If 0.15 m aqueous solution of KCI freezes at − 0.5 1 ∘ C , calculate van't Hoff factor of KCI (cryoscopic constant of water is 1.86 K kg mol − 1 )

∴ ∴ Δ T f = T f 0 − T f = 0 − ( − 0.5 1 ∘ C ) = 0.51 ∘ C = 0.51 K Δ T f = iK f m i = K f m Δ T = 1.86 K kg mol − 1 × 0.15 mol kg − 1 0.51 K = 1.82

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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2023

ChemistryEasysingle choice

If $0.15 m$ aqueous solution of $KCI$ freezes at $- 0.5 1^{\circ} C$ , calculate van't Hoff factor of $KCI$ (cryoscopic constant of water is $1.86 K kg mol^{- 1}$ )

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Solution:

$∴ ∴ Δ T_{f} = T_{f}^{0} - T_{f} = 0 - (- 0.5 1^{\circ} C) = 0.51^{\circ} C = 0.51 K Δ T_{f} = iK_{f} m i = \frac{Δ T}{K _{f} m} = \frac{0.51 K}{1.86 K kg mol ^{- 1} \times 0.15 mol kg ^{- 1}} = 1.82$

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

⏱ 2mℹ️ Source: PYQ_2023

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