Azeotropes are formed by the solutions which show deviations from ideal behaviour.

$$\eqalign{ & {\text{Molarity}} \cr & = \frac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution (in }}mL{\text{)}}}} \times 1000 \cr & = \frac{{25.3 \times 1000}}{{106 \times 250}} \cr & = 0.9547 \approx 0.955\,M \cr} $$
$$N{a_2}C{O_3}$$  in aqueous solution remains dissociated as
$$\mathop {N{a_2}C{O_3}}\limits_x \rightleftharpoons \mathop {2\,N{a^ + }}\limits_{2x} + \mathop {CO_3^{2 - }}\limits_x $$
Since, the molarity of $$N{a_2}C{O_3}$$  is $$0.955\,M,$$  the molarity of $$CO_3^{2 - }$$  is also $$0.955\,M$$  and that of $$N{a^ + }$$  is $$2 \times 0.955 = 1.910\,M$$

According to Henry's law,
$${x_{{N_2}}} \times {K_H} = {p_{{N_2}}}$$     ( $${p_{{N_2}}} = $$   Partial pressure of $${N_2}$$  )
Given, total pressure $$ = 5\,atm,$$  mole fraction of $${N_2} = 0.8$$
∴ Partial pressure of $${N_2} = 0.8 \times 5 = 4$$
$$ \Rightarrow {x_{{N_2}}} \times 1 \times {10^5} = 4 \Rightarrow {x_{{N_2}}} = 4 \times {10^{ - 5}}$$
no. of moles of $${H_2}O,{n_{{H_2}O}} = 10$$
no. of moles of $${N_2},{n_{{N_2}}} = ?$$
$$\eqalign{ & \frac{{{n_{{N_2}}}}}{{{n_{{N_2}}} + {n_{{H_2}O}}}} = {x_{{N_2}}} = 4 \times {10^{ - 5}} \cr & \Rightarrow \frac{{{n_{{N_2}}}}}{{10 + {n_{{N_2}}}}} = 4 \times {10^{ - 5}} \cr & \Rightarrow {n_{{N_2}}} = 4 \times {10^{ - 4}}\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,{n_{{N_2}}} < < < 10} \right] \cr} $$

Total vapour pressure of mixture
= Vapour pressure of pentane in mixture + vapour pressure of hexane in mixture
Since, the ratio of pentane to hexane = 1 : 4
$$\therefore $$  $$Mole$$  fraction of pentane $$ = \frac{1}{5}$$
$$Mole$$  fraction of hexane $$ = \frac{4}{5}$$
= ( $$mole$$  fraction of pentane × vapour pressure of pentane ) + ( $$mole$$  fraction of hexane × vapour pressure of hexane)
$$\eqalign{ & = \left( {\frac{1}{5} \times 440 + \frac{4}{5} \times 120} \right) \cr & = 184\,mm \cr} $$
$$\because $$   Vapour pressure of pentane in mixture
= Vapour pressure of mixture × $$mole$$  fraction of pentane in vapour phase
$$88 = 184 \times mole$$    fraction of pentane in vapour phase
$$\therefore $$  $$Mole$$  fraction of pentane in vapour phase
$$ = \frac{{88}}{{184}} = 0.478$$

For an ideal solution $$\Delta H$$  and $$\Delta V$$  for mixing should be zero. $${P_{{\text{Total}}}} = {p_A} + {p_B}$$    and $$A - A,B - B$$    and $$A - B$$   interactions are nearly same.

$$\eqalign{ & {\rm{Let}}\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{solution}}\,\,{\rm{be}}\,\,{\rm{100 g}}{\rm{.}} \cr & {\rm{Volume}}\,\,{\rm{of}}\,\,{\rm{solution}} \cr & = {{{\rm{Mass}}} \over {{\rm{Density}}}} \cr & = {{100} \over {1.04}} \cr & = 96.154\,mL \cr & = 0.096\,L \cr} $$
$${w_B} = 3.5\,g;{M_B} = 58.5\,g\,mo{l^{ - 1}}$$      ( molar mass of $$NaCl$$  )
For complete ionisation of $$NaCl,$$  van't Hoff factor will be $$'2'.$$
$$\eqalign{ & i = 2, \cr & {\rm{we}}\,\,{\rm{know,}} \cr & \pi = iCRT \cr & \,\,\,\, = i{n \over V}RT \cr & \,\,\, = 2 \times {{3.5 \times 0.0821 \times 293} \over {58.5 \times 0.096}} \cr & \,\,\, = 29.98\,atm \cr} $$

As both the solutions are isotonic hence there is no net movement of the solvent through the semipermeable membrane between two solutions.

Urea is non-electrolyte, hence will not dissociate to give ions.

$$V.P.$$  of solution at $${t^ \circ }C = 760\,mm$$
[ at $$b.p.,$$  $$V.P.$$  of solution = atompheric pressure ]
$${\text{Thus}} = P_A^ \circ .\,{X_A} + P_B^ \circ .\,{X_B}$$
$${\text{or}}\,P = P_A^ \circ .\,{X_A} + P_B^ \circ .\left( {1 - {X_A}} \right)$$       $$\left[ {\because \,{X_A} + {X_B} = 1} \right]$$
$${\text{or}}\,\,760 = 400{X_A} + 800\left( {1 - {X_A}} \right)$$       $$\left[ {\because P = 760\,mm\,Hg} \right]$$
$$\eqalign{ & {\text{or}}\,\, - 800 + 760 = - 400\,{X_A} \cr & {\text{or}}\,\, - 40 = - 400\,{X_A} \cr & {\text{or}}\,\,{X_A} = \frac{{40}}{{400}} = 0.1 \cr} $$
Thus mole fraction of in solution is 0.1

When acetone and chloroform are mixed together, a hydrogen bond is formed between them which increases intermolecular interactions. Hence, $$A - B$$  interactions are stronger than $$A - A$$  and $$B - B$$  interactions.