The vapour pressure in mm of $Hg$, of an aqueous solution obtained by adding 18 g of glucose ($C_6{H}_{12}{O}_6$) to 180 g of water at (${100}^{\circ }C$) is:

The correct option is D 752.4
At ${100}^{o}C$, the vapour pressure of water,
$P^0=760$ mm Hg $=$ atmospheric pressure.

The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.

Number of moles of glucose $=\frac{18}{180}=0.1$

Number of moles of water $=\frac{180}{18}=10$

Mole fraction of glucose $X=\frac{0.1}{10+0.1}=0.0099$

Let P mm be the vapour pressure of solution
$\frac{P^o-P}{P^o}=X$

$\frac{760-P}{760}=0.0099$

$760-P=7.524$

$P=752.4$ mm Hg