Question

Calculate the molecular mass of a substance, 1.0 gram of which when dissolved in 100 gram of solvent gace an elecation of 0.307 K in the boiling point. (Molar elevation constant `K_(b)=1.84 K kg mol^(-1)`).

Answer 1

Correct Answer - 59.93 g `mol^(-1)`
`W_(B)=1.0g, W_(A)=0.1 kg, DeltaT_(b)=0.307 K, K_(b)=1.84 K//m, M_(B)=?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.84" K kg mol"^(-1))(1.0g))/((0.307 K)xx(0.1 kg))=59.93" g mol"^(-1)`.