A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now `105^(@)C`. Calculate mass of solute present in `500g` of solvent. (Given molar mass of solute =`120`g/mol,`K_(b)=0.512^(@)Cm^(-1),K_(f)` `=1.86^(@)Cm^(-1))`

Calculate the freezing point and the boiling point at 1 atmosphere of a solution containing 20g cane sugar (molecular mass 343) and 150g water. Given : K_(b) = 0.513 and K_(f)= 1.86 .

A hypothetical solute AB is 50% dissociated at freezing point and 80% dissociated at boiling point of solution.If 2 mole of solute is dissolved in 500 g solvent and the difference in boiling point and freezing point of solvent is 100K then calculate the difference in boiling point and freezing point of solution. [Given K_(b)= 0.5Kkgmol^(-1) K_(f)=2.0Kkgmol^(-1) ]

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.

Calculate the freezing point depression and boiling point elevation of a solution of 10.0 g of urea (M_(B)=60) in 50.0 g of water at 1 atm . pressure. K_(b) and K_(f) for water 0.52^(@)C m^(-1) and 1.86^(@)C m^(-1) respectively.