Given below are two statements: Statement (I): NaCl is added to the ice at 0circC$0^{circ}C$, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circC$0^{circ}C$, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

### Related Formula Δ Tb = Kb × m$$\Delta T_b = K_b \times m$$ m = Weight of solute (g)Molar mass of solute × 1000Weight of solvent (g)$$m = \frac{\text{Weight of solute (g)}}{\text{Molar mass of solute}} \times \frac{1000}{\text{Weight of solvent (g)}}$$ ### Core Logic For compound PQ: (Δ Tb)PQ = Kb · m$$(\Delta \mathrm{T_b})_{\mathrm{PQ}} = \mathrm{K_b} \cdot m$$ 1.176 = 5 × 1M₁ × (1000)/(50)$$1.176 = 5 \times \frac{1}{\mathrm{M}_1} \times \frac{1000}{50}$$ M₁ = (5 × 20)/(1.176) = 85.03 g/mol$$\mathrm{M}_1 = \frac{5 \times 20}{1.176} = 85.03 \text{ g/mol}$$ For compound PQ₂$PQ_2$: (Δ Tb)PQ₂ = 5 × 1M₂ × (1000)/(50) = 0.689$$(\Delta \mathrm{T_b})_{\mathrm{PQ}_2} = 5 \times \frac{1}{\mathrm{M}_2} \times \frac{1000}{50} = 0.689$$ M₂ = (5 × 20)/(0.689) = 145.13 g/mol$$\mathrm{M}_2 = \frac{5 \times 20}{0.689} = 145.13 \text{ g/mol}$$ Let molar mass of P & Q be MP$\mathrm{M}_P$ and MQ$\mathrm{M}_Q$ respectively: MP + MQ = 85.03 --- (1)$$\mathrm{M}_P + \mathrm{M}_Q = 85.03 \quad \text{--- (1)}$$ MP + 2MQ = 145.13 --- (2)$$\mathrm{M}_P + 2\mathrm{M}_Q = 145.13 \quad \text{--- (2)}$$ Subtracting (1) from (2): MQ = 145.13 - 85.03 = 60.1 ≈ 60 g/mol$$\mathrm{M}_Q = 145.13 - 85.03 = 60.1 \approx 60 \text{ g/mol}$$ Substituting back into (1): MP + 60.1 = 85.03 MP = 24.93 ≈ 25 g/mol$$\mathrm{M}_P + 60.1 = 85.03 \implies \mathrm{M}_P = 24.93 \approx 25 \text{ g/mol}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula π = i C R T$\pi = i C R T$ ### Core Logic Given: - π = 12 atm$\pi = 12 \text{ atm}$ - T = 300 K$T = 300 \text{ K}$ - i = 2$i = 2$ (for complete dissociation of NaCl$\text{NaCl}$) - R = 0.08 L atm K⁻¹ mol⁻¹$R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$ 12 = 2 × C × 0.08 × 300 12 = 48C C = 0.25 mol/L$$12 = 2 \times C \times 0.08 \times 300 \implies 12 = 48C \implies C = 0.25 \text{ mol/L}$$ ### Step 1: Calculating Strength Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol$\text{NaCl} = 23 + 35.5 = 58.5 \text{ g/mol}$. Strength = 0.25 × 58.5 = 14.625 g/L ≈ 15 g/L$= 0.25 \times 58.5 = 14.625 \text{ g/L} \approx 15 \text{ g/L}$. ### Pattern Recognition Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor i = 2$i = 2$ for NaCl$\text{NaCl}$. ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Δ Tb = Kb · m, P° - PₛP° = χsolute$$\Delta T_b = K_b \cdot m, \quad \frac{P^{\circ} - P_s}{P^{\circ}} = \chi_{\text{solute}}$$ ### Core Logic Using boiling point elevation: Δ Tb = Kb × m 0.5 = 5.0 × m m = 0.1 mol/kg$$\Delta T_b = K_b \times m \implies 0.5 = 5.0 \times m \implies m = 0.1 \text{ mol/kg}$$ Moles of solvent Y = (150)/(300) = 0.5 mol$= \frac{150}{300} = 0.5 \text{ mol}$. Moles of solute X = 0.1 × 0.150 = 0.015 mol$= 0.1 \times 0.150 = 0.015 \text{ mol}$. Relative lowering of vapour pressure = χsolute = (0.015)/(0.015 + 0.5) ≈ (0.015)/(0.5) = 0.03 = 3 × 10⁻²$= \chi_{\text{solute}} = \frac{0.015}{0.015 + 0.5} \approx \frac{0.015}{0.5} = 0.03 = 3 \times 10^{-2}$. ### Step 1: Final Calculation Thus, the nearest integer value is 3. ### Pattern Recognition Sees: boiling point elevation combined with relative lowering of vapour pressure calculation. Trap: Confusing solvent mass with solution mass when calculating mole fraction. ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula P = KH · X$$P = K_{H} \cdot X$$ Taking logarithm on both sides: P = KH + X$$\log P = \log K_{H} + \log X$$ ### Core Logic From Henry's Law: P(g) = X(g) + KH$$\log P(g) = \log X(g) + \log K_{H}$$ This equation represents a straight line of the form y = mx + c$y = mx + c$, where: y = P(g)$y = \log P(g)$ x = X(g)$x = \log X(g)$ Slope (m$m$) = +1 y-intercept (c$c$) = KH$\log K_{H}$ (which is positive). The plot of P$\log P$ versus X$\log X$ is a straight line with a positive slope and a positive intercept on the y-axis. ### Step 1: Conclusion Graph 3 corresponds to a straight line with a positive slope and a positive y-intercept. ### Pattern Recognition When variables are multiplied (P = KH X$P = K_{H} X$), their log plot is always a straight line with a slope of +1 and an intercept equal to the log of the constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Core Logic Statement I: KH$K_H$ is independent of concentration as long as the solution behaves ideally (is very dilute). This matches the physical definition of Henry's Law constant. (True) Statement II: Henry's Law constant (KH$K_H$) depends on the nature of the gas, the nature of the solvent, and the temperature. Therefore, it absolutely differs for the same gas dissolved in different solvents. (False) ### Step 1: Final Conclusion Statement I is true but Statement II is false. ### Pattern Recognition The constants for solubility laws (like KH$K_H$) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions