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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. - Chemistry

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प्रश्न

Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

संख्यात्मक

उत्तर

Let the vapour pressure of pure octane be `"p"_1^0`.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is `80/100 "p"_1^0 = 0.8  "p"_1^0`

Molar mass of solute, M2 = 40 g mol−1

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1

= 114 g mol−1

Applying the relation,

`("p"_1^0 - "p"_1)/("p"_1^0) = ("w"_2xx"M"_1)/("M"_2xx"w"_1)`

`=>("p"_1^0 - 0.8  "p"_1^0)/("p"_1^0)`  = `("w"_2xx114)/(40xx114)`

`=> (0.2  "p"_1^0)/"p"_1^0 = "w"_2/40`

`=> 0.2 = "w"_2/40`

`=> "w"_2 = 8  "g"`

Hence, the required mass of the solute is 8 g.

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पाठ 2: Solutions - Exercises [पृष्ठ ६०]

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एनसीईआरटी Chemistry [English] Class 12
पाठ 2 Solutions
Exercises | Q 18 | पृष्ठ ६०

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