frac{p^0 - p}{p^0} = x_{\text{urea}} = \frac{0.833}{48.053} = 0.0173$$

Answer: Vapour pressure of solution = 23.4 mm Hg; Relative lowering = 0.0173

Given:
- Normal boiling point of water = 99.63°C
- Desired boiling point = 100°C
- Elevation in boiling point: $\Delta T_b = 100 - 99.63 = 0.37°C = 0.37$ K
- $K_b$ for water = 0.52 K kg mol⁻¹
- Mass of water = 500 g = 0.5 kg
- Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 g mol⁻¹

Using: $\Delta T_b = K_b \times m$
$$0.37 = 0.52 \times m$$
$$m = \frac{0.37}{0.52} = 0.7115\,\text{mol kg}^{-1}$$

Moles of sucrose needed:
$$n = m \times \text{mass of solvent (kg)} = 0.7115 \times 0.5 = 0.3558\,\text{mol}$$

Mass of sucrose:
$$= 0.3558 \times 342 = 121.67\,\text{g}$$

Answer: Mass of sucrose to be added = 121.67 g

Given:
- Depression in freezing point: $\Delta T_f = 1.5$ K
- $K_f = 3.9$ K kg mol⁻¹
- Mass of acetic acid (solvent) = 75 g = 0.075 kg
- Molar mass of ascorbic acid (C₆H₈O₆) = $6(12) + 8(1) + 6(16) = 72 + 8 + 96 = 176$ g mol⁻¹

Using: $\Delta T_f = K_f \times m$
$$1.5 = 3.9 \times m$$
$$m = \frac{1.5}{3.9} = 0.3846\,\text{mol kg}^{-1}$$

Moles of ascorbic acid:
$$n = m \times \text{mass of solvent (kg)} = 0.3846 \times 0.075 = 0.02885\,\text{mol}$$

Mass of ascorbic acid:
$$= 0.02885 \times 176 = 5.077\,\text{g}$$

Answer: Mass of ascorbic acid required = 5.077 g

Given:
- Mass of polymer = 1.0 g
- Molar mass = 185,000 g mol⁻¹
- Volume = 450 mL = 0.450 L
- Temperature = 37°C = 310 K
- $R = 8.314$ Pa m³ mol⁻¹ K⁻¹ = 8.314 J mol⁻¹ K⁻¹

Moles of polymer:
$$n = \frac{1.0}{185000} = 5.405 \times 10^{-6}\,\text{mol}$$

Concentration:
$$C = \frac{n}{V} = \frac{5.405 \times 10^{-6}}{0.450 \times 10^{-3}} = 1.201 \times 10^{-2}\,\text{mol m}^{-3}$$

Osmotic pressure:
$$\pi = CRT = 1.201 \times 10^{-2} \times 8.314 \times 310$$
$$\pi = 1.201 \times 10^{-2} \times 2577.3 = 30.95 \approx 30.96\,\text{Pa}$$

Answer: Osmotic pressure = 30.96 Pa

Definition of Solution:
A solution is a homogeneous mixture of two or more chemically non-reacting substances. The component present in the largest amount is called the solvent and the other component(s) are called solute(s).

Types of Solutions:
Depending on the physical states of solute and solvent, nine types of solutions are possible:

| S.No. | Type of Solution | Solute | Solvent | Example |
|---|---|---|---|---|
| 1 | Gas in Gas | Gas | Gas | Air (mixture of O₂, N₂, etc.) |
| 2 | Gas in Liquid | Gas | Liquid | CO₂ dissolved in water (soda water) |
| 3 | Gas in Solid | Gas | Solid | H₂ dissolved in palladium |
| 4 | Liquid in Gas | Liquid | Gas | Water vapour in air (humidity) |
| 5 | Liquid in Liquid | Liquid | Liquid | Ethanol in water |
| 6 | Liquid in Solid | Liquid | Solid | Mercury in zinc (amalgam) |
| 7 | Solid in Gas | Solid | Gas | Camphor vapours in N₂ |
| 8 | Solid in Liquid | Solid | Liquid | Salt (NaCl) in water |
| 9 | Solid in Solid | Solid | Solid | Copper dissolved in gold (alloy) |

The most common type of solution is solid in liquid (e.g., salt in water).

Answer:
An example of a solid solution in which the solute is a gas is hydrogen dissolved in palladium (Pd).

In this solution, hydrogen gas (solute) is dissolved in solid palladium (solvent). Palladium can absorb large volumes of hydrogen gas to form a solid solution.

Another example: dissolved gases in minerals (e.g., O₂ or N₂ dissolved in certain minerals).

(i) Mole Fraction:
Mole fraction of a component in a solution is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the solution.

For a binary solution with components A and B:
$$x_A = \frac{n_A}{n_A + n_B}, \quad x_B = \frac{n_B}{n_A + n_B}$$

Note: $x_A + x_B = 1$. Mole fraction is dimensionless and independent of temperature.

(ii) Molality:
Molality (m) is defined as the number of moles of solute dissolved per kilogram of solvent.
$$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$$
Unit: mol kg⁻¹. It is independent of temperature.

(iii) Molarity:
Molarity (M) is defined as the number of moles of solute dissolved per litre (dm³) of solution.
$$M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}$$
Unit: mol L⁻¹ (M). It depends on temperature since volume changes with temperature.

(iv) Mass Percentage:
Mass percentage of a component in a solution is defined as the mass of that component present per 100 g of the solution.
$$\text{Mass\%} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$
It is dimensionless and independent of temperature.

Given:
- Mass percentage of HNO₃ = 68%
- Density of solution = 1.504 g mL⁻¹
- Molar mass of HNO₃ = 1 + 14 + 48 = 63 g mol⁻¹

Step 1: Consider 1 L (1000 mL) of solution.

