1.

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

Answer»

`1.01`
10
`10.1`
100

Solution :`m=(K_(b)xxw XX 1000)/(Delta T_(b)xx W)=(2.16xx0.15xx1000)/(0.216xx15)=100`.


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