Solutions and Colligative Properties 1 Question 24
23. The vapour pressure of a dilute aqueous solution of glucose $\left(\mathrm{C}{6} \mathrm{H}{12} \mathrm{O}_{6}\right)$ is $750 \mathrm{~mm}$ of mercury at $373 \mathrm{~K}$. Calculate (i) molality and (ii) mole fraction of the solute.
$(1989,3 \mathrm{M})$
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Solution:
- At $373 \mathrm{~K}$ (bp) of $\mathrm{H}_{2} \mathrm{O}$, Vapour pressure $=760 \mathrm{~mm}$
VP of solution at $373 \mathrm{~K}=750 \mathrm{~mm}$
$$ \Rightarrow \quad p=p_{0} \chi_{1} $$
or $\quad 750=760 \chi_{1}$
$\Rightarrow \quad \chi_{1}=\frac{75}{76}=$ mole fraction of $\mathrm{H}_{2} \mathrm{O}$
$\Rightarrow \quad \chi_{2}=1-\frac{75}{76}=\frac{1}{76}=$ mole fraction of solute
Now $\quad \frac{n_{2}}{n_{1}+n_{2}}=\frac{1}{76}$
$\Rightarrow \quad \frac{n_{1}}{n_{2}}=75$
$\Rightarrow \quad$ Molality $=\frac{n_{2}}{n_{1} M_{1}} \times 1000=\frac{1000}{75 \times 18}=0.74$ molal
