Solutions and Colligative Properties 2 Question 28
28. What is the molarity and molality of a $13 %$ solution (by weight) of sulphuric acid with a density of $1.02 \mathrm{~g} / \mathrm{mL}$ ? To what volume should $100 \mathrm{~mL}$ of this solution be diluted in order to prepare a $1.5 \mathrm{~N}$ solution?
$(1978,2 \mathrm{M})$
(a) $0.027 \mathrm{mmHg}$
(b) $0.031 \mathrm{mmHg}$
(c) $0.017 \mathrm{mmHg}$
(d) $0.028 \mathrm{mmHg}$
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Solution:
- Molar mass of solute $\left(M_{B}\right)=\frac{1000 \times K_{f} \times W_{B}}{W_{A} \times \Delta T_{f}}$
$$ \begin{aligned} \Rightarrow \quad & M_{B}=\frac{1000 \times 14 \times 75.2}{1000 \times 7} \ M_{B} & =150.4 \mathrm{~g} \mathrm{per} \mathrm{mol} \end{aligned} $$
Actual molar mass of phenol $=94 \mathrm{~g} / \mathrm{mol}$
Now, van’t Hoff factor, $i=\frac{\text { Calculated molar mass }}{\text { Observed molar mass }}$
$$ \therefore \quad i=\frac{94}{150.4}=0.625 $$
Dimerisation of phenol can be shown as :
| $2 \mathrm{C}{6} \mathrm{H}{5} \mathrm{OH}$ | $\rightleftharpoons\left(\mathrm{C}{6} \mathrm{H}{5} \mathrm{OH}\right)_{2}$ | |
|---|---|---|
| Initial | 1 | 0 |
| At equilibrium | $1-\alpha$ | $\frac{\alpha}{2}$ |
Total number of moles at equilibrium, $i=1-\alpha+\frac{\alpha}{2}$
$$ i=1-\frac{\alpha}{2} $$
But $i=0.625$, thus,
$$ \begin{aligned}
