Question

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

• A. 0.11
• B. 0.021
• C. 0.004
• D. 0.222

Answer 1

Answer: (A)

0.11

Answer 2

$\frac{P_{0}-P}{P_{0}}=\frac{W_{2}}{M_{2}}\times\frac{M_{1}}{W_{1}}$
$0.002=\frac{W_{2}}{M_{2}}\times\frac{18}{1000}$
$\frac{w_{2}}{M_{2}}=0.11\, mole$
$Molality=\frac{w_{2}}{M_{2}}\times\frac{1000}{W_{1}}=0.11\times\frac{1000}{1000}=0.11 m$