Question

A solution of sucrose (molar mass $=342$ ) is prepared by dissolving $68.4\, g$ in $1000 \, g$ of water. Calculate
The vapour pressure of solution at $293\, K$

• A. $0.0229$
• B. $0.4$
• C. $0.6$
• D. $0.9$

Answer 1

Answer: (A)

Vapour Pressure of solution can be calculated from the relation:
$\frac{P_{ A }{ }^{\circ}-P_{ A }}{P_{ A }^{\circ}}=\chi_{ B }=\frac{W_{ B } / M w_{ B }}{W_{ A } / M w_{ A }}$
Here $W_{ A }=1000 \,g , W_{ A }=68.4 \,g , P_{ A }^{\circ}=0.023 \,atm$,
$M w_{ A }=18, M w_{ B }=342$
Substituting the values, we get
$\frac{0.023-P_{ A }}{0.023}=\frac{68.4 / 342}{1000 / 18}$
or $\frac{0.023-P_{ A }}{0.023}=\frac{68.4 \times 18}{342 \times 100}$
or $\frac{0.023-P_{ A }}{0.023}=0.0036$
or $0.023-P_A=0.0036 \times 0.023=0.000083$
$P_{ A }=0.023-0.00008=0.02292 \,atm$
Vapor pressure of solution $=0.02292 \,atm$