Step-I : Calculation for concentration of the solution.
According to Van't Hoff's equation
p = CRT or C = \(\frac { \pi }{ RT } \)
p = 8.8 atm, T = 40\(°\)C = 313 K
R = 0.0821 L atm K^-1 mol^-1
\(C=\frac { 8.8 }{ 0.0821\times 313 } \)=0.34 molL^-1(M)
Step-II: Calculation of freezing point of the solution depression in freezing point of solution
\(\Delta\)T_f = K_f \(\times\) m
m = 0.34 mol kg^-1
(same as molarity as given)
K_f = 1.86 K kg mol^-1
\(\Delta\)T_f = 1.86 \(\times\) 0.84 (\(\because\) \(\Delta\) T_f = K_f m)
 = 0.63 K = 0.63\(°\)C
Freezing point of the solution = 0 - 0.63 \(°\)C
= - 0.63\(°\) C

Molar mass of KCI = 39 + 35.5 = 74.35 g mol^-1
A KCI dissociates completely, number of ions produced are 2.There, Van't Hoff factor, i = 2
Mass of KCI solution = 1000 \(\times\)1.04 = 1040 g
Mass of solvent = 1040 - 74.5 = 965.5 g = 0.9655 kg^1/2
Molality of the solution :
\(\\ \frac { No.\quad of\quad moles\quad of\quad solute }{ Mass\quad of\quad solvent\quad in\quad kg } =\frac { 1\quad mol }{ 0.9655\quad kg } =1.0357\quad m\)
T_b = i \(\times\)K_b \(\times\)m
= 2 \(\times\) 0.52 \(\times\)1.0357 = 1.078 \(°\) C
Therefore, boiling point of solution
= 100 = 1.078 = 101.078\(°\)C

\({ \pi }_{ cane\ sugar }={ \pi }_{ X }\)
Therefore, c_{cane sugar}=c_X
(where c is molar concentration)
\(\frac { { W }_{ cane\ sugar } }{ { M }_{ cane\ sugar } } =\frac { { W }_{ X } }{ { M }_{ X } } \)
\(\frac { 5g }{ 342 \ g \ { mol }^{ -1 } } =\frac { 0.877 }{ { M }_{ X } } \)
\(\\ \Rightarrow { M }_{ X }=\frac { 0.877\times 342 }{ 5 } g\ { mol }^{ -1 }\)
\(\Rightarrow { M }_{ X }=59.9 \ or \ 60 \ g{ \ { mol }^{ -1 } }\)

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol \(=\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
\(\text {Moles of ethylene glycol }=\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.73 \mathrm{~mol}\)
\(\text {Mass of water in } \mathrm{kg}=\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}=0.6 \mathrm{~kg}\)
\(\text {Hence molality of ethylene glycol }=\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}=1.2 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Therefore freezing point depression,
ÄT_f = 1.86 K kg mol^–1 x 1.2 mol kg^–1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

\(\Delta\)T_b = iK_bm
i = 2
\(\Delta\)T_b = i \(\times\)K_b \(\times\)\(\frac { { W }_{ 2 }\times 1000 }{ M\times { W }_{ 1 } } \)
\(=2\times 0.52 \ K \ kg \ { mol }^{ -1 }\times \frac { 4g\times 1000 \ g/kg }{ 120 \ g/mol\times 100g } \)
\(=\frac { 2\times 0.52 }{ 3 } =0.346\ K\)
Boiling point of solution=\(\frac { 373.15\quad K }{ 373\quad K } \)
T_b = \({ \Delta T }_{ b }^{ o }+\Delta { T }_{ b }\)
\(=\frac { 373.15+0.346\quad K }{ 373\quad K+\quad 0.346\quad K } \)
\(=\frac { 373.496\ K }{ 373.346\ K } \)

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
M = 256
S \(\times \) x = 256
32 \(\times \) x = 256
x = 8
(b) (i) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink.
(ii) It we place blood cells in solution containing 0.4% sodium chloride solution, they swell.

(a) \(\Delta { T }_{ f }=i\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(\Delta { T }_{ f }=3\times \left( \frac { 1.86\times 1.9 }{ 95\times 50 } \right) \times 1000\)
= 2.23 K
\(\Delta { T }_{ f }-{ \Delta T }_{ f }^{ ' }=\frac { 273.15-2.23 }{ 273-2.23 } \)
\({ T }_{ f }^{ ' }\) = 270.92 K or 270.77 K
(b) (i) 2 M glucose has a higher boiling point because higher the number of particles lesser is the vapour pressure.
(ii) Reverse osmosis.

(a) (i) The ratio of number of moles of one component to the total number of moles of solution is known as mole fraction.
(ii) The solution which follows Raoult's law over entire range of concentrations at specific temperature is called ideal solution.
(b) W_B = 15 g, W_A = 450 g, \(\Delta\)T_f = 0.34 K, M_B = ?
\({ M }_{ B }=\frac { 1000\times { K }_{ f }\times { W }_{ B } }{ \Delta { T }_{ f }\times { W }_{ A } } \)
\(=\frac { 1000\times 1.86 \ K \ kg \ { mol }^{ -1 }\times 15 \ g }{ 0.34K\times 450 \ g } \)
= 182.35 g/mol

(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B  = 10g, W_s = 100 g, W_A = 90 g, M_B = 180 g.mol and d = 1.2 g/mL
\(M=\frac { Wt%\times density\times 10 }{ Mol.wt } \)
\(M=\frac { 10\times 1.2\times 10 }{ 180 } \)
= 0.66 M or 0.66 mol/L
\(m=\frac { { W }_{ B }\times 1000 }{ { M }_{ B }\times { W }_{ A }(in\quad g) } \)
\(m=\frac { 10\times 1000 }{ 180\times 90 } \)
= 0.61 m or 0.61 mol/kg
(or any other suitable method)