For a dilute solution containing 2.5 g of non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K_b = 0.76 kg mol^–1)

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718

724

736

740

answer is A.

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Detailed Solution

$\begin{array}{l} {ΔT}_{b} = K_{b} \cdot \frac{wt}{Gmt} \cdot \frac{1000}{M \cdot wt of solvent} \Rightarrow Gmt = 9 .5 \\ \frac{P^{0} - P}{P^{0}} = \frac{W}{m} \cdot \frac{M}{W} \Rightarrow 760 - P = 36 \Rightarrow P = 724 \end{array}$

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