Question

In ideal solution of non volatile solute $B$ in solvent $A$ in $2 : 5$ molar ratio has vapour pressure $250\, mm$ . If an another solution in ratio $3 : 4$ prepared then vapour pressure above this solution

• A. 200 mm
• B. 250 mm
• C. 350 mm
• D. 400 mm

Answer 1

Answer: (A)

$\frac{n_{B}}{n_{A}}=\frac{2}{5}, $
$P_{s}=250, $
$ \frac{n_{B}}{n_{A}}=\frac{3}{4} $
$ P_{s}=?$
$ P_{s}=p_{A}+\left(p_{e}\right) \longrightarrow $ non-volatile solute $ p_{e}=0$
$P_{s}=p_{A}+p_{x}^{\circ} \times x_{A}$
$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{5}{5+2}=\frac{5}{7}\,\,\,...(i)$
$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{4}{4+3}=\frac{4}{7}\,\,\,...(ii)$
$250=P_{s}=p_{A}^{\circ} \times \frac{5}{7}\,\,\,...(iii)$
$P_{s}=p_{A}^{\circ} \times \frac{4}{7}\,\,\,...(iv)$
$\frac{1}{2} \rightarrow \frac{250=p_{A}^{\circ} \times \frac{5}{7}}{P_{s}=p_{A}^{\circ} \times \frac{4}{7}}$
$\frac{250}{P_{s}}=\frac{5}{4}$
$P_{s}=\frac{250 \times 4}{5}=200\, mm$