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)` :

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))

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)

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)

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)

What is the boiling point of 1 molal aqueous solution of NaCl (K_b)=0.52 K molal^-1]

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)= 1.86K kg mol^(-1) and K_(b)= 0.512 K kg mol^(-1))

AN aqueous solution of a substacne freezes at -0.186^(@)C. The boiling point of the same solution (k _(f) =1.86 Km^(-1), K_(b)=0.512 Km^(-1)) is :

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) )