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Vapour pressure of pure water at 298 K is 23.8 mm Hg. 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. - Chemistry

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प्रश्न

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 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.

संख्यात्मक
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उत्तर

Given, P0 = 23.8 mm Hg

w1 = 850 g, M1 (water) = 18 g mol−1

w2 = 50 g, M2 (urea) = 60 g mol−1

Ps = ?, `("P"^0 - "P"_"s")/"P"^0` = ?

∴ n1 = `850/18` = 47.22

∴ n2 = `50/60` = 0.83

Applying Raoult's law, `("P"^0 - "P"_"s")/"P"^0 = "n"_2/("n"_1+"n"_2)`

or, `("P"^0 - "P"_"s")/"P"^0 = 0.83/(47.22 + 0.83)`

`("P"^0 - "P"_"s")/"P"^0 = 0.83/48.05`

`("P"^0 - "P"_"s")/"P"^0` = 0.017

Thus, the relative lowering of vapour pressure = 0.017

Again, `(Δ"P")/"P"^0` = 0.017

∴ ΔP = 0.017 × 23.8

or, P0 − Ps = 0.017 × 23.8

or, Ps = 23.8 − (0.017 × 23.8)

or, Ps = 23.4 mm Hg

Thus, the vapour pressure of water in the solution is 23.4 mm Hg.

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पाठ 1: Solutions - Intext Questions [पृष्ठ २३]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 1 Solutions
Intext Questions | Q 1.9 | पृष्ठ २३

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