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Twenty grams of a solute are added to `100g` of water at `25^(@)C`. The vapour pressure of pure water is `23.76 mmHg`, the vapour pressure of the solution is `22.41` Torr. (a) Calculate the molar mass of the solute. (b) What mass of this solute is required in `100g` of water of reduce the vapour pressure ot one-half the value for pure water? |
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Answer» Correct Answer - (a) `60 g//"mol"`, (b) `333.6g` (a) Relative lowering of vapour pressure `=` mole fraction of solute `(23.76-22.41)/(23.76)=(20//M)/(20//M+100//18)=0.0568` so `M=60 g//"mol"` (b) Let `w` be the mass of solute reqd. `& P=(P^(@))/(2)` so `(P^(@)-P)/(P^(@))=(w//60)/(w//60+10//18)` or `(P^(@)-P^(@)//2)/(P^(@))=(w//60)/(w//60+5.56)` `therefore w=333.6gm` |
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