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`)

A mixture of an organic liquid `A` and water distilled under one atmos

1.	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» 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`.

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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`)

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