The vapour pressure of $$30\%$$ (w/v), aqueous solution of glucose is _____ mm Hg at $$25°$$C.
[Given: The density of 30% (w/v), aqueous solution of glucose is $$1.2$$ g cm$$^{-3}$$ and vapour pressure of pure water is $$24$$ mm Hg.]
(Molar mass of glucose is 180 g mol$$^{-1}$$)

30% (w/v) glucose solution means 30 g glucose in 100 mL solution.

Density = 1.2 g/cm³, so mass of 100 mL solution = 120 g.

Mass of water (solvent) = 120 - 30 = 90 g.

Moles of glucose = $$\frac{30}{180} = \frac{1}{6}$$ mol

Moles of water = $$\frac{90}{18} = 5$$ mol

Using Raoult's law:

$$\frac{P^0 - P}{P^0} = x_{solute} = \frac{1/6}{1/6 + 5} = \frac{1/6}{31/6} = \frac{1}{31}$$

$$P = P^0\left(1 - \frac{1}{31}\right) = 24 \times \frac{30}{31} = \frac{720}{31} \approx 23.23$$

Nearest integer: $$P \approx 23$$ mm Hg

This matches the answer key value of $$\mathbf{23}$$.