Q54 / 160 · JIPMER 2010 · Chemistry

The vapour pressure of 100 g water reduces from $3000 \mathrm{Nm}^{-2}$ to $2985 \mathrm{Nm}^{-2}$ when 5 g of substance ' $X$ ' is dissolved in it. Substance ' $X$ ' is

  • A
    methanol
  • Correct answer
    B
    glucose
  • C
    carbon dioxide
  • D
    Cannot predict

Solution

$\begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n}{n+N} & =\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}} \\ \because \quad \frac{w}{m} & \lt \frac{W}{M} \\ \therefore \quad \frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w / m}{W / M}=\frac{w}{m} \times \frac{M}{W} \\ \frac{3000-2985}{3000} & =\frac{5}{m} \times \frac{18}{100} \\ m & =\frac{5 \times 18 \times 3000}{15 \times 100} \\ m & =180\end{aligned}$ 180 is the molecular weight of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$, thus the substance ' $X$ ' is glucose.

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

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