Question

A liquid is immiscible in water was steam distilled at 95.2^oC at a pressure of 0.983 atm. What is the mass of the liquid present per gram of water in the distillate. Molar mass of the liquid is 134.3 g/mol and the vapour pressure of water is 0.84 atm. Also, Vapour pressure of pure liquid is 0.143 atm.

• A. 1 g
• B. 1.27 g
• C. 0.787 g
• D. 13.43 g

Answer 1

Answer: (B)

${\text{p}}_{\text{T}}=\text{0.983 atm}$

${\text{p}}_{\text{water}}^{{}^{\circ }}=\text{0.84 atm}$

$\therefore$ ${\text{p}}_{\text{liquid}}^{{}^{\circ }}=\text{0.143 atm}$

Apply Dalton's law of vapour phase

$\frac{{\text{p}}^{\text{o}}\text{water}}{{\text{p}}^{\text{o}}\text{liquid}}=\frac{\text{0.84}}{\text{0.143}}=\frac{n_{H_{2}O}}{n_{Liquid}}$

$\frac{\text{0.84}}{\text{0.143}}=\frac{\frac{\text{1}}{\text{18}}}{\frac{{\text{w}}_{\text{e}}}{\text{134.3}}}=\frac{\text{134.3}}{\text{18}\times {\text{w}}_{\text{e}}}$

${\text{w}}_{\text{e}}=\frac{\text{134.3}}{\text{18}}\times \frac{\text{0.143}}{\text{0.84}}=\text{1.27 g}$