Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Vapour pressure of water, p_{1}^{0}
= 17.535 mm of Hg
Mass of glucose, w_{2} = 25 g
Mass of water, w_{1} = 450 g
We know that
Molar mass of glucose (C_{6}H_{12}O_{6}), M_{2}
= 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^{1}
Molar mass of water, M_{1}
= 18 g mol^{1}
Then, number of moles of glucose n_{2}
= \(\frac{25}{180 \ g \ mol^{1}}\)
= 0.139 mol
And, number of moles of water, n_{1} = \(\frac{450g}{18 \ g \ mol^{1}}\)
= 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  p_{1} = \(\frac{0.139 \times 17.535}{25.139}\)
⇒ 17.535  p_{1} = 0.097
⇒ p_{1} = 17.44 mm of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.

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