\({ \pi }_{ cane\ sugar }={ \pi }_{ X }\)
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

From Henry's law, m = K_p. If m₁ is the solubility of a gas at pressure
P₁ and m₂ is its solubility at pressure P₂'\({m_1\over m_2}={P_1\over P_2}\)
Thus, knowing m₁ at P₁- m₂ at P₂ can be calculated. The equation, m = KP also suggests that plot of pressure, P vs solubility m will be a straight line passing through the origin with slope = K. Greater the value of K, greater is the solubility. 

ΔT_b = 100.0832 - 100.0 = 0.0832°C,
K_b = 0.52 K kg mol^-1,
W_A = 250 g, W_B = 3.12 g,
M_B = 137 + 71 = 208 9 mol^-1
Now \(ΔT_b=iK_b\times{W_B\over M_B}\times{1000\over W_A}\)
⇒ \(0.0832 = i \times 0.52 \times{3.12\over4 6.489}\times{1000\over 250}\)
\(⇒ i={0.0832\times208\over 4\times3.12\times0.52}={17.30\over 6.489}=2.66\)
\(\alpha={i-1\over n-1}={2.66-1\over 3-1}={1.66\over 2}=0.83\)
Degree of dissociation (α) = 83%

\(\triangle { T }_{ f }={ K }_{ f }\times m\)
\( \Rightarrow \triangle { T }_{ f }={ K }_{ f }\times \frac { { W }_{ B } }{ { M }_{ B } } \times \frac { 1000 }{ { W }_{ A } }\)
\( \Rightarrow \triangle { T }_{ f }=1.86\times \frac { 54 }{ 180 } \times \frac { 1000 }{ 250 } \)
\( \Rightarrow \triangle { T }_{ f }=1.86\times \frac { 3 }{ 10 } \times 4=\frac { 22.32 }{ 10 } \\ =2.232 \ K\)
\( Freezing \ point=273K-2.232\)
\( =270.768 \ K\)
\(or \ F.pt.=273.15 \ K-2.23 \ K=270.7 \ K\)

\(\text { Using relation, } \pi=i C R T=i \frac{n}{V} R T\)
\(n=\frac{\pi V}{i R T}=\frac{0.75 \times 2 \cdot 5}{2 \cdot 47 \times 0 \cdot 0821 \times 300}=0.0308 \text { mole }\)
\( \text { Molar mass of } \mathrm{CaCl}_{2}=40+2 \times 35 \cdot 5=111 \mathrm{~g} \mathrm{~mol}^{-1} \)
\(\therefore \text { Amount of } \mathrm{CaCl}_{2} \text { dissolved }\)
\(=0.0308 \times 111=3.42 \mathrm{~g}\)

Vant's Hoff equation is \(\pi V=c R T\) Berkley and Hartley method is commonly used for measuring osmotic pressure.

\(x_{\text {water }}=\frac{\text { Moles of water }}{\text { Moles of water }+ \text { Moles of methanol }}\)
Now, moles of water = Moles of methanol = 1(say)
\(x_{\text {water }}=\frac{1}{1+1}=0.5\)

\(4 A \longrightarrow A_4\)
1 mole of A after association gives \(\frac{1}{4}\) mole of A₄. Hence, \(i=\frac{1}{4}=0.25\)

i > 1 when the solute undergoes dissociation in the solution.

8.199 g

Warming up of the solution means that the process of mixing is exothermic, i.e., \(\Delta H_{ mixing }=-ve\). This implies that the solution shows a negative deviation.

In C₂H₅OH, i = 1, because it is not an electrolyte.
In Ba₃(PO₄)₂ i = 5; In Na₂SO₄, i = 3;
In KCI, i = 2; In Li₃PO₄, i = 4
Greater the value of i, greater will be ΔT_b' and higher will be b.pt. of solution.C2HsOH, KCI, Na₂SO₄, Li₃PO₄ and Ba₃(PO₄)₂ is the increasing order of b.pt. Greater the value of i, greater will be ΔT_f lower will be freezing point of solution.Therefore, Ba₃(PO₄)₂ Li₃PO₄, Na₂SO₄, KCI and C₂H₅OH is the increasing order of freezing point.

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
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) (i) Henry's Law: It states the partial vapour pressure of gas in vapour phase is directly proportional to mole fraction of the gas in the solution.
p = K_H.x
where K_H is Henry's law constant, 'x' is mole fraction of gas in solution and p is partial vapour pressure of the gas in solution.
(ii) Raoult's Law for solution of non-volatile solute: The relative lowering of vapour pressure for a solution is equal to the mole fraction of solute when solvent alone is volatile.
\({p_A^0-p_A\over p_A^0}=x_B\)
where p⁰ is vapour pressure of pure component 'A', p_A is vapour pressure of component 'A' in solution, x_B is mole fraction of solute, P_A⁰ - P_A is lowering of vapour pressure and \(p_A^0-p_A\over p_A^0\) is relative lowering of vapour pressure.
(b) πV = nRT
\(\Rightarrow\ \pi V={W_B\over M_B}\times R \times T\)
\(\Rightarrow\ {0.335\over 760}\times{35\over 1000}={8.95\times10^{-3}\times0.0821\times 2988\over M_B}\)
\(\Rightarrow\ M_B={760 \times 1000 \times 8.95 \times 10-3 \times 0.0821\times 298\over 0.335\times35}\)
\(={166.416\times10^3\over11.725}=14.19\times10^3\)
= 1.419 x 10⁴ g mol^-1

\(\pi =CRT\\ C=0.002\quad mol\quad { dm }^{ -3 }=2\quad mol\quad { m }^{ -3 }\\ R={ 8.314JK }^{ -1 }{ mol }^{ -1 },T=293K\\ \pi =2\times 80314\times 293=4872\quad Pa\)

Methanol being more volatile than water, the solution of methanol in water will have more vapour pressure than water (a case of positive deviation).