Solutions — NCERT 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:

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