Question

18 g glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to 178.2 g water. The vapor presure of water (in torr) for this aqueous solution is:

• A. 76.0
• B. 752.4
• C. 759.0
• D. 7.6

Answer 1

The correct option is B

752.4

$\frac{p^{\circ }-p_S}{p_S}=\frac{n_{\text{solute}}}{n_{\text{solvent}}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{18}{17.82}$

At normal boiling point of water V.P. = $p^{\circ }=760torr$.

$\therefore \frac{760-p_S}{p_S}=\frac{18}{1782}$

or, 1800 ps= $760\times 1782$

ps= 752.4 torr