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The vapour pressure of water is 12.3 k `P_(a)` at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it. |
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Answer» Given molality of solution = 1m Vapour pressure of water `P_(0)=12.3" kP"_(a)` Number of moles of solute `=1.(n_(s))` Number of moles of water `=(1000)/(18)` `=55.55(n_(0))` Mole fraction of solute `(X_(s))=(n_(s))/(n_(0)+n_(s))` `=(1)/(55.55+1)` `=(1)/(56.55)=0.0177` Mole fraction of water `(X_(0))=1-X_(s)` `=1-0.0177=0.9823` Vapour pressure of solution `(P_(s))=P_(0)X_(0)` `=12.3xx0.9823=12.08" kP"_(a)` |
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