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If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by `0.216^(@)C` than that of the pure solvent. The molecular weight of the substance (molal elevation constant for the solvent is `2.16^(@)C` kg/mol) is
A. `1.01`
B. 10
C. `10.1`
D. 100

1 Answer

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Correct Answer - D
`m=(K_(b)xxw xx 1000)/(Delta T_(b)xx W)=(2.16xx0.15xx1000)/(0.216xx15)=100`.

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