Question

A mixture of an organic liquid `A` and water distilled under one atmospheric pressure at `99.2^(@)C`. How many grams of steam will be condensed to obtain `1.0 g` of liquid `A` in the distillate? (Vapour pressure of water `99.2^(@)C` is `739 mm Hg`. Molecular weight of `A=123`)

Answer 1

Correct Answer - `5.15g`
When the liquid `A-` water mixture distills, the pressure `=760 mm Hg`.
`therefore` partial pressure of `A=760-739=21 mm Hg`.
Mole fraction of `A` in the distillate `=(21)/(760)=0.02764`
Mole fraction of water in the distillate `=(739)/(760)=0.9723`
`therefore` mass of `A` in the distillate `=0.02764xx123=3.399g`
Mass of steam condensed per `g` of liquid `A=(17.5)/(3.399)=5.15g`.