Calculate the osmotic pressure in $atm$ at $0\,^oC$ of a $0.18\, m$ aqueous solution of $KCl$ which has a freezing point of $-0.68\,^oC$. Assume the volume of the solution is equal to the volume of pure water. $(K_f = 1.86\,^oC\, m^{-1})$

The freezing point of an aqueous solution is $-0.186^oC$. If the molal elevation constant and molal depression constant of the solvent are $0.512$ and $1.86$ respectively,then the elevation in boiling point is .......... $^oC$.

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:

At $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:

When $1.685 \ g$ of an alkali metal chloride is dissolved in $200 \ g$ water,the boiling point of the solution is measured to be $100.051 \ ^\circ C$. If the ionic solid has a crystal lattice with cation and anion radius $1.70 \ \mathring{A}$ and $1.80 \ \mathring{A}$ respectively,find the edge length of the solid assuming no defect in the crystal. Given: $K_b(H_2O) = 0.51 \ K \ kg \ mol^{-1}$,$N_A = 6 \times 10^{23}$,atomic masses: $[Li = 7, Na = 23, K = 39, Rb = 85.5, Cs = 133, Cl = 35.5]$.

At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)