1. Use Raoult’s law for a non-volatile solute

Since glucose is a non-volatile solute, only water contributes to vapour pressure.

$P_{\text{solution}} = X_{\text{water}}\, P^\circ_{\text{water}}$

At $100^\circ\text{C}$, pure water has vapour pressure:

$P^\circ_{\text{water}} = 760\ \text{Torr}$

2. Calculate moles of glucose

Molar mass of glucose, $C_6H_{12}O_6$:

$6\times 12 + 12\times 1 + 6\times 16 = 72 + 12 + 96 = 180\ \text{g mol}^{-1}$

So,

$n_{\text{glucose}} = \frac{18}{180} = 0.1\ \text{mol}$

3. Calculate moles of water

Molar mass of water = $18\ \text{g mol}^{-1}$

$n_{\text{water}} = \frac{178.2}{18} = 9.9\ \text{mol}$

4. Calculate mole fraction of water

$X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{glucose}}} = \frac{9.9}{9.9+0.1} = \frac{9.9}{10.0} = 0.99$

5. Calculate vapour pressure of solution

$P_{\text{solution}} = 0.99 \times 760 = 752.4\ \text{Torr}$

6. Match with the options

$752.4\ \text{Torr}$

So the correct option is:

D: 752.40 Torr

7. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They match.