Mass of 1 L solution = $1000 \times 1.504 = 1504$ g

Step 2: Mass of HNO₃ in 1504 g solution:
$$= \frac{68}{100} \times 1504 = 1022.72\,\text{g}$$

Step 3: Moles of HNO₃:
$$= \frac{1022.72}{63} = 16.23\,\text{mol}$$

Step 4: Molarity:
$$M = \frac{16.23\,\text{mol}}{1\,\text{L}} = 16.23\,\text{mol L}^{-1}$$

Answer: Molarity of concentrated HNO₃ = 16.23 M

Given:
- 10% w/w glucose solution → 10 g glucose in 100 g solution
- Mass of water = 100 − 10 = 90 g = 0.090 kg
- Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹
- Molar mass of water = 18 g mol⁻¹
- Density = 1.2 g mL⁻¹

Moles:
$$n_{\text{glucose}} = \frac{10}{180} = 0.0556\,\text{mol}$$
$$n_{\text{water}} = \frac{90}{18} = 5.0\,\text{mol}$$

(a) Molality:
$$m = \frac{0.0556}{0.090} = 0.617\,\text{mol kg}^{-1}$$

(b) Mole fractions:
$$x_{\text{glucose}} = \frac{0.0556}{0.0556 + 5.0} = \frac{0.0556}{5.0556} = 0.011$$
$$x_{\text{water}} = \frac{5.0}{5.0556} = 0.989$$

(c) Molarity:

Volume of 100 g solution:
$$V = \frac{100}{1.2} = 83.33\,\text{mL} = 0.08333\,\text{L}$$

$$M = \frac{0.0556}{0.08333} = 0.667\,\text{mol L}^{-1}$$

Answer: Molality = 0.617 mol kg⁻¹; $x_{\text{glucose}}$ = 0.011, $x_{\text{water}}$ = 0.989; Molarity = 0.667 mol L⁻¹

Given:
- 1 g mixture of Na₂CO₃ and NaHCO₃ in equimolar amounts
- Concentration of HCl = 0.1 M

Molar masses:
- Na₂CO₃ = 106 g mol⁻¹
- NaHCO₃ = 84 g mol⁻¹

Let moles of each = $n$ (equimolar)

Total mass: $106n + 84n = 1$
$$190n = 1 \Rightarrow n = \frac{1}{190} = 5.263 \times 10^{-3}\,\text{mol}$$

Reactions with HCl:
$$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$
$$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$

Moles of HCl required:
- For Na₂CO₃: $2 \times 5.263 \times 10^{-3} = 0.01053$ mol
- For NaHCO₃: $1 \times 5.263 \times 10^{-3} = 0.005263$ mol
- Total moles of HCl = $0.01053 + 0.005263 = 0.01579$ mol

Volume of 0.1 M HCl:
$$V = \frac{0.01579}{0.1} = 0.1579\,\text{L} = 157.9\,\text{mL}$$

Answer: Volume of 0.1 M HCl required = 157.9 mL

Given:
- Solution 1: 300 g of 25% (by mass)
- Solution 2: 400 g of 40% (by mass)

Mass of solute in Solution 1:
$$= \frac{25}{100} \times 300 = 75\,\text{g}$$

Mass of solute in Solution 2:
$$= \frac{40}{100} \times 400 = 160\,\text{g}$$

Total mass of solute:
$$= 75 + 160 = 235\,\text{g}$$

Total mass of solution:
$$= 300 + 400 = 700\,\text{g}$$

Mass percentage of resulting solution:
$$= \frac{235}{700} \times 100 = 33.57\%$$

Answer: Mass percentage of the resulting solution = 33.57%

Given:
- Mass of ethylene glycol = 222.6 g
- Molar mass of C₂H₆O₂ = 2(12) + 6(1) + 2(16) = 62 g mol⁻¹
- Mass of water = 200 g = 0.200 kg
- Density of solution = 1.072 g mL⁻¹

Moles of ethylene glycol:
$$n = \frac{222.6}{62} = 3.590\,\text{mol}$$

(a) Molality:
$$m = \frac{3.590}{0.200} = 17.95 \approx 17.95\,\text{mol kg}^{-1}$$

(b) Molarity:

Total mass of solution = 222.6 + 200 = 422.6 g

Volume of solution:
$$V = \frac{422.6}{1.072} = 394.2\,\text{mL} = 0.3942\,\text{L}$$

$$M = \frac{3.590}{0.3942} = 9.11\,\text{mol L}^{-1}$$

Answer: Molality = 17.95 mol kg⁻¹; Molarity = 9.11 mol L⁻¹

Given:
- Contamination level = 15 ppm (by mass)
- 15 ppm means 15 g of CHCl₃ per 10⁶ g of solution
- Molar mass of CHCl₃ = 12 + 1 + 3(35.5) = 119.5 g mol⁻¹

(i) Percent by mass:
$$\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$$
$$\text{Mass\%} = \frac{15}{10^6} \times 100 = 1.5 \times 10^{-3}\%$$

(ii) Molality:

In 10⁶ g of solution: 15 g CHCl₃ and (10⁶ − 15) ≈ 10⁶ g water ≈ 1000 kg water

Moles of CHCl₃:
$$= \frac{15}{119.5} = 0.1255\,\text{mol}$$

Molality:
$$m = \frac{0.1255}{1000} = 1.255 \times 10^{-4}\,\text{mol kg}^{-1}$$

Answer: (i) 1.5 × 10⁻³ % (ii) Molality = 1.255 × 10⁻⁴ mol kg⁻¹

Answer:

Pure water molecules are held together by strong hydrogen bonds (O–H···O). Pure alcohol molecules are also held together by hydrogen bonds, but these are weaker than those in water.

When alcohol and water are mixed:
- New hydrogen bonds form between alcohol (–OH) and water (H₂O) molecules.
- However, the new interactions (alcohol–water) are weaker than the original water–water hydrogen bonds.
- This results in positive deviation from Raoult's law — the vapour pressure of the solution is higher than expected.
- The solution shows an increase in volume (ΔV > 0) and is endothermic (ΔH > 0) on mixing.

