InterviewSolution
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InterviewSolution
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(a) `10 g` of a certain non-volatile solute were dissolved in `100 g` water at `20^(@) C`. The vapour pressure was lowered from `17.3555 mm` to `17.2350 mm`, calculate `m`. wt. of solute. (b) The vapour pressure of pure water at `25^(@)C` is `23.62 mm`. What will be the vapour pressure of a solution of `1.5 g` of urea in `50 g` of water? |
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Answer» (a) `(P^(@)-P_(S))/(P_(S)) = (n)/(N) = (w)/(m) xx (M)/(W)` Given that, `P^(@) = 17.3555 mm, P_(S) = 17.2350 mm`, `w = 10 g, W = 100g, M = 18` `(17.3555-17.2350)/(17.2350) = (10 xx 18)/(m xx 100)` `:. m = 257.45` (b) `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` `(23.62-P_(S))/(P_(S)) = (1.5 xx 18)/(60 xx 50)` `:. P_(S) = 23.41 mm` |
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