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The vapour pressure of water is `12.3 kPa` at `300 K`. Calculate vapour pressure of `1` molal solution of a solute in it. |
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Answer» Correct Answer - `P_(s)=12.08 kPa` `p^(@)=12.3kPa,p_(S)=?` `M=1(1 mol e` of solute is dissolved in `1 L` of solution `)` Molarity of water `=(1000)/(18)=55.56` `(` or number of moles of water `=55.56)` `(p^(@)-p_(S))/(p^(@))=CHMi_(2)=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))(` for dilute solution `)` `(12.3-p_(S))/(12.3)=(1)/(55.56)` `12.30p_(S)=(12.3)/(55.56)=0.2213` `p_(S)=12.3-0.2213=12.078~~12.08kPa` |
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