At T( ~K ) , the vapour pressure of pure benzene is 0.85 bar. A non-volatile, non-electrolyte substance weighing 0.5 ~g when added to 39 ~g of benzene, the vapour pressure of the solution is 0.845 bar. The molar mass (in g mol ⁻¹ ) of the s

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At T( ~K ) , the vapour pressure of pure benzene is 0.85 bar. A non-volatile, non-electrolyte substance weighing 0.5 ~g when added to 39 ~g of benzene, the vapour pressure of the solution is 0.845 bar. The molar mass (in g mol ⁻¹ ) of the substance is

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Correct answer: D. 169

If, p^ = vapour pressure of pure benzene. p= vapour pressure of solution. aligned & p^ -p p = n( solute ) n( solvent = n₁ n₂ & = 0.85-0.845 0.845 = w₁ M₁ M₂ w₂ aligned where, w₁ and w₂ are masses of solute respectively and M₁, M₂ are molar masses of solute and solvent respectively. aligned & M₁= 0.5 78 0.845 0.05 39 & M₁=169 ~g ~mol ⁻¹ aligned

At T( ~K ) , the vapour pressure of pure benzene is 0.85 bar. A non-volatile, non-electrolyte substance weighing 0.5 ~g when added to 39 ~g of benzene, the vapour pressure of the solution is 0.845 bar. The molar mass (in g mol ⁻¹ ) of the substance is

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