Q27 / 105 · JEE Main 2008 · Chemistry

The vapour pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{~mm} \mathrm{Hg}$. If $18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water at $20^{\circ} \mathrm{C}$, the vapour pressure of the resulting solution will be

  • A
    $17.675 \mathrm{~mm} \mathrm{Hg}$
  • B
    $15.750 \mathrm{~mm} \mathrm{Hg}$
  • C
    $16.500 \mathrm{~mm} \mathrm{Hg}$
  • Correct answer
    D
    $17.325 \mathrm{~mm} \mathrm{Hg}$

Solution

$$ \begin{aligned} & \frac{P^0-P_s}{P_s}=X_{\text {solute }} \\ & \frac{17.5-P_s}{P_s}=\frac{0.1}{10} \\ & \frac{17.5-P_s}{P_s}=0.01 \\ & \Rightarrow P_s=17.325 \mathrm{~mm} \mathrm{Hg} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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