In summary, the molecular interactions between alcohol and water are weaker than those in pure components, leading to positive deviation from ideal behaviour.

Answer:

The dissolution of a gas in a liquid is an exothermic process:
$$\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat}$$

According to Le Chatelier's Principle, when temperature is increased, the equilibrium shifts in the direction that absorbs heat, i.e., in the reverse direction (towards the gas phase).

This means that at higher temperatures, more gas molecules escape from the solution back into the gas phase, thereby decreasing the solubility of the gas.

Also, at higher temperatures, gas molecules have greater kinetic energy and can overcome the intermolecular forces holding them in solution, making them less soluble.

Conclusion: Gases are always less soluble in liquids as temperature increases because dissolution is exothermic and Le Chatelier's principle favours the reverse reaction at higher temperatures.

Henry's Law:
At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically:
$$p = K_H \cdot x$$
where $p$ = partial pressure of the gas, $x$ = mole fraction of the gas in solution, and $K_H$ = Henry's law constant (characteristic of the gas-liquid system at a given temperature).

Important Applications:

1. Carbonated beverages: CO₂ is dissolved in soft drinks under high pressure. When the bottle is opened, pressure decreases and CO₂ escapes, causing fizzing.

2. Scuba diving (Bends): At high pressure underwater, N₂ dissolves in the blood. If a diver ascends too quickly, N₂ comes out of solution rapidly forming bubbles in the blood, causing a painful and dangerous condition called 'bends'. To avoid this, scuba tanks are filled with air diluted with helium.

3. Anoxia at high altitudes: At high altitudes, partial pressure of O₂ is low, so less O₂ dissolves in blood, causing anoxia (oxygen deficiency), making climbers weak.

4. Respiration: Oxygen dissolves in blood in the lungs (high O₂ pressure) and is released in tissues (low O₂ pressure).

Given:
- Case 1: mass of ethane = $6.56 \times 10^{-3}$ g, partial pressure $p_1 = 1$ bar
- Case 2: mass of ethane = $5.00 \times 10^{-2}$ g, partial pressure $p_2 = ?$

Using Henry's Law: $p = K_H \cdot x$

At constant temperature, for the same solvent, the mole fraction (and hence mass, for dilute solutions) is proportional to partial pressure.

Since the solvent amount is the same:
$$\frac{p_2}{p_1} = \frac{\text{mass}_2}{\text{mass}_1}$$
$$p_2 = p_1 \times \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}}$$
$$p_2 = 1 \times \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}} = \frac{0.0500}{0.00656} = 7.62\,\text{bar}$$

Answer: Partial pressure of ethane = 7.62 bar

Raoult's Law for ideal solutions:
$$p_{\text{total}} = p_1^0 x_1 + p_2^0 x_2$$

Positive Deviation from Raoult's Law:
- The observed vapour pressure is greater than predicted by Raoult's law.
- This occurs when solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions.
- The molecules escape more easily into vapour phase.
- \Delta_{\text{mix}}H > 0 (endothermic mixing) — heat is absorbed.
- \Delta_{\text{mix}}V > 0 — volume increases on mixing.
- Example: ethanol + water, acetone + carbon disulphide.

Negative Deviation from Raoult's Law:
- The observed vapour pressure is less than predicted by Raoult's law.
- This occurs when solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions.
- Molecules are held more tightly in solution.
- \Delta_{\text{mix}}H < 0 (exothermic mixing) — heat is released.
- \Delta_{\text{mix}}V < 0 — volume decreases on mixing.
- Example: chloroform + acetone, HCl + water.

Relation of $\Delta_{\text{mix}}H$ to deviations:
- Positive deviation → \Delta_{\text{mix}}H > 0 (endothermic)
- Negative deviation → \Delta_{\text{mix}}H < 0 (exothermic)

Given:
- 2% (w/w) non-volatile solute → 2 g solute in 100 g solution → 98 g water
- Vapour pressure of solution, $p_s = 1.004$ bar
- At normal boiling point of water, $p_{\text{H}_2\text{O}}^0 = 1.013$ bar (1 atm)
- Molar mass of water = 18 g mol⁻¹

Using Raoult's Law:
$$\frac{p^0 - p_s}{p^0} = x_{\text{solute}}$$
$$\frac{1.013 - 1.004}{1.013} = x_{\text{solute}}$$
$$x_{\text{solute}} = \frac{0.009}{1.013} = 0.00888$$

Moles of water:
$$n_{\text{H}_2\text{O}} = \frac{98}{18} = 5.444\,\text{mol}$$

Let molar mass of solute = $M$ g mol⁻¹
$$x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{H}_2\text{O}}}$$

Since $x_{\text{solute}}$ is small:
$$x_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{H}_2\text{O}}} = \frac{2/M}{5.444}$$
$$0.00888 = \frac{2}{5.444 \times M}$$
$$M = \frac{2}{5.444 \times 0.00888} = \frac{2}{0.04834} = 41.37\,\text{g mol}^{-1}$$

Answer: Molar mass of solute ≈ 41.37 g mol⁻¹

Given:
- $p_{\text{heptane}}^0 = 105.2$ kPa, $p_{\text{octane}}^0 = 46.8$ kPa
- Mass of heptane = 26.0 g, Molar mass of heptane (C₇H₁₆) = 100 g mol⁻¹
- Mass of octane = 35 g, Molar mass of octane (C₈H₁₈) = 114 g mol⁻¹

Moles:
$$n_{\text{heptane}} = \frac{26.0}{100} = 0.26\,\text{mol}$$
$$n_{\text{octane}} = \frac{35}{114} = 0.307\,\text{mol}$$

Mole fractions:
$$x_{\text{heptane}} = \frac{0.26}{0.26 + 0.307} = \frac{0.26}{0.567} = 0.4586$$
$$x_{\text{octane}} = 1 - 0.4586 = 0.5414$$

