Vapour pressure of water at 293 Kis 17.535 mm Hg.

Question:

Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.

Solution:

Vapour pressure of water, $p_{1}^{0}=17.535 \mathrm{~mm}$ of $\mathrm{Hg}$

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

= 0.139 mol

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

= 25 mol

We know that,

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{1}}{n_{2}+n_{1}}$

$\Rightarrow \frac{17.535-p_{1}}{17.535}=\frac{0.139}{0.139+25}$

$\Rightarrow 17.535-p_{1}=\frac{0.139 \times 17.535}{25.139}$

⇒ 17.535 − p1 = 0.097

⇒ p1 = 17.44 mm of Hg

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

 

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