Question

18 g of glucose $(C_6H_{12}O_6)$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100$^\circ$C

• A. 76.00 Torr
• B. 752.40 Torr
• C. 759.00 Torr
• D. 7.60 Torr

Answer 1

Answer: (B)

752.40 Torr

Answer 2

$\frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1}$
$\therefore$ $\frac{760 - p_s}{p_s} = \frac{18/180}{178.2/18}$ or $\frac{760}{p_s} - 1 = \frac{18}{1782}$
or $\frac{760}{p_s} = \frac{18}{1782} + 1 = \frac{1820}{1782}$ or $p_s = 752.4$ torr