The relative lowering of vapour pressure caused by dissolving 71.3 g of a substance in 1000 g of water is 7.13 × 1 0 - 3 . The molecular mass of the substance is (consider the solution is highly diluted)

Δ P P 0 = X B Δ P P 0 ≈ n B n A = w B × m A m B × w A 7.13 × 1 0 - 3 = 71 . 3 × 18 m B × 1000 m B = 180 g / mol .

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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2020

ChemistryEasysingle choice

The relative lowering of vapour pressure caused by dissolving 71.3 g of a substance in 1000 g of water is $7.13 \times 1 0^{- 3}$ . The molecular mass of the substance is (consider the solution is highly diluted)

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Answer:

Solution:

$\frac{Δ P}{P_{0}} = X_{B}$
$\frac{Δ P}{P_{0}} \approx \frac{n_{B}}{n_{A}} = \frac{w_{B} \times m_{A}}{m_{B} \times w_{A}}$
$7.13 \times 1 0^{- 3} = \frac{71.3 \times 18}{m_{B} \times 1000}$
$m_{B} = 180 g / mol$ .

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Liquid SolutionSubtopic:Abnormal Colligatiove Properties

⏱ 2mℹ️ Source: PYQ_2020

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