Question

The vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate the value to the nearest integer of $\frac{\text{molality}}{10}$.

Answer 1

At 373 K, i.e., b.pt. of $H_{2}O,P_{H_{2}O}^0=76cm$
$\because (P^0-P_S)/P_S=\left(w/m\right)\times \left(M/W\right)$
$\therefore Molarity=\frac{w}{m\times W}\times 1000=\frac{P^0-P_S}{P_S}\times \frac{1}{M}\times 1000$
$=\frac{760-750}{750}\times \frac{1000}{18}$
$=0.741mol/kgsolvent$
Hence, the molality of the solution is $0.741$ mol/kg.