If boiling point of urea solution is 100 ⋅ 1 8 ∘ C and kb for water is 0.512 K kg mol − 1 , molality of solution is? (Boiling point of water = 10 0 ∘ C )

(D) Δ T b = ( 100.18 + 273 ) − ( 100 + 273 ) = 0.18 K K b = 0.512 K kg mol − 1 , m = ? Δ T b = K b × m ∴ m = K b Δ T b ∴ m = 0.512 K kg mol − 1 0.18 K = 0.35 mol kg − 1

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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesMedium2 minPYQ_2020

ChemistryMediumsingle choice

If boiling point of urea solution is $100 \cdot 1 8^{\circ} C$ and $kb$ for water is $0.512 K kg mol^{- 1}$ , molality of solution is? (Boiling point of water $= 10 0^{\circ} C$ )

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Answer:

Solution:

(D) $Δ T_{b} = (100.18 + 273) - (100 + 273) = 0.18 K$ $K_{b} = 0.512 K kg mol^{- 1}, m = ?$ $Δ T_{b} = K_{b} \times m ∴ m = \frac{Δ T _{b}}{K _{b}}$ $∴ m = \frac{0.18 K}{0.512 K kg mol ^{- 1}} = 0.35 mol kg^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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