$$12\,g$$  glucose $$\left( {{C_6}{H_{12}}{O_6}} \right)$$   is added to $$178.2 g$$  water. The vapour pressure of water (in torr) for this aqueous solution is :

Solution :
According to Raoult's Law
$$\frac{{{P^ \circ } - {P_s}}}{{{P_s}}} = \frac{{{W_B} \times {M_A}}}{{{M_B} \times {W_A}}}\,\,\,\,\,\,....({\text{i}})$$
Here $${P^o}{\text{ = }}$$  Vapour pressure of pure solvent,
$${P_s} = $$  Vapour pressure of solution
$${W_B} = $$  Mass of solute, $${W_A} = $$  Mass of solvent
$${M_B} = $$  Molar mass of solute, $${M_A} = $$  Molar Mass of solvent
Vapour pressure of pure water at $${100^ \circ }C$$  ( by assumption = 760 $$torr$$  )
By substituting values in equation (i) we get,
$$\frac{{760 - {P_s}}}{{{P_s}}} = \frac{{18 \times 18}}{{180 \times 178.2}}\,\,\,...{\text{(ii)}}$$
On solving (ii) we get
$${P_s} = 752.4\,torr$$