`0.15` g of a substance dissolved in `15` g of solvent boiled at a temperature higher at `0.216^(@)`C than that of the pure solvent. Calculate the molecular weight of the substance. Molal elevation constant for the solvent is `2.16^(@)`C

Explanation

Given `K_(b) = 2/16^(@)C, w = 0.15 g, DeltaT_(b) = 0.216^(@)C, W = 15g`
`DeltaT_(b) =` molality `xx K_(b)`
`DeltaT_(b) = (w)/(m xx W) xx 1000 xx K_(b)`
`0.216 = (0.15)/(m xx 15) xx 1000 xx 2.16`
`m = (0.15 xx 1000 xx 2.16)/(0.216 xx 15) = 100`