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1.5 g of a non-volatile, non-electrolyte...

1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene (`K_b = 2.5 kg mol^(-1)`). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g `mol^(-1)` is :

A

200

B

50

C

75

D

100

Text Solution

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The correct Answer is:
D
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Knowledge Check

  • If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent ( K_(b)=x "K kg mol"^(-1) ) is y K, then the depression in freezing point of the same concentration would be ( K_(f) of the solvent = z "K kg mol"^(-1) )

    A
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    B
    `(yz)/(2)`
    C
    `(xz)/(y)`
    D
    `(xz)/(2y)`

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