Total vapour pressure (Raoult's Law):
$$p_{\text{total}} = p_{\text{heptane}}^0 \cdot x_{\text{heptane}} + p_{\text{octane}}^0 \cdot x_{\text{octane}}$$
$$= 105.2 \times 0.4586 + 46.8 \times 0.5414$$
$$= 48.24 + 25.34 = 73.58\,\text{kPa}$$

Answer: Vapour pressure of the mixture = 73.58 kPa

Given:
- $p_{\text{H}_2\text{O}}^0 = 12.3$ kPa
- Molality = 1 mol kg⁻¹ → 1 mol solute in 1000 g (1 kg) of water
- Molar mass of water = 18 g mol⁻¹

Moles of water:
$$n_{\text{H}_2\text{O}} = \frac{1000}{18} = 55.56\,\text{mol}$$

Mole fraction of solute:
$$x_{\text{solute}} = \frac{1}{1 + 55.56} = \frac{1}{56.56} = 0.01768$$

Mole fraction of water:
$$x_{\text{H}_2\text{O}} = 1 - 0.01768 = 0.9823$$

Vapour pressure of solution:
$$p = x_{\text{H}_2\text{O}} \times p_{\text{H}_2\text{O}}^0 = 0.9823 \times 12.3 = 12.08\,\text{kPa}$$

Answer: Vapour pressure of the solution = 12.08 kPa

Given:
- Vapour pressure is reduced to 80% → $p_s = 0.80\, p^0$
- Molar mass of solute = 40 g mol⁻¹
- Mass of octane = 114 g, Molar mass of octane (C₈H₁₈) = 114 g mol⁻¹

Relative lowering of vapour pressure:
$$\frac{p^0 - p_s}{p^0} = \frac{p^0 - 0.80p^0}{p^0} = 0.20$$

Using Raoult's Law:
$$\frac{p^0 - p_s}{p^0} = x_{\text{solute}}$$
$$x_{\text{solute}} = 0.20$$

Moles of octane:
$$n_{\text{octane}} = \frac{114}{114} = 1\,\text{mol}$$

Let moles of solute = $n$
$$x_{\text{solute}} = \frac{n}{n + 1} = 0.20$$
$$n = 0.20(n + 1)$$
$$n = 0.20n + 0.20$$
$$0.80n = 0.20$$
$$n = 0.25\,\text{mol}$$

Mass of solute:
$$= 0.25 \times 40 = 10\,\text{g}$$

Answer: Mass of non-volatile solute = 10 g

Given:
- Case 1: 30 g solute + 90 g water, $p_1 = 2.8$ kPa
- Case 2: 30 g solute + (90+18) = 108 g water, $p_2 = 2.9$ kPa
- Let molar mass of solute = $M$ g mol⁻¹, $p^0$ = vapour pressure of pure water

Moles:
$$n_{\text{solute}} = \frac{30}{M}$$

Case 1: $n_{\text{water}} = \frac{90}{18} = 5$ mol

Case 2: $n_{\text{water}} = \frac{108}{18} = 6$ mol

Applying Raoult's Law:

Case 1:
$$\frac{p^0 - 2.8}{p^0} = \frac{30/M}{30/M + 5} \quad \cdots (1)$$

Case 2:
$$\frac{p^0 - 2.9}{p^0} = \frac{30/M}{30/M + 6} \quad \cdots (2)$$

Let $n = 30/M$ (moles of solute)

From (1): $\frac{p^0 - 2.8}{p^0} = \frac{n}{n+5}$ → $1 - \frac{2.8}{p^0} = \frac{n}{n+5}$

From (2): $\frac{p^0 - 2.9}{p^0} = \frac{n}{n+6}$ → $1 - \frac{2.9}{p^0} = \frac{n}{n+6}$

Subtracting equation (2) from (1):
$$\frac{2.9}{p^0} - \frac{2.8}{p^0} = \frac{n}{n+5} - \frac{n}{n+6}$$
$$\frac{0.1}{p^0} = n\left[\frac{1}{n+5} - \frac{1}{n+6}\right] = n \cdot \frac{1}{(n+5)(n+6)}$$
$$\frac{0.1}{p^0} = \frac{n}{(n+5)(n+6)} \quad \cdots (3)$$

From equation (1):
$$\frac{2.8}{p^0} = 1 - \frac{n}{n+5} = \frac{5}{n+5}$$
$$p^0 = \frac{2.8(n+5)}{5} \quad \cdots (4)$$

Substituting (4) into (3):
$$\frac{0.1 \times 5}{2.8(n+5)} = \frac{n}{(n+5)(n+6)}$$
$$\frac{0.5}{2.8(n+5)} = \frac{n}{(n+5)(n+6)}$$
$$\frac{0.5}{2.8} = \frac{n}{n+6}$$
$$0.5(n+6) = 2.8n$$
$$0.5n + 3 = 2.8n$$
$$3 = 2.3n$$
$$n = \frac{3}{2.3} = 1.304\,\text{mol}$$

(i) Molar mass of solute:
$$M = \frac{30}{n} = \frac{30}{1.304} = 23.0\,\text{g mol}^{-1}$$

(ii) Vapour pressure of water:
From (4):
$$p^0 = \frac{2.8(1.304 + 5)}{5} = \frac{2.8 \times 6.304}{5} = \frac{17.65}{5} = 3.53\,\text{kPa}$$

Answer: (i) Molar mass of solute = 23.0 g mol⁻¹ (ii) Vapour pressure of water = 3.53 kPa

Given:
- 5% cane sugar solution: freezing point = 271 K
- Freezing point of pure water = 273.15 K
- $\Delta T_f$ (cane sugar) = 273.15 − 271 = 2.15 K
- Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g mol⁻¹
- Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹

