An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$:

For $0.1 \ M$ solutions of urea and $Al_2(SO_4)_3$,which of the following statements is correct? $(a)$ $i$ for urea is $1$ and $i$ for $Al_2(SO_4)_3$ is $5$. $(b)$ $Al_2(SO_4)_3$ has a higher elevation in boiling point. $(c)$ $Al_2(SO_4)_3$ has a higher depression in freezing point. $(d)$ Urea has a higher vapor pressure.

Match List-$I$ with List-$II$.

List-$I$	List-$II$
$A$. van't Hoff factor,$i$	$I$. Cryoscopic constant
$B$. $k_{f}$	$II$. Isotonic solutions
$C$. Solutions with same osmotic pressure	$III$. $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$D$. Azeotropes	$IV$. Solutions with same composition of vapour above it

Choose the correct answer from the options given below:

Which of the following is $CORRECT$ with respect to the property mentioned against it?

An aqueous solution freezes at $-0.186\,^\circ C$ ($K_f = 1.86\, K\, kg\, mol^{-1}$; $K_b = 0.512\, K\, kg\, mol^{-1}$). What is the elevation in boiling point?