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