5% solution means: 5 g solute in 95 g water (= 0.095 kg)

Molality of cane sugar:
$$m_{\text{sugar}} = \frac{5/342}{0.095} = \frac{0.01462}{0.095} = 0.1539\,\text{mol kg}^{-1}$$

Finding $K_f$:
$$K_f = \frac{\Delta T_f}{m} = \frac{2.15}{0.1539} = 13.97\,\text{K kg mol}^{-1}$$

Molality of glucose:
$$m_{\text{glucose}} = \frac{5/180}{0.095} = \frac{0.02778}{0.095} = 0.2924\,\text{mol kg}^{-1}$$

Depression in freezing point for glucose:
$$\Delta T_f = K_f \times m = 13.97 \times 0.2924 = 4.085\,\text{K}$$

Freezing point of glucose solution:
$$T_f = 273.15 - 4.085 = 269.065 \approx 269.07\,\text{K}$$

Answer: Freezing point of 5% glucose solution = 269.07 K

Given:
- Solvent: benzene, mass = 20 g = 0.020 kg
- $K_f = 5.1$ K kg mol⁻¹
- For AB₂: 1 g, $\Delta T_f = 2.3$ K
- For AB₄: 1 g, $\Delta T_f = 1.3$ K

Molar mass from $\Delta T_f = K_f \times m$:

For AB₂:
$$m = \frac{\Delta T_f}{K_f} = \frac{2.3}{5.1} = 0.4510\,\text{mol kg}^{-1}$$
$$\text{Moles of AB}_2 = 0.4510 \times 0.020 = 9.02 \times 10^{-3}\,\text{mol}$$
$$M_{\text{AB}_2} = \frac{1}{9.02 \times 10^{-3}} = 110.86 \approx 110.9\,\text{g mol}^{-1}$$

For AB₄:
$$m = \frac{1.3}{5.1} = 0.2549\,\text{mol kg}^{-1}$$
$$\text{Moles of AB}_4 = 0.2549 \times 0.020 = 5.098 \times 10^{-3}\,\text{mol}$$
$$M_{\text{AB}_4} = \frac{1}{5.098 \times 10^{-3}} = 196.15 \approx 196.2\,\text{g mol}^{-1}$$

Setting up equations:
Let atomic mass of A = $a$, atomic mass of B = $b$
$$a + 2b = 110.9 \quad \cdots (1)$$
$$a + 4b = 196.2 \quad \cdots (2)$$

Subtracting (1) from (2):
$$2b = 85.3 \Rightarrow b = 42.65 \approx 42.7$$

From (1):
$$a = 110.9 - 2(42.65) = 110.9 - 85.3 = 25.6$$

Answer: Atomic mass of A = 25.6 g mol⁻¹ and atomic mass of B = 42.7 g mol⁻¹

Given:
- Solution 1: 36 g glucose in 1 L, $\pi_1 = 4.98$ bar, T = 300 K
- Solution 2: $\pi_2 = 1.52$ bar, T = 300 K, concentration = ?

Using: $\pi = CRT$

From Solution 1:
$$C_1 = \frac{36/180}{1} = 0.2\,\text{mol L}^{-1}$$
$$R = \frac{\pi_1}{C_1 T} = \frac{4.98}{0.2 \times 300} = 0.083\,\text{bar L mol}^{-1}\text{K}^{-1}$$

For Solution 2:
$$C_2 = \frac{\pi_2}{RT} = \frac{1.52}{0.083 \times 300} = \frac{1.52}{24.9} = 0.061\,\text{mol L}^{-1}$$

Answer: Concentration of the second solution = 0.061 mol L⁻¹

(i) n-hexane and n-octane:
Both are non-polar hydrocarbons. The dominant interaction is London dispersion forces (van der Waals forces).

(ii) I₂ and CCl₄:
Both are non-polar molecules. The dominant interaction is London dispersion forces (induced dipole–induced dipole interactions).

(iii) NaClO₄ and water:
NaClO₄ is an ionic compound and water is a polar molecule. The dominant interaction is ion-dipole interaction. NaClO₄ dissociates into Na⁺ and ClO₄⁻ ions which are stabilised by water molecules through ion-dipole forces.

(iv) Methanol and acetone:
Methanol has an –OH group capable of hydrogen bonding. Acetone has a C=O group (hydrogen bond acceptor). The dominant interaction is hydrogen bonding between the –OH of methanol and the C=O of acetone.

(v) Acetonitrile (CH₃CN) and acetone (C₃H₆O):
Both are polar molecules. The dominant interaction is dipole–dipole interaction between the polar C≡N group of acetonitrile and the polar C=O group of acetone.

n-Octane is a non-polar solvent. The principle is 'like dissolves like' — non-polar solutes dissolve well in non-polar solvents.

Analysis of each solute:
- KCl: Ionic compound, highly polar. Very poor solubility in non-polar n-octane.
- CH₃OH (Methanol): Polar, capable of hydrogen bonding. Poor solubility in n-octane.
- CH₃CN (Acetonitrile): Polar (has C≡N dipole) but less polar than methanol. Slightly better solubility than methanol but still poor.
- Cyclohexane: Non-polar molecule. Completely miscible with n-octane (like dissolves like).

Order of increasing solubility in n-octane:
\text{KCl} < \text{CH}_3\text{OH} < \text{CH}_3\text{CN} < \text{Cyclohexane}

Explanation: KCl is ionic and has the strongest polarity, hence least soluble. Methanol is polar and forms H-bonds, so less soluble than acetonitrile (which is polar but cannot form H-bonds as effectively). Cyclohexane is non-polar like n-octane, so it is most soluble.

Water is a polar solvent. Polar and hydrogen-bonding compounds dissolve well; non-polar compounds do not.

(i) Phenol (C₆H₅OH):
Has –OH group (can form H-bonds with water) but also a large non-polar benzene ring. → Partially soluble in water.

