Boiling point of benzene is 353.23 K . When 1.8 g of non-volatile solute is dissolved in 90 g of benzene. Then boiling point is raised to 354.11 K . Given K b (benzene) = 2.53 kg mol − 1 . The molec

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Boiling point of benzene is 353.23 K . When 1.8 g of non-volatile solute is dissolved in 90 g of benzene. Then boiling point is raised to 354.11 K . Given K b ​ (benzene) = 2.53 kg mol − 1 . The molecular mass of non-volatile substance is

TestHub · Accepted Answer

T b ∘ ​ = 353.23 K , W B ​ = 1.8 g , W A ​ = 90 g ′ ​ T b ​ = 354.11 K , K b ​ = 2.53 kg mol − 1 Δ T b ​ = T b ​ − T b ​ ∘ = 354.11 − 353.23 = 0.88 K M B ​ = Δ T b ​ × W A ​ W B ​ × K b ​ × 1000 ​ = 0.88 × 90 1.8 × 2.53 × 1000 ​ = 57.5 ≈ 58 g mol − 1

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