Ncert Exercises & Solutions

2.34 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, 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, $n_{2} = \frac{25}{180 g {mol}^{- 1}} = 0.139 mol$

And, number of moles of water, $n_{1} = \frac{450 g}{18 g {mol}^{- 1}} = 25 mol$

We know that,

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{n_{1}}{n_{2} + n_{1}}$
$\Rightarrow \frac{17.535 - p_{1}}{17.535} = \frac{0.139}{0.139 + 25}$
$\Rightarrow 17.535 - p_{1} = \frac{0.139 \times 17.535}{25.139}$
$\Rightarrow 17.535 - p_{1} = 0.097$
$\Rightarrow p_{1} = 17.44 mm of Hg$

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

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