(ii) Toluene (C₆H₅CH₃):
Non-polar hydrocarbon. No H-bonding possible. → Insoluble in water.

(iii) Formic acid (HCOOH):
Small molecule, highly polar, forms strong H-bonds with water. → Highly soluble in water.

(iv) Ethylene glycol (HOCH₂CH₂OH):
Two –OH groups, forms strong H-bonds with water. → Highly soluble in water.

(v) Chloroform (CHCl₃):
Slightly polar but cannot form H-bonds with water effectively. → Insoluble (or very slightly soluble) in water.

(vi) Pentanol (C₅H₁₁OH):
Has –OH group but the long non-polar carbon chain dominates. → Partially soluble in water.

Given:
- Density of lake water = 1.25 g mL⁻¹
- Na⁺ ions = 92 g per kg of water
- Molar mass of Na⁺ = 23 g mol⁻¹

Moles of Na⁺ per kg of water:
$$n_{\text{Na}^+} = \frac{92}{23} = 4\,\text{mol}$$

Mass of solution containing 1 kg water and 92 g Na⁺:
$$= 1000 + 92 = 1092\,\text{g}$$

Volume of this solution:
$$V = \frac{1092}{1.25} = 873.6\,\text{mL} = 0.8736\,\text{L}$$

Molarity of Na⁺:
$$M = \frac{4}{0.8736} = 4.58\,\text{mol L}^{-1}$$

Answer: Molarity of Na⁺ ions = 4.58 mol L⁻¹

Given:
- $K_{sp}$ of CuS = $6 \times 10^{-16}$

Dissolution of CuS:
$$\text{CuS} \rightleftharpoons \text{Cu}^{2+} + \text{S}^{2-}$$

Let solubility = $s$ mol L⁻¹
$$[\text{Cu}^{2+}] = s, \quad [\text{S}^{2-}] = s$$
$$K_{sp} = s \times s = s^2$$
$$s = \sqrt{K_{sp}} = \sqrt{6 \times 10^{-16}}$$
$$s = 2.449 \times 10^{-8}\,\text{mol L}^{-1}$$

Answer: Maximum molarity of CuS = 2.45 × 10⁻⁸ mol L⁻¹

Note: The formula given in the question is C₉H₆O₄ (aspirin = acetylsalicylic acid).

Given:
- Mass of aspirin = 6.5 g
- Mass of acetonitrile = 450 g
- Total mass of solution = 6.5 + 450 = 456.5 g

Mass percentage of aspirin:
$$= \frac{6.5}{456.5} \times 100 = 1.424\%$$

Answer: Mass percentage of aspirin = 1.424%

Given:
- Dose of nalorphene = 1.5 mg = $1.5 \times 10^{-3}$ g
- Molality of solution = $1.5 \times 10^{-3}$ m = $1.5 \times 10^{-3}$ mol kg⁻¹
- Molar mass of nalorphene (C₁₉H₂₁NO₃) = 19(12) + 21(1) + 14 + 3(16) = 228 + 21 + 14 + 48 = 311 g mol⁻¹

Molality means: $1.5 \times 10^{-3}$ mol nalorphene per kg of water

Mass of nalorphene per kg of water:
$$= 1.5 \times 10^{-3} \times 311 = 0.4665\,\text{g per kg water}$$

Mass of solution per kg water:
$$= 1000 + 0.4665 = 1000.4665\,\text{g}$$

To get 1.5 × 10⁻³ g of nalorphene:
$$\text{Mass of solution} = \frac{1000.4665}{0.4665} \times 1.5 \times 10^{-3}$$
$$= \frac{1000.4665 \times 1.5 \times 10^{-3}}{0.4665} = \frac{1.5007}{0.4665} = 3.216\,\text{g}$$

Answer: Mass of solution required = 3.216 g ≈ 3.22 g

Given:
- Volume of solution = 250 mL = 0.250 L
- Molarity = 0.15 M
- Molar mass of benzoic acid (C₆H₅COOH) = 7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122 g mol⁻¹

Moles of benzoic acid required:
$$n = M \times V = 0.15 \times 0.250 = 0.0375\,\text{mol}$$

Mass of benzoic acid:
$$= 0.0375 \times 122 = 4.575\,\text{g}$$

Answer: Mass of benzoic acid required = 4.575 g

Answer:

The depression in freezing point is a colligative property that depends on the number of solute particles in solution. Greater the number of particles (ions/molecules), greater the depression.

All three acids ionise in water:
$$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$$

The degree of ionisation (acid strength) determines the number of particles:

- Acetic acid (CH₃COOH): Weakest acid among the three. Least ionised → fewest particles → least depression in freezing point.

- Trichloroacetic acid (CCl₃COOH): Three electronegative Cl atoms withdraw electron density from the –COOH group, making it a stronger acid than acetic acid. More ionised → more particles → greater depression.

- Trifluoroacetic acid (CF₃COOH): Three highly electronegative F atoms (more electronegative than Cl) cause even greater electron withdrawal, making it the strongest acid. Most ionised → most particles → greatest depression in freezing point.

Order of depression: Acetic acid < Trichloroacetic acid < Trifluoroacetic acid

This is due to the inductive effect of halogens: F > Cl in electronegativity, so CF₃COOH is a stronger acid than CCl₃COOH.

