IF 0.15 g of a solute, dissolved in 15g of solvent is boiled at a temperature higher by `0.126^@C`, than that of the pure solvent , The molecular weight of the substance, (Molal elevation constant for the solvent is `2.16^@C`) is

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`