Q. Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is

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Solution:

$\frac{P_{0}-P}{P_{0}}=\frac{W_{2}}{M_{2}}\times\frac{M_{1}}{W_{1}}$

$0.002=\frac{W_{2}}{M_{2}}\times\frac{18}{1000}$

$\frac{w_{2}}{M_{2}}=0.11\, mole$

$Molality=\frac{w_{2}}{M_{2}}\times\frac{1000}{W_{1}}=0.11\times\frac{1000}{1000}=0.11 m$

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