At 1 atmospheric pressure the boiling point of mixture is $${\text{8}}{{\text{0}}^ \circ }C.$$
At boiling point the vapour pressure of mixture, $${P_T} = 1$$  atmosphere $$ = 760\,mm\,Hg.$$
Using the relation,
$$\eqalign{ & {P_T} = P_A^ \circ {x_A} + P_B^ \circ {x_B},{\text{we get}} \cr & {{\text{P}}_T} = 520{x_A} + 1000\left( {1 - {x_A}} \right) \cr & \left\{ {\because \,P_A^ \circ = 520\,mm\,Hg,} \right. \cr & P_B^ \circ = 1000\,mm\,Hg;\,{x_A} + {x_B} = 1 \cr & {\text{or}}\,\,760 = 520{x_A} + 1000 - 1000{x_A}\,{\text{or}}\,\,{\text{480}}{{\text{x}}_A} = 240 \cr & {\text{or}}\,\,{x_A} = \frac{{240}}{{480}} = \frac{1}{2}\,\,{\text{or}}\,\,50\,mol\,{\text{percent}} \cr} $$

$$\eqalign{ & P = {P_A}{X_A} + {P_B}{X_B} \cr & = {P_A}{X_A} + {P_B}\left( {1 - {X_A}} \right) \cr & \Rightarrow {P_A}{X_A} + {P_B} - {P_B}{X_A} \cr & \Rightarrow {P_B} + {X_A}\left( {{P_A} - {P_B}} \right) \cr} $$

Acetonitrile $$\left( {\mathop {C{H_3}}\limits^{\delta + } - C \equiv \,\mathop N\limits^{\delta - } } \right)$$    and acetone both are polar molecules, hence dipole-dipole interaction exist between them. Between $$KCl$$  and water ion-dipole interaction is found and in Benzene ethanol and Benzene-Carbon tetra chloride dispersion force is present

$$\eqalign{ & A{l_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons 2\,A{l^{3 + }} + 3SO_4^{2 - } \cr & {\text{Value of van't Hoff's factor}}\,\,\left( i \right) = 5 \cr} $$
$$\left( {\text{A}} \right){K_2}S{O_4} \rightleftharpoons 2{K^ + } + SO_4^{2 - }\,\,\,\left( {i = 3} \right)$$
$$\left( {\text{B}} \right){K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftharpoons $$      $$3\,{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\,\,\,\left( {i = 4} \right)$$
$$\left( {\text{C}} \right)Al{\left( {N{O_3}} \right)_3} \rightleftharpoons $$     $$A{l^{3 + }} + 3NO_3^ - \,\,\,\left( {i = 4} \right)$$
$$\left( {\text{D}} \right){K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \rightleftharpoons $$      $$4\,{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\,\,\left( {i = 5} \right)$$
Therefore, $${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$   has same value of $$i$$ that of $$A{l_2}{\left( {S{O_4}} \right)_3}$$   i.e. $$i=5.$$

Osmotic pressure is a colligative property which is used to find the molecular weight of polymer because other colligative properties give so low value of molecular weight that it cannot be measured accurately.

Using relation,
$$\frac{{{p^ \circ } - {p_s}}}{{{p_s}}} = \frac{{{w_2}{M_1}}}{{{w_1}{M_2}}}$$
where $${w_1},{M_1} = $$   mass in $$g$$ and $$mol.$$  mass of solven
$${w_2},{M_2} = $$   mass in $$g$$ and $$mol.$$  mass of solute
Let $${{\text{M}}_2} = x$$
$${p^ \circ } = 185\,torr\,;\,{p_s} = 183\,torr$$
$$\frac{{185 - 183}}{{183}} = \frac{{1.2 \times 58}}{{100x}}$$     ( $$Mol.$$  mass of acetone = 58 )
$$x = 64$$
∴ Molar mass of substance $${\text{ = 64}}$$

Acetone + ethanol is an example of solutions showing positive deviation from Raoult's law. Since acetone-ethanol attractions are weaker than acetone-acetone and ethanol-ethanol attractions.

$$\eqalign{ & {\text{We know,}}\,\,{P_{{\text{Total}}}} = {p_1} + {p_2} \cr & {\text{Also}}\,\,{p_1} = p_1^ \circ \times {x_1}\,\,{\text{and}}\,\,{p_2} = p_2^ \circ \times {x_2} \cr & {x_1}{\text{(mole fraction of}}\,\,C{H_3}OH) \cr & = \frac{{\frac{{40}}{{32}}}}{{\frac{{40}}{{32}} \times \frac{{60}}{{46}}}} = 0.49 \cr & \left[ {Mol.wt.\,{\text{of}}\,\,C{H_3}OH = 32,Mol.wt.\,{\text{of}}\,\,{C_2}{H_5}OH = 46} \right] \cr & {x_2}\left( {{\text{Mole fraction of}}\,\,{C_2}{H_5}OH} \right) \cr & = \frac{{\frac{{60}}{{40}}}}{{\frac{{40}}{{32}} + \frac{{60}}{{46}}}} = 0.51 \cr & {p_1} = {\text{Partial vapour pressure of}} \cr & C{H_3}OH = 88.7 \times 0.49 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 43.46\,\,mm\,\,Hg \cr & {p_2} = {\text{Partial vapour pressure of}} \cr & {C_2}{H_5}OH = 44.5 \times 0.51 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 22.69\,\,mm\,\,Hg \cr & {p_{{\text{Total}}}} = {p_1} + {p_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {43.46 + 22.69} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 66.15\,\,mm\,\,Hg \cr & {\text{Mole fraction of}}\,\,C{H_3}OH\,\,{\text{in vapour}} \cr & = \frac{{43.46}}{{66.15}} \cr & = 0.657 \cr} $$

$$\eqalign{ & {\text{For }}{10^{ - 4}}\,M\,NaCl\,\,\,\,\,\,\,\,\,i = 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{10^{ - 4}}\,M\,\,Urea\,\,\,\,\,\,\,\,\,\,i = 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{10^{ - 3}}\,M\,MgC{l_2}\,\,\,\,\,i = 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{10^{ - 2}}\,M\,NaCl\,\,\,\,\,\,\,\,i = 2 \cr} $$
More the value of $$i, C,$$  more will be the elevation in boiling point hence increasing order of boiling point is
$${10^{ - 4}}\,M\,\,Urea < {10^{ - 4}}\,M\,NaCl < {10^{ - 3}}\,M\,MgC{l_2} < {10^{ - 2}}\,M\,NaCl$$

At equilibrium,
rate of dissolution = rate of crystallisation.