The vapour pressure of water at 293 k when 25 g of glucose is dissolved in 450 gram of water is

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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

= 25 mol

We know that,

(p_1^0-p_1)/p_1^0

Molar mass of water, M1 = 18 g mol−1

Then, number of moles of glucose, `n_2 = 25/(180 g mol^(-1))`

= 0.139 mol

And, number of moles of water `n_1= (450g)/18 g mol^(-1)`

= 25 mol

We know that,

`(p_1^0-p_1)/p_1^0 = n_1/(n_2+n_1)`

`=>(17.535 - p_1)/17.535 = 0.139/(0.139+25)`

`=>17.535 - p_1 = (0.139 xx 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.

Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g

Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2

= 60 g mol–1
According to Raoult's law, p0-pp0=ωMWm

p=p0-w×Mm×W×p°

p=23.8-50×1860×850

=23.8-0.017=23.78

Hence, 23.78 mm Hg. Ans.