Question

$36\, g$ glucose $\left( C _{6} H _{12} O _{6}\right)$ is added to $176.4\, g$ water. The vapour pressure of water (in torr) for this aqueous solution at $100^{\circ} C$ is_______.

Answer 1

Molecular weight of glucose $\left( C _{6} H _{12} O _{6}\right)=180$
$\therefore $ Number of moles of glucose $\left(n_{B}\right)=\frac{36}{180}=0.2$
Molecular weight of water $\left( H _{2} O \right)=18$
$\therefore $ Number of moles of water $\left( n _{ A }\right)=\frac{176.4}{18}=9.8$
$\therefore $ Mole fraction of water $\left(x_{ H _{2} O }\right)=\frac{ n _{ A }}{ n _{ A }+ n _{ B }}$
$=\frac{9.8}{9.8+0.2}$
$=\frac{9.8}{10}$
$\because p _{ H _{2} O }=x_{ H _{2} O } p _{ H _{2} O }^{\circ}$
$\therefore p _{ H _{2} O }=\frac{9.8}{10} \times 760=744.8$ torr
(At boiling point of water, it vapour pressure is $1\, atm =760\, mm\, Hg$ )