Due to higher pressure exerted by steam than in the open pan.

01 M BaCI₂ solution will have higher boiling point because of large number of particles on dissociation.
\( \mathrm{NaCI} \longrightarrow \mathrm{Na}^{+}+\mathrm{CI}^{-}(2 \text { particles }) \\ \mathrm{BaCI}_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{CI}^{-}(3 \text { particles }) \)

The Henry's law states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure of the gas in equilibrium with the solution.
\( m \propto p \\ \mathrm{m}=\mathrm{kp} \)
where k is proportionality constant.

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

Define Raoult's law for binary solutions may be defined as: at a given temperature, for a solution of two volatile liquids the partial vapour pressure of each component of the solution is equal to the product of the vapour pressure of the pure component and its mole fraction in solution. Thus,
\(P_A=P_A \quad{ }^{\circ} \times x_a \text { and } P_B=P_B \quad{ }^{\circ} \times x_B\)

The solubility first first increases upto 32.4^oC (called transition temperature) and then decreases.

Molality \(=\frac{10 / 98}{90} \times 1000=1.13 \mathrm{~m}\)

Molality of a solution involves masses of the solute and solvent which do not change with temperature.

x₁ + x₂ + x₃ = 1

Normality = 2 × Molarity.

ΔT_b = 100.13°C - 100°C = 0.13 °C,
ΔT_b = K_b x m => 0.13 = 0.52 x m
=> m = 0.25 m/kg
Also, ΔT_f = K_f x m = 1.86 x 0.25 = 0.465 K
Freezing point of solution
= 273 - 0.465
= 272.535 K

Raoult's law states that at a given temperature, for a solution of volatile liquids, the partial pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction.
For example, for a binary solution of two volatile liquids A and B having mole fractions \({ x }_{ A } \ and \ { x }_{ B },\)
\({ p }_{ A }={ p }_{ A }^{ \circ }{ x }_{ A }and{ p }_{ A }={ p }_{ B }^{ \circ }{ x }_{ B }\)
where \({ p }_{ A }\ and\ { p }_{ B }\)  are the vapour pressures of the components in solutions and \({ p }_{ A }^{ \circ }\ and\ { p }_{ B }^{ \circ }\) are vapour pressure of pure components.
According to Henry's law for a gas dissolved in a liquid, the pressure of the gas is directly proportional to mole fraction i.e.
 \(p=Kx\)
where K is a proportionality constant known as Henry's constant.
But, Raoult's law states that
 \(p={ p }^{ \circ }x\\ \therefore K={ p }^{ \circ }\)
This means that Raoult's law is a special case of Henry's law.

(i) Mole fraction (x): It is the ratio of number of moles of a particular component to the total number of moles of all the components.
For example, mole fraction of a component A,
\({ X }_{ A }=\frac { { n }_{ A } }{ { n }_{ A }+{ n }_{ B } } \)
where \({ n }_{ A }\ and \ { n }_{ B }\)  are the number of moles of components 'A' and 'B' respectively.
(ii) Isotonic solutions: Two solutions are said to be isotonic when they exert the same osmotic pressure because they have same molar concentration. All intravenous injections must be isotonic with body fluids.
(iii) van't Hoff factor (i): The ratio of the experimental value of a colligative property to the theoretical value (calculated on the basis of normal behaviour of solute) is known as van't Hoff factor.
Experimentally determined
 \(i=\frac { value \ of \ the \ colligative \ property }{ \ \ Calculated \ value \ of\ the\\ \ \ colligative \ property } \)
(iv) Ideal solution: Those solutions which obey Raoult's law are called ideal solutions. When the forces of attraction between A-A, B-B are similar to A-B, then A and B will form ideal solution.

Molar mass of solute,
 \({ M }_{ B }=\frac { { K }_{ f }\times{ W }_{ B }\times1000 }{ { W }_{ A }\times \triangle { T }_{ f } } \)
\( { W }_{ B }=1.0g\)
\( { W }_{ A }=50.0g\)
\( \triangle { T }_{ f }=0.40K\)
\( { K }_{ f }=5.12 \ K \ kg \ { mol }^{ -1 }\)
\(\\ { M }_{ B }=\frac { 5.12\times1.0\times1000 }{ 50\times0.40 } \)
\(=256 \ g/mol\)

Osmotic pressure is directly proportional to the number of moles of solute.

Henry's law states that partial pressure of gas is directly proportional to its mole fraction.

It means solution shows -ve deviation from Raoult's law due to increase in force of attraction, volume decreases.

Molality is defined as the number of moles of dissolved per kg of solvent, whereas molarity is defined as the number of moles of solute dissolved per litre of solution.

\( \alpha=\frac{i-1}{n-1} \Rightarrow 0.5=\frac{i-1}{2-1} \Rightarrow i=1.5 \\ \left[\alpha=50=0.5, n=2 \quad \because \quad A B \longrightarrow A^{+}+B^{-}\right] \)

Molality is more accurate than molarity because molarity does not depend on temperatures mass does not change with temperature.

Reverse Osmosis: If extra pressure is applied on the solution side and exceeds the osmotic pressure, the osmosis can be reversed. That is, pure water can be forced out of the solution to pass through the pores of the membrane in the opposite direction. This is called reverse osmosis.

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

\(\pi V\)= nRT = \(\frac{{W}^{B}}{{M}^{B}}\times R \times T\)

\(\Rightarrow \)  \(\pi \times1 = \frac{68.4}{342}\times0.0821 \times 300 K \)

\(\Rightarrow \) \(\pi =\frac{24.63}{5} = 4.92 \ atm\)

Given that Henry’s law constant
K_H = 4.27× 10⁵ mmHg
p = 760mmHg
If X is the molar fraction then
According to Henry’s law,
p = K_HX
760 = 4.27×10⁵×X 
Divide by 4.27 × 105  , we get
Molar fraction, X 
\(=\frac{760}{4.27\times10_5}\)
= 178×10^-5 
So that, the molar fraction of methane in benzene = 1.78×10^-3 

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

Higher the solubility, lower is K_H. As CO₂ has maximum solubility, its K_H is lowest.

\({ N }_{ 1 }{ V }_{ 1 }+{ N }_{ 2 }{ V }_{ 2 }+{ N }_{ 3 }{ V }_{ 3 }={ N }_{ 4 }{ V }_{ 4 }={ N }_{ 4 }({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 })\)

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
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.