Find the molar mass of solute when 2 gram dissolved in 60 gram benzene at 3 0 ∘ C and relative lowering of vapour pressure is 0.06 . (Molar mass of benzene is 78 g mol )

P A ∘ Δ P = M B W B × W A M A 0.06 = M B 2 × 60 78 M B = 30 × 0.06 78 = 43.3 g / mol

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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2022

ChemistryEasysingle choice

Find the molar mass of solute when 2 gram dissolved in 60 gram benzene at $3 0^{\circ} C$ and relative lowering of vapour pressure is 0.06 . (Molar mass of benzene is $78 g mol$ )

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Answer:

Solution:

$\frac{Δ P}{P _{A}^{\circ}} = \frac{W _{B}}{M _{B}} \times \frac{M _{A}}{W _{A}}$ $0.06 = \frac{2}{M _{B}} \times \frac{78}{60}$ $M_{B} = \frac{78}{30 \times 0.06} = 43.3 g / mol$

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

⏱ 2mℹ️ Source: PYQ_2022

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