Q24 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Chemistry
A mixture of 3.0 moles of $\mathrm{Na}_2 \mathrm{O}$ and $1.5 \mathrm{~mol}$ of $\mathrm{KO}_2$ is dissolved in $1000 \mathrm{~mL}$ of water. The vapour pressure of the solution in Torr, at $100{ }^{\circ} \mathrm{C}$ is
- A740
- B760
- C580
- Correct answerD608
Solution
$\mathrm{Na}_2 \mathrm{O}$ and $\mathrm{KO}_2$, both are ionic compounds and are completely ionised as follows :
(i) $\mathrm{Na}_2 \mathrm{O} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{O}^{2-}(i=3)$
(ii) $\mathrm{KO}_2 \longrightarrow \mathrm{K}^{+}+2 \mathrm{O}^{2-}(i=3)$
Also,
$\because \quad \frac{p^{\circ}-p_s}{p^{\circ}}=i \cdot \chi_{\mathrm{Na}_2 \mathrm{O}}+i \cdot \chi_{\mathrm{KO}_2}$
$\text { and moles of } \mathrm{H}_2 \mathrm{O}=\frac{1000}{18}=55.5$
where, $p^{\circ}=$ atmospheric pressure
$p_s=\text { vapour pressure of solution }$
$ p^{\circ}=760$
$\therefore \frac{p^{\circ}-p_s}{p^{\circ}}=3 \times \frac{3}{3+3+55.5}+3 \times \frac{1.5}{1.5+1.5+55.5}$
$=\frac{9}{61.5}+\frac{1.5 \times 3}{58.5}=\frac{9}{61.5}+\frac{4.5}{58.5}$
$\frac{760-p_s}{760}=0.14+0.07=0.21 \approx 0.20$
$p_s=608$
Hence, the correct answer is option: 'D'.