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# 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.

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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.

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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.

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