18g glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ is added to 178.2g water. The vapor pressure of water (in torr) for this aqueous solution is ______ .

Vapour pressure of water $\left(p°\right)=760$ torr

Number of moles of glucose
$=\frac{\text{ Mas }\left(\mathrm{g}\right)}{\text{ Molecular mass }(\mathrm{gmo}}$
$=\frac{18 \mathrm{g}}{180 \mathrm{g}{ \mathrm{mol}}^{-1}}=0.1$

We know that the value is 18 g/mol for molar mass of water
Water mass (given) = 178.2 g
number of water moles

$=\frac{\text{ Mass of water }}{\text{ Molar mass of water }}=\frac{178.2}{18 \mathrm{g}}=9.9 \mathrm{mol}$

Total moles = (0.1 + 9.9) = 10 moles
The mole fraction of glucose in solution is now equal to the pressure change in relation to the initial pressure.
i.e. $\frac{\Delta p}{\Delta p^0}=\frac{0.1}{10}$
or $\Delta \mathrm{p}=0.01{\mathrm{p}}^0=0.01\times 760=7.6$ torr

Consequently, the solution's vapour pressure is (760-7.6) torr.

=752.4 torr

hence, the correct answer 752.4 torr.