Question

# The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (Kb for benzene is 2.53Kkgmol−1)

A

58.0gmol1

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B

68.0gmol1

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C

78.0gmol1

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D

48.0gmol1

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Solution

## The correct option is A 58.0gmol−1 Given values are: T∘benzene=353.00K;Kb=2.53Kkgmol−1 Tb(solution)=354.00k Wsolute=1.80g Wsolvent=90g The elevation in boiling point,ΔTb=Tb(solution)−T∘b(solvent) =354.11−353.23 =0.88K Molar mass of solute is given as Mwsolute=Kb×1000×WsolventΔTb×Wsolvent Mwsolute=2.53×1000×1.800.88×90=58.0gmol−1 Hence, the molar mass of solute is 58.0gmol−1

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