Given:
- Mass of CH₃CH₂CHClCOOH (β-chlorobutyric acid) = 10 g
- Mass of water = 250 g = 0.250 kg
- $K_a = 1.4 \times 10^{-3}$
- $K_f = 1.86$ K kg mol⁻¹
- Molar mass of CH₃CH₂CHClCOOH = 3(12) + 6(1) + 35.5 + 12 + 2(16) + 1 = 36 + 6 + 35.5 + 12 + 32 + 1 = 122.5 g mol⁻¹

Moles of acid:
$$n = \frac{10}{122.5} = 0.08163\,\text{mol}$$

Molality:
$$m = \frac{0.08163}{0.250} = 0.3265\,\text{mol kg}^{-1}$$

Degree of dissociation (α):

For weak acid HA ⇌ H⁺ + A⁻:
$$K_a = \frac{c\alpha^2}{1-\alpha} \approx c\alpha^2 \quad (\text{if } \alpha \ll 1)$$

Here $c = 0.3265$ mol kg⁻¹ (using molality as approximation for concentration):
$$\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.3265}} = \sqrt{4.289 \times 10^{-3}} = 0.06549$$

Van't Hoff factor:
$$i = 1 + \alpha = 1 + 0.06549 = 1.0655$$

Depression in freezing point:
$$\Delta T_f = i \times K_f \times m = 1.0655 \times 1.86 \times 0.3265$$
$$= 1.0655 \times 0.6073 = 0.6470 \approx 0.647\,\text{K}$$

Answer: Depression in freezing point = 0.647 K

Given:
- Mass of CH₂FCOOH = 19.5 g
- Mass of water = 500 g = 0.500 kg
- $\Delta T_f = 1.0$ K
- $K_f = 1.86$ K kg mol⁻¹
- Molar mass of CH₂FCOOH = 12 + 2 + 19 + 12 + 32 + 1 = 78 g mol⁻¹

Moles of CH₂FCOOH:
$$n = \frac{19.5}{78} = 0.25\,\text{mol}$$

Molality:
$$m = \frac{0.25}{0.500} = 0.5\,\text{mol kg}^{-1}$$

Calculated $\Delta T_f$ (without dissociation):
$$\Delta T_f^{\text{calc}} = K_f \times m = 1.86 \times 0.5 = 0.93\,\text{K}$$

Van't Hoff factor:
$$i = \frac{\Delta T_f^{\text{obs}}}{\Delta T_f^{\text{calc}}} = \frac{1.0}{0.93} = 1.0753$$

Degree of dissociation (α):

For HA ⇌ H⁺ + A⁻: $i = 1 + \alpha$
$$\alpha = i - 1 = 1.0753 - 1 = 0.0753$$

Dissociation constant $K_a$:

Concentration of acid $c = 0.5$ mol kg⁻¹ ≈ 0.5 mol L⁻¹ (dilute solution)

$$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} = \frac{0.5 \times 0.005670}{0.9247} = \frac{0.002835}{0.9247} = 3.066 \times 10^{-3}$$

Answer: Van't Hoff factor $i$ = 1.0753; Dissociation constant $K_a$ = 3.07 × 10⁻³

Given:
- $p_{\text{H}_2\text{O}}^0 = 17.535$ mm Hg
- Mass of glucose = 25 g, Molar mass of glucose = 180 g mol⁻¹
- Mass of water = 450 g, Molar mass of water = 18 g mol⁻¹

Moles:
$$n_{\text{glucose}} = \frac{25}{180} = 0.1389\,\text{mol}$$
$$n_{\text{H}_2\text{O}} = \frac{450}{18} = 25\,\text{mol}$$

Mole fraction of water:
$$x_{\text{H}_2\text{O}} = \frac{25}{25 + 0.1389} = \frac{25}{25.1389} = 0.9945$$

Vapour pressure of solution:
$$p = x_{\text{H}_2\text{O}} \times p^0 = 0.9945 \times 17.535 = 17.44\,\text{mm Hg}$$

Answer: Vapour pressure of water in the solution = 17.44 mm Hg

Given:
- $K_H = 4.27 \times 10^5$ mm Hg
- Partial pressure of methane = 760 mm Hg

Henry's Law: $p = K_H \cdot x$
$$x_{\text{CH}_4} = \frac{p}{K_H} = \frac{760}{4.27 \times 10^5} = 1.78 \times 10^{-3}$$

This is the mole fraction of methane in benzene.

Mole fraction of methane = $1.78 \times 10^{-3}$

This means: in 1 mol of solution, $1.78 \times 10^{-3}$ mol is methane.

Moles of benzene ≈ $1 - 1.78 \times 10^{-3} \approx 1$ mol = 78 g

Moles of methane = $1.78 \times 10^{-3}$ mol

Molality:
$$m = \frac{1.78 \times 10^{-3}\,\text{mol}}{0.078\,\text{kg}} = 0.0228\,\text{mol kg}^{-1}$$

Answer: Mole fraction of methane = 1.78 × 10⁻³; Molality = 0.0228 mol kg⁻¹

Given:
- Mass of A = 100 g, $M_A = 140$ g mol⁻¹
- Mass of B = 1000 g, $M_B = 180$ g mol⁻¹
- $p_B^0 = 500$ torr
- $p_{\text{total}} = 475$ torr

Moles:
$$n_A = \frac{100}{140} = 0.7143\,\text{mol}$$
$$n_B = \frac{1000}{180} = 5.556\,\text{mol}$$

Mole fractions:
$$x_A = \frac{0.7143}{0.7143 + 5.556} = \frac{0.7143}{6.270} = 0.1139$$
$$x_B = 1 - 0.1139 = 0.8861$$

Partial pressure of B:
$$p_B = p_B^0 \cdot x_B = 500 \times 0.8861 = 443.1\,\text{torr}$$

Partial pressure of A:
$$p_A = p_{\text{total}} - p_B = 475 - 443.1 = 31.9\,\text{torr}$$

Vapour pressure of pure A:
$$p_A = p_A^0 \cdot x_A$$
$$p_A^0 = \frac{p_A}{x_A} = \frac{31.9}{0.1139} = 280.1\,\text{torr}$$

Answer: Vapour pressure of pure liquid A = 280.1 torr; Vapour pressure of A in solution = 31.9 torr

Given:
- $p_{\text{acetone}}^0 = 741.8$ mm Hg
- $p_{\text{chloroform}}^0 = 632.8$ mm Hg

Ideal solution calculations (using Raoult's law: $p = p^0 x$):

