A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 text g mol^-1) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is \_\_\_\_ times 10^-2. (Nearest integer) [Given: K_b of the solvent = 5.0 text K kg mol^-1] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

### Related Formula Delta T_b = K_b cdot m, quad fracP^circ - P_sP^circ = chi_textsolute ### 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 Moles of solvent Y = frac150300 = 0.5 text mol. Moles of solute X = 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. ### 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

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