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A solution of a non-volatile solute in water freezes at `-0.30^(@)C`. The vapour pressure of pure water at `298 K` is `23.51 mm Hg` and `K_(f)` for water is `1.86 degree//molal`. Calculate the vapour pressure of this solution at `298 K`. |
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Answer» Correct Answer - `P_(s)=23.442 mm Hg` We know that `DeltaT_(f)=K_(f) xx "Molality of solution"` From given data, `0.3=1.86xx"Molality of solution"` Hence, molality of solution =`0.3/1.86=0.161 m` Lowering in vapour pressure (for dilute solution) `(P^(@)-P_(S))/P^(@)=(W_(2)xxMw_(1))/(Mw_(1) xx W_(1))` =`(W_(2)/((Mw_(2))xxW_(1))xx(Mw_(1))/1000)/1000` `rArr P^(@)-P_(S))/P^(@)` =`"Molality of solution" xx ("Molecular weight of solvent")/1000` or `23.51-P_(S)/23.51=0.61 xx 18/1000` By solving this equation, we get Vapour pressure of solution `(P_(S))=23.44 mm` |
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