For ideal solution:
$$p_{\text{acetone}}^{\text{ideal}} = 741.8 \times x_{\text{acetone}}$$
$$p_{\text{chloroform}}^{\text{ideal}} = 632.8 \times (1 - x_{\text{acetone}})$$
$$p_{\text{total}}^{\text{ideal}} = 741.8\,x_{\text{acetone}} + 632.8(1 - x_{\text{acetone}}) = 632.8 + 109\,x_{\text{acetone}}$$

Selected ideal values:

| $x_{\text{acetone}}$ | $p_{\text{acetone}}^{\text{ideal}}$ | $p_{\text{chloroform}}^{\text{ideal}}$ | $p_{\text{total}}^{\text{ideal}}$ |
|---|---|---|---|
| 0 | 0 | 632.8 | 632.8 |
| 0.2 | 148.4 | 506.2 | 654.6 |
| 0.4 | 296.7 | 379.7 | 676.4 |
| 0.6 | 445.1 | 253.1 | 698.2 |
| 0.8 | 593.4 | 126.6 | 720.0 |
| 1.0 | 741.8 | 0 | 741.8 |

Experimental total pressures (from given data):

| $x_{\text{acetone}}$ | $p_{\text{acetone}}^{\text{exp}}$ | $p_{\text{chloroform}}^{\text{exp}}$ | $p_{\text{total}}^{\text{exp}}$ |
|---|---|---|---|
| 0 | 0 | 632.8 | 632.8 |
| 0.118 | 54.9 | 548.1 | 603.0 |
| 0.234 | 110.1 | 469.4 | 579.5 |
| 0.360 | 202.4 | 359.7 | 562.1 |
| 0.508 | 322.7 | 257.7 | 580.4 |
| 0.582 | 405.9 | 193.6 | 599.5 |
| 0.645 | 454.1 | 161.2 | 615.3 |
| 0.721 | 521.1 | 120.7 | 641.8 |

Observation: The experimental total vapour pressures are lower than the ideal values calculated from Raoult's law.

Conclusion: The acetone–chloroform system shows negative deviation from Raoult's law.

Reason: Chloroform and acetone form hydrogen bonds (C–H···O=C) with each other. These interactions are stronger than the interactions in pure components, so molecules are held more tightly in solution, reducing vapour pressure below the ideal value. \Delta_{\text{mix}}H &lt; 0 (exothermic mixing).

Given:
- $p_{\text{benzene}}^0 = 50.71$ mm Hg
- $p_{\text{toluene}}^0 = 32.06$ mm Hg
- Mass of benzene = 80 g, Molar mass = 78 g mol⁻¹
- Mass of toluene = 100 g, Molar mass = 92 g mol⁻¹

Moles:
$$n_{\text{benzene}} = \frac{80}{78} = 1.026\,\text{mol}$$
$$n_{\text{toluene}} = \frac{100}{92} = 1.087\,\text{mol}$$

Mole fractions in liquid phase:
$$x_{\text{benzene}} = \frac{1.026}{1.026 + 1.087} = \frac{1.026}{2.113} = 0.4856$$
$$x_{\text{toluene}} = 1 - 0.4856 = 0.5144$$

Partial pressures:
$$p_{\text{benzene}} = 50.71 \times 0.4856 = 24.62\,\text{mm Hg}$$
$$p_{\text{toluene}} = 32.06 \times 0.5144 = 16.49\,\text{mm Hg}$$

Total pressure:
$$p_{\text{total}} = 24.62 + 16.49 = 41.11\,\text{mm Hg}$$

Mole fraction of benzene in vapour phase:
$$y_{\text{benzene}} = \frac{p_{\text{benzene}}}{p_{\text{total}}} = \frac{24.62}{41.11} = 0.5989 \approx 0.599$$

Answer: Mole fraction of benzene in vapour phase = 0.599

Given:
- Total pressure = 10 atm = $10 \times 760 = 7600$ mm Hg
- Volume % of O₂ = 20% → mole fraction in air = 0.20
- Volume % of N₂ = 79% → mole fraction in air = 0.79
- $K_H(\text{O}_2) = 3.30 \times 10^7$ mm Hg
- $K_H(\text{N}_2) = 6.51 \times 10^7$ mm Hg

Partial pressures:
$$p_{\text{O}_2} = 0.20 \times 7600 = 1520\,\text{mm Hg}$$
$$p_{\text{N}_2} = 0.79 \times 7600 = 6004\,\text{mm Hg}$$

Mole fractions in water (Henry's Law: $p = K_H \cdot x$):
$$x_{\text{O}_2} = \frac{p_{\text{O}_2}}{K_H(\text{O}_2)} = \frac{1520}{3.30 \times 10^7} = 4.61 \times 10^{-5}$$
$$x_{\text{N}_2} = \frac{p_{\text{N}_2}}{K_H(\text{N}_2)} = \frac{6004}{6.51 \times 10^7} = 9.22 \times 10^{-5}$$

Answer: Mole fraction of O₂ in water = 4.61 × 10⁻⁵; Mole fraction of N₂ in water = 9.22 × 10⁻⁵

Given:
- $\pi = 0.75$ atm
- $T = 27°C = 300$ K
- $V = 2.5$ L
- $i = 2.47$
- $R = 0.0821$ L atm mol⁻¹ K⁻¹
- Molar mass of CaCl₂ = 40 + 2(35.5) = 111 g mol⁻¹

Using: $\pi = iCRT = i \cdot \frac{n}{V} \cdot RT$
$$n = \frac{\pi V}{iRT} = \frac{0.75 \times 2.5}{2.47 \times 0.0821 \times 300}$$
$$= \frac{1.875}{60.85} = 0.03082\,\text{mol}$$

Mass of CaCl₂:
$$= 0.03082 \times 111 = 3.42\,\text{g}$$

Answer: Amount of CaCl₂ required = 3.42 g