Question

The vapour pressure of $100g$ of water reduces from $17.53mm$ to $17.22mm$ when $17.10g$ of substance $X$ is dissolved in it. Substance $X$ can be

• A. methanol
• B. glucose
• C. carbon dioxide
• D. cannot predict

Answer 1

Answer: (B)

Given $p^{\circ }=17.53,p_s=17.22$ and $W=17.10$
$\frac{p^0-p_s}{p^0}=\frac{n}{n+N}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}$
$\because \frac{w}{m}<\frac{W}{M}$
$\therefore \frac{p^0-p_s}{p^0}=\frac{w/m}{W/M}=\frac{w}{m}\times \frac{M}{W}$
$\frac{17.53-17.22}{17.53}=\frac{17.10}{m}\times \frac{18}{100}$
$\Rightarrow m=\frac{17.10\times 18\times 17.53}{0.31\times 100}=174.05$
174 is nearest to the molecular weight of glucose $\left(C_6{H}_{12}{O}_6\right)$, thus the substance $X$ can be glucose.