What is the depression in freezing point of a solution of non-electrolyte if elevation in boiling point is `0.13K` ? `(k_(b)=0.52 mol^(-1) kg, k_(f)= 1.86 mol^(-1) kg)`

The freezing point of 0.05 molal solution of non- electrolyte in water is _____. (K_(f) = 1.86 ^(@)Ckg 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))

Elevation in boiling point of an aqueous solution of a non-electroluyte solute is 2.01^(@) . What is the depression in freezing point of the solution ? K_(b)(H_(2)O)=0.52^(@)mol^(-1)kg K_(f)(H_(2)O)=1.86^(@)mol^(-1)kg

If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent ( K_(b)=x "K kg mol"^(-1) ) is y K, then the depression in freezing point of the same concentration would be ( K_(f) of the solvent = z "K kg mol"^(-1) )

if the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associanting solute in solvent ( K_(b) =x K kg "mol"^(-1) )is yK,then the depression in freezing point of solution of same concentration would be ( K_(f) )of the sovent = zk.kg"mol"^(-1) )

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .