0 .15 g of a substance dissolved in 15 g of solvent boiled at a temperature higher by 0.21 6 o C than that of the pure solvent. The molecular weight of the substance is [Given: Molal elevation constant for the solvent is 2 .16 Kkg mol − 1 ]

Step-by-step solution

Here it is given that w   =   0 .15   g,     ΔT b   =   0 .216 o C W   =   15   g,   K b   =   2 .16   Kkg   mol − 1 Substituting values in the expression, m = 1000 × K b × w Δ T b × W m   =   1000×216×0 .15 0 .216×15 =100   g/mol