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The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it. - Chemistry

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प्रश्न

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

संख्यात्मक

उत्तर

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol−1

∴ Number of moles present in 1000 g of water = `1000/18`

= 55.56 mol

Therefore, the mole fraction of the solute in the solution is

`"x"_2 = 1/(1+55.56) = 0.0177`

Vapour pressure of water, `"p"_1^0` = 12.3 kPa

Applying the relation, `("p"_1^0-"p"_1)/"p"_1^0 = "x"_2`

`=> (12.3- "p"_1)/12.3 = 0.0177`

`=>12.3 - "p"_1 =  0.2177`

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

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अध्याय 2: Solutions - Exercises [पृष्ठ ६०]

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एनसीईआरटी Chemistry [English] Class 12
अध्याय 2 Solutions
Exercises | Q 17 | पृष्ठ ६०

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