18 g glucose ($$C_6H_{12}O_6$$) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is:

First, we recall Raoult’s Law for a non-volatile solute: the vapour pressure of the solvent in an ideal dilute solution is directly proportional to its mole fraction. Mathematically,

$$P_{\text{solution}} = x_{\text{solvent}}\;P^{\circ}_{\text{solvent}},$$

where $$P^{\circ}_{\text{solvent}}$$ is the vapour pressure of the pure solvent and $$x_{\text{solvent}}$$ is the mole fraction of the solvent in the solution.

We start by calculating the moles of each component. The molar mass of glucose $$(C_6H_{12}O_6)$$ is obtained by adding the atomic masses:

$$M_{\text{glucose}} = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180\;\text{g mol}^{-1}.$$

The given mass of glucose is 18 g, so the number of moles of glucose is

$$n_{\text{glucose}} = \frac{18\;\text{g}}{180\;\text{g mol}^{-1}} = 0.10\;\text{mol}.$$

Next, the molar mass of water $$(H_2O)$$ is

$$M_{\text{water}} = 2(1) + 16 = 18\;\text{g mol}^{-1}.$$

The mass of water taken is 178.2 g, so its moles are

$$n_{\text{water}} = \frac{178.2\;\text{g}}{18\;\text{g mol}^{-1}} = 9.90\;\text{mol}.$$

We now find the total moles present in the solution:

$$n_{\text{total}} = n_{\text{glucose}} + n_{\text{water}} = 0.10 + 9.90 = 10.00\;\text{mol}.$$

Thus, the mole fraction of water is

$$x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{total}}} = \frac{9.90}{10.00} = 0.99.$$

The vapour pressure of pure water at the stated temperature is taken as $$P^{\circ}_{\text{water}} = 760\;\text{torr}.$$ Substituting into Raoult’s Law, we get

$$P_{\text{solution}} = x_{\text{water}}\;P^{\circ}_{\text{water}} = 0.99 \times 760\;\text{torr}.$$

Carrying out the multiplication,

$$P_{\text{solution}} = 752.4\;\text{torr}.$$

Hence, the correct answer is Option A.