'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g text mol^-1$g \text{ mol}^{-1}$ when dissolved in 100 text mL$100 \text{ mL}$ water, decreases vapour pressure of water from 640 text mm Hg$640 \text{ mm Hg}$ to 600 text mm Hg$600 \text{ mm Hg}$. If aqueous solution of the electrolyte boils at 375 text K$375 \text{ K}$ and K_b$K_b$ for water is 0.52 text K kg mol^-1$0.52 \text{ K kg mol}^{-1}$, then the mole fraction of the electrolyte solute (X_2)$(X_2)$ in the solution can be expressed as (Given : density of water = 1 text g/mL$1 \text{ g/mL}$ and boiling point of water = 373 text K$373 \text{ K}$)

### Related Formula Delta T_b = K_b times m$$\Delta T_b = K_b \times m$$ m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight 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: (Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m$$(\Delta \mathrm{T_b})_{\mathrm{PQ}} = \mathrm{K_b} \cdot m$$ 1.176 = 5 times frac1mathrmM_1 times frac100050$$1.176 = 5 \times \frac{1}{\mathrm{M}_1} \times \frac{1000}{50}$$ mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol$$\mathrm{M}_1 = \frac{5 \times 20}{1.176} = 85.03 \text{ g/mol}$$ For compound PQ_2$PQ_2$: (Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689$$(\Delta \mathrm{T_b})_{\mathrm{PQ}_2} = 5 \times \frac{1}{\mathrm{M}_2} \times \frac{1000}{50} = 0.689$$ mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol$$\mathrm{M}_2 = \frac{5 \times 20}{0.689} = 145.13 \text{ g/mol}$$ Let molar mass of P & Q be mathrmM_P$\mathrm{M}_P$ and mathrmM_Q$\mathrm{M}_Q$ respectively: mathrmM_P + mathrmM_Q = 85.03 quad text--- (1)$$\mathrm{M}_P + \mathrm{M}_Q = 85.03 \quad \text{--- (1)}$$ mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2)$$\mathrm{M}_P + 2\mathrm{M}_Q = 145.13 \quad \text{--- (2)}$$ Subtracting (1) from (2): mathrmM_Q = 145.13 - 85.03 = 60.1 approx 60 text g/mol$$\mathrm{M}_Q = 145.13 - 85.03 = 60.1 \approx 60 \text{ g/mol}$$ Substituting back into (1): mathrmM_P + 60.1 = 85.03 implies mathrmM_P = 24.93 approx 25 text 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 pi = i C R T$\pi = i C R T$ ### Core Logic Given: - pi = 12 text atm$\pi = 12 \text{ atm}$ - T = 300 text K$T = 300 \text{ K}$ - i = 2$i = 2$ (for complete dissociation of textNaCl$\text{NaCl}$) - R = 0.08 text L atm K^-1 text mol^-1$R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$ 12 = 2 times C times 0.08 times 300 implies 12 = 48C implies C = 0.25 text 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 textNaCl = 23 + 35.5 = 58.5 text g/mol$\text{NaCl} = 23 + 35.5 = 58.5 \text{ g/mol}$. Strength = 0.25 times 58.5 = 14.625 text g/L approx 15 text 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 textNaCl$\text{NaCl}$. ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Delta T_b = K_b cdot m, quad fracP^circ - P_sP^circ = chi_textsolute$$\Delta T_b = K_b \cdot m, \quad \frac{P^{\circ} - P_s}{P^{\circ}} = \chi_{\text{solute}}$$ ### Core Logic Using boiling point elevation: Delta T_b = K_b times m implies 0.5 = 5.0 times m implies m = 0.1 text 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 = frac150300 = 0.5 text mol$= \frac{150}{300} = 0.5 \text{ mol}$. Moles of solute X = 0.1 times 0.150 = 0.015 text mol$= 0.1 \times 0.150 = 0.015 \text{ mol}$. Relative lowering of vapour pressure = chi_textsolute = frac0.0150.015 + 0.5 approx frac0.0150.5 = 0.03 = 3 times 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 = K_H cdot X$$P = K_{H} \cdot X$$ Taking logarithm on both sides: log P = log K_H + log X$$\log P = \log K_{H} + \log X$$ ### Core Logic From Henry's Law: log P(g) = log X(g) + log K_H$$\log P(g) = \log X(g) + \log K_{H}$$ This equation represents a straight line of the form y = mx + c$y = mx + c$, where: y = log P(g)$y = \log P(g)$ x = log X(g)$x = \log X(g)$ Slope (m$m$) = +1 y-intercept (c$c$) = log K_H$\log K_{H}$ (which is positive). The plot of log P$\log P$ versus log 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 = K_H 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: K_H$K_H$ 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 (K_H$K_H$) 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 K_H$K_H$) 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