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. 0.004
• B. 0.222
• C. 0.111
• D. 0.021

Answer 1

The correct option is C 0.111

Here, glucose (solute) is dissolved in water (solvent). Let us take the mass of water to be 1000 g, or 1 kg and calculate the number of moles of solute in this. This would give us the molality of the solution.
Moles of water can be calculated as -
$n_w=\frac{W_w}{M_w}$
$n_w=\frac{1000}{18}$
$n_w$ = 55.56 mol
Relative lowering of vapor pressure is equal to mole fraction of solute.
$\frac{\Delta P}{P°}=X_g=\frac{n_g}{(n_g+n_w)}$
0.002 = $\frac{n_g}{(n_g+55.56)}$
$n_g$ = 0.111 mol