18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100° C is

Question

18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100° C is (A) 759.2 torr (B) 760.5 torr (C) 76.8 torr (D) 752.40 torr

Tardigrade Admin · Accepted Answer

(po-ps/po)=(n/N) (760-ps/760)=((18/180)/(178.2)18)=(0.1/9.9) 760-ps=(1/99)× 760 ps=760-7.6=752.4

Q. 18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at $100^{\circ} C$ is

759.2 torr

760.5 torr

76.8 torr

752.40 torr

$\frac{p ^{o} - p _{s}}{p ^{o}} = \frac{n}{N}$
$\frac{760 - p _{s}}{760} = \frac{\frac{18}{180}}{\frac{178.2}{18}} = \frac{0.1}{9.9}$
$760 - p_{s} = \frac{1}{99} \times 760$
$p_{s} = 760 - 7.6 = 752.4$

Q. 18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100∘C is