Vapour pressure of water, p₁⁰

= 17.535 mm of Hg

Mass of glucose, w₂ = 25 g

Mass of water, w₁ = 450 g

We know that

Molar mass of glucose (C₆H₁₂O₆), M₂

= 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol^-1

Molar mass of water, M₁

= 18 g mol^-1

Then, number of moles of glucose n₂

= \(\frac{25}{180 \ g \ mol^{-1}}\)

= 0.139 mol

And, number of moles of water, n₁ = \(\frac{450g}{18 \ g \ mol^{-1}}\)

= 25 mol

We know that

\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{n_1}{n_2 + n_1}\)

⇒ \(\frac{17.535 - p_1}{17.535}\) = \(\frac{0.139}{0.139 + 25}\)

⇒ 17.535 - p₁ = \(\frac{0.139 \times 17.535}{25.139}\)

⇒ 17.535 - p₁ = 0.097

⇒ p₁ = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.