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

$\frac{\text{p}^{\text{o}} \text{water}}{\text{p}^{\text{o}} \text{liquid}} = \frac{\text{0.84}}{\text{0.143}} = \frac{n_{H_{2} O}}{n_{L i q u i d}}$

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

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