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`)

An aqueous solution boils at 100.50^(@)C .The freezing point of the solution would be (K_(b) for water =0.51^(@)C//m),(K_(f) for water =1.86^(@)C//m ) [no association or dissociation]

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)

The measured freezing point depression of a non-volatile solute in aqueous solution is 0.20^(@) C . The elevation in boiling point of the same solution will be [ K_(f) = 1.86 K/m, K_(b) = 0.52 K/m]

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))

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 :