Question

18 g of glucose $(C_6{H}_{12}{O}_6)$ 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)

$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n_2}{n_1}=\frac{w_2/M_2}{w_1/M_1}$
$\therefore$ $\frac{760-p_s}{p_s}=\frac{18/180}{178.2/18}$ 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