Question

QUESTION 2.34

Vapour pressure of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of water at 293K when 25g of glucose is dissolved in 450 g of water.

Answer 1

$W_B\left(solute\right)=25g;M_B=180gmo{l}^{-1};p_A^{\circ }=17.535mm$
$W_A\left(water\right)=450g;M_A=18gmol;p_s=?$
$\frac{p_A^{\circ }-p_s}{p_A^{\circ }}=x_B=\frac{n_B}{n_B+n_A}$
$n_B=\frac{W_B}{M_B}=\frac{25}{180}=0.1389n_A=\frac{W_A}{M_A}=\frac{450}{18}=25\frac{17.535-p_s}{17.535}=\frac{0.1389}{0.1389÷25}\frac{17.535-p_s}{17.535}=0.00552p_s=17.535-0.097p_s=17.438mmHg$