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)`

An aqueous solution freezes at 272.814 K . The boiling point of the same solution is __________. (K_(r)=1.86 K m^(-1), K_(b) = 0.512 K m^(-1) )

Freezing point of an aqueous solution is -0.186^(@)C . Elevation of boiling point of the same solution is ……..if K_(b)=0.512 K "molality"^(-1) and K_(f)=1.86K "molality"^(-1) :

An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )

The depression of freezing point of watar for a particular solution is 0.186 K. The boiling point of the same solution is _______. (K_(f)= 1.863= K kg = 0.512 k Kg mol mol^(-1))

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]