An aqueous solution containing $20\%$ by weight of liquid 'A' ($\text{Mol.wt} = 140$) has vapour pressure of $160\text{ mm of Hg}$ at $57^\circ\text{C}$. Find the vapour pressure of pure A if that of water is $150\text{ mm of Hg}$ at this temperature.

$472.58\text{ mm Hg}$

The correct answer is Option (3) → $472.58\text{ mm Hg}$ ##

Number of moles of A dissolved $= \frac{20}{140}$

$= 0.143$

Number of moles of B $= \frac{80}{18}$

$= 4.44$

Mole fraction of A ($\chi_{\text{A}}$) $= \frac{0.143}{0.143 + 4.44}$

$= 0.031$

Mole fraction of B ($\chi_{\text{B}}$) $= \frac{4.44}{0.143 + 4.44}$

$= 0.969$

Total pressure $= 160\text{ mm}$

$= p_{\text{A}} \times \chi_{\text{A}} + p_{\text{B}} \times \chi_{\text{B}}$

$160 = p_{\text{A}} \times 0.031 + 150 \times 0.969$

$∴p_{\text{A}} = 472.58\text{ mm Hg}$