⇒Solution

Correct Option is (D)

$\mathrm{NaCl}⟶{\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}$

Total ions $=0.1+0.1=0.2$ ions

${\mathrm{KNO}}_3⟶{\mathrm{K}}^{+}+{\mathrm{NO}}_3^{-}$

Total ions $=0.2+0.2=0.4$ ions

${\mathrm{Na}}_2{\mathrm{SO}}_4⟶2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}$

Total ions $=0.2+0.1=0.3$ ions

${\mathrm{CaCl}}_2⟶{\mathrm{Ca}}^{2+}+2{\mathrm{Cl}}^{-}$

Total ions $=0.05+0.1=0.15$ ions

$0.05\mathrm{m}{\mathrm{CaCl}}_2$ solution has minimum ions in solution, so it shows minimum boiling point elevation.

Correct Option is (D)

The aqueous solution contains $36\mathrm{g}$ glucose per ${\mathrm{dm}}^3$, so mass of solute ${\mathrm{W}}_2$ is $36\mathrm{g}$.

Assuming that the density of solution is $1\mathrm{g}/{\mathrm{dm}}^3$, the mass of solvent (water) is $1000\mathrm{g}$.

$\begin{array}{rl}& \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=\frac{1000{\mathrm{K}}_{\mathrm{b}}{\mathrm{W}}_2}{{\mathrm{M}}_2{\mathrm{W}}_1}\\ & \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=\frac{1000\mathrm{g}{\mathrm{kg}}^{-1}\times {\mathrm{K}}_{\mathrm{b}}\times 36\mathrm{g}}{180\mathrm{g}\times 1000\mathrm{g}}\\ & \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=\frac{2{\mathrm{K}}_{\mathrm{b}}}{10}\\ & \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\circ }\\ & {\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}^{\circ }+\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}\\ \therefore & {\mathrm{T}}_{\mathrm{b}}={(100+\frac{2{\mathrm{K}}_{\mathrm{b}}}{10})}^{\circ }\mathrm{C}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& {\mathrm{n}}_2=1\mathrm{mol}\\ & {\mathrm{n}}_1=\frac{36}{18}=2\mathrm{mol}\end{array}$

Relative lowering of vapour pressure

$=\frac{{\mathrm{P}}_1^0-{\mathrm{P}}_1}{{\mathrm{P}}_1^0}=x_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

$\begin{array}{ll}\therefore & \frac{32\mathrm{mm}\mathrm{Hg}-{\mathrm{P}}_1}{32\mathrm{mm}\mathrm{Hg}}=\frac{1}{3}\\ & 96\mathrm{mm}\mathrm{Hg}-3{\mathrm{P}}_1=32\mathrm{mm}\mathrm{Hg}\\ & 64\mathrm{mm}\mathrm{Hg}=3{\mathrm{P}}_1\\ \therefore & {\mathrm{P}}_1=21.33\mathrm{mm}\mathrm{Hg}\approx 21.44\mathrm{mm}\mathrm{Hg}\end{array}$

Correct Option is (C)

${\mathrm{M}}_2=\frac{{\mathrm{K}}_{\mathrm{b}}\times {\mathrm{W}}_2\times 1000}{\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}\times {\mathrm{W}}_1}=\frac{2.5\times 3.5\times 1000}{0.35\times 100}=250\mathrm{g}{\mathrm{mol}}^{-1}$

Correct Option is (B)

For glucose solution, $\mathrm{m}=0.1\mathrm{m},{\mathrm{T}}_{\mathrm{b}}={100.16}^{\circ }\mathrm{C}$

$\begin{array}{rl}\therefore \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}& ={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}{}^{\circ }\\ & =100.16-100={0.16}^{\circ }\mathrm{C}\end{array}$

${\mathrm{K}}_{\mathrm{b}}=\frac{\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}}{\mathrm{m}}=\frac{{0.16}^{\circ }\mathrm{C}}{0.1\mathrm{m}}={1.6}^{\circ }\mathrm{C}/\mathrm{m}$

For sucrose solution, $\mathrm{m}=0.5\mathrm{m},{\mathrm{K}}_{\mathrm{b}}={1.6}^{\circ }\mathrm{C}/\mathrm{m}$

$\begin{array}{rl}\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}& ={\mathrm{K}}_{\mathrm{b}}\times \mathrm{m}\\ & ={1.6}^{\circ }\mathrm{C}/\mathrm{m}\times 0.5\mathrm{m}={0.80}^{\circ }\mathrm{C}\\ \mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}& ={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\circ }\\ {\mathrm{T}}_{\mathrm{b}}& =\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}+{\mathrm{T}}_{\mathrm{b}}^{\circ }\\ & =0.80+100={100.80}^{\circ }\mathrm{C}\end{array}$