5% solution of a substnace in water has freezing point 269.06K. Calculate molar mass of solute. Freezing point of pure water 273.15K. [`K_(f)=14K.kg" "mol^(-1)]`.

(a) A 5% solution (by mass) of cane - sugar in water has freezing point of 271 K. Calculate the frttzing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K. [Molecular masses : Glucose C_(6)H_(12)O_(6) : 180 amu, Cane - sugar C_(12)H_(12)O_(11) : 342 amu] (b) State Henry's law and mention two of its important applications.

a) 24 g of a non-volatile, non-electrolyte solute is added to 600 g of water. The boiling point of the resulting solution is 373.35K. Calculate the molar mass of the solute ("Given boiling point of pure water = 373 K and Kb for water=0.52 K kg mol"^(-1)) .

A solution containing 12.5 g of non-electrolyte solution in 175g of water gave a boiling point elevation of 0.7 k. calculate the molar mass of the solute if K_(b) for water is 0.52 K kg mol^(-1) .

A solution containing 18g of non - volatile non - electrolyte solute is dissolved in 200g of water freezes at 272.07K. Calculate the molecular mass of solute. Given K_(f)=1.86kg//mol and freezing point of water = 273K

A solution containg 12 g of a non-electrolyte substance in 52 g of water gave boiling point elevation of 0.40 K . Calculate the molar mass of the substance. (K_(b) for water = 0.52 K kg mol^(-1))

A solution of urea in water has boiling point of 100.15^(@)C . Calculate the freezing point of the same solution if K_(f) and K_(b) for water are 1.87 K kg mol^(-1) and 0.52 K kg mol^(-1) , respectively.

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^(@)C . Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol^(-1))

What is the freezing point of solution containing 3g of a non-volatile solute in 20g of water. Freezing point of pure water is 273K, K_(f) of water = 1.86 Kkg/mol. Molar mass of solute is 300 g/mol. T^(@)-T=K_(f)m