The vapor pressure of water at 23$°$C is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. The vapor pressure (in mm) of the resultant solution is

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{Glucose} {\mathrm{n}}_{\mathrm{B}}=0.1,$

$\text{Number of moles of Water }{n}_A=\frac{178.2}{18}=9.9$

${\mathrm{x}}_{\mathrm{A}}=\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}=\frac{9.9}{9.9+0.1}=0.99$

$\mathrm{The} \mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{resultant} \mathrm{solution} \mathrm{p}={\mathrm{p}}_0{\mathrm{x}}_{\mathrm{A}}$

$=19.8\times 0.99=19.602 \mathrm{mm}$