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(D) Given,Vapour pressure of pure water $(p^{\circ})$ $= 23 	ext{ mmHg}$. $10 \%$ mass per cent of solute '$A$' means $10 	ext{ g}$ of '$A$' in $100

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$0.028$

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(D) Given,Vapour pressure of pure water $(p^{\circ})$ $= 23 	ext{ mmHg}$. $10 \%$ mass per cent of solute '$A$' means $10 	ext{ g}$ of '$A$' in $100 	ext{ g}$ of solution. Mass of solute '$A$' $= 10 	ext{ g}$, Mass of water $= 90 	ext{ g}$. Molecular weight of '$A$' $= 50 	ext{ g/mol}$, Molecular weight of water $= 18 	ext{ g/mol}$. Moles of '$A$' $(n_A)$ $= \frac{10}{50} = 0.2 	ext{ mol}$. Moles of water $(n_w)$ $= \frac{90}{18} = 5 	ext{ mol}$. Mole fraction of solvent $(x_w)$ $= \frac{n_w}{n_w + n_A} = \frac{5}{5 + 0.2} = \frac{5}{5.2} \approx 0.9615$. Vapour pressure of solution $(p_s)$ $= x_w 	imes p^{\circ} = 0.9615 	imes 23 \approx 22.11 	ext{ mmHg}$. Converting to $	ext{atm}$: $p_s = \frac{22.11}{760} \approx 0.029 	ext{ atm} \approx 0.028 	ext{ atm}$.

The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution, which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

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