The freezing point of an aqueous solution is – 0.385^@C. If K_f = 3.85 K kg mol^(-1) \n K_b= 0.712 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.

The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,K_(f) = 1.86 K kg mol^(-1))

The boiling point of an aqueous solution of a non - electrolyte is 100.52^@C . Then freezing point of this solution will be [ Given : k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1) for water]

Elevation in boiling point of a solution of non-electrolyte in CCl_(4) is 0.60^(@)C . What is the depression in freezing point for the same solution? K_(f)(CCl_(4)) = 30.00 K kg mol^(-1), k_(b)(CCl_(4)) = 5.02 k kg mol^(-1)

Elevation in boiling point of a solution of non-electrolyte in CCl_(4) is 0.60^(@)C . What is the depression in freezing point for the same solution? K_(f)(CCl_(4)) = 30.00 K kg mol^(-1), k_(b)(CCl_(4)) = 5.02 k kg mol^(-1)

How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104^(@)C between boiling point and freezing point ? ( K_(f) = 1.86 K kg mol^(-1) , K_(b) = 0.52 K kg mol^(-1) )

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)