Question

Calculate the vapour pressure of an aqueous solution of 1.0 molal glucose solution at ${100}^{o}C.$

Answer 1

Molality $=\frac{w}{m\times W}\times 1000$
where, w = mass of solute in grams;
W = mass of solvent in grams
$1.0=\frac{w}{m\times W}\times 1000$
or $\frac{w}{m\times W}=\frac{1.0}{1000}=0.001$
Applying Raoult's law for dilute solution,
$\frac{P_0-P_s}{P_0}=\frac{w}{m\times W}\times M(M=18)$
$\frac{760-P_s}{760}=0.001\times 18(P_0=760mmat{100}^{o}C)$
or $P_s=760-760\times 0.001\times 18$
$=760-13.68$
$=746.32mm$