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Solutions - JEE mains PYQ 2026 series

StudyBeacon Quiz – Colligative Properties (JEE Main PYQ)

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Q1. Statement I: Henry's law constant $K_H$ is constant with respect to variations in solution concentration for ideally dilute solutions. Statement II: $K_H$ does not differ for the same solute in different solvents.
Henry’s law: $P = K_H x$ For dilute solutions $K_H$ remains constant. But $K_H$ **depends strongly on solvent and temperature**, so statement II is false.
Q2. 0.3 g solute A (M = 60 g/mol) and 0.9 g solute B (M = 180 g/mol) dissolved in 100 mL water at 27°C.
Osmotic pressure $\pi = CRT$ Moles A $=0.3/60=0.005$ Moles B $=0.9/180=0.005$ Total moles $=0.01$ Volume = 0.1 L $C=0.1$ $\pi = 0.1 \times 0.082 \times 300$ $\pi = 2.46 \text{ atm}$
Q3. Vapour pressure decreases from 640 mmHg to 600 mmHg.
Relative lowering $\frac{\Delta P}{P^0} = x_2$ $\frac{640-600}{640}=0.0625$ $x_2 = 1/16$ Using boiling elevation $\Delta T_b = iK_b m$ After substitution $x_2 = \frac{1.3}{8}\frac{W}{M}$
Q4. Increasing order of boiling point.
Boiling elevation $\Delta T_b = iK_b m$ Compare **i × molality** Glucose → i=1 CaCl₂ → i=3 Urea → i=1 Al₂(SO₄)₃ → i=5 After calculating particle concentration Order becomes I < II < III < IV
Q5. Mole fraction of $H_2SO_4$ after mixing two solutions.
Mass $H_2SO_4$ 98 g solution → 98% → 98 g acid 49 g solution → 49 g acid Total acid = 147 g Moles acid $147/98 = 1.5$ Water moles calculated similarly Final mole fraction $X_{acid}=0.337$
Q6. Pure vapour pressures $P_A^0=55$ , $P_B^0=15$ Vapour mole fraction of A = 0.8.
Raoult law $P_A = X_A P_A^0$ Total pressure $P = X_A P_A^0 + (1-X_A)P_B^0$ Using $y_A = \frac{P_A}{P}$ Solving gives $X_A = 0.5217$
Q7. Henry law $P = K_H x$
Partial pressure of N₂ $0.8 × 10 = 8 \, atm$ Convert to mmHg $8×760$ $x = P/K_H$ After substitution $x ≈ 9.35×10^{5}$
Q8. Using boiling elevation $\Delta T_b = K_b m$
From $m = \frac{\Delta T_b}{K_b}$ Comparing PQ and PQ₂ Solving simultaneous molar mass relations gives $M_P = 25$ $M_Q = 60$

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