Question

18 gm of glucose is dissolved in 90 gm of water. The relative lowering of vapour pressure of the solution is equal to:

• A. $6$
• B. $0.2$
• C. $5.1$
• D. $0.02$

Answer 1

The correct option is D $0.02$

Relative lowering of vapour pressure is given as,

$\frac{P_o-P_s}{P_o}=\frac{w}{m}\times \frac{M}{W}$

Where w = mass of the solute (glucose)

M = Molar mass of solvent (Water)

m = molar mass of solute (Glucose)

W = mass of solvent (Water)

$\frac{P_o-P_s}{P_o}=\frac{(18\times 18)}{(90\times 180)}=\frac{1}{50}=0.02$