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**Q1.**The vapour pressure of a solution is proportional to :

Solution

From Raoult’s law : 𝑃

From Raoult’s law : 𝑃

_{0}−𝑃_{s}/𝑃_{0}= 𝑁_{1}/(𝑁_{1}+𝑁_{2})1 − 𝑃

_{s}/𝑃_{0}= 𝑁_{1}/ (𝑁_{1}+ 𝑁_{2})or
𝑃

_{s}/𝑃_{0}= 𝑁_{2}/(𝑁_{1}+𝑁_{2}) 𝑖.𝑒.,
𝑃

_{s}= 𝑃_{0}=𝑁_{2}/(𝑁_{1}+ 𝑁_{2})𝑃

_{s}= 𝑃

_{0}× mole fraction of solvent

**Q2.**At low concentrations, the statements that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for

Solution

Equal osmatic pressure only applicable to nonelectrolytes solution at low concentration

Equal osmatic pressure only applicable to nonelectrolytes solution at low concentration

**Q3.**When a solute is added in two immiscible solvents, it distributes itself between two liquids so that its concentration in first liquid is 𝑐

_{1}and that in the second liquid is 𝑐

_{2}. If the solute forms a stable trimer in the first liquid, the distribution law suggests that :

Solution

This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent.

This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent.

**Q4.**Solution A contains 7 g/L of 𝑀𝑔𝐶𝑙

_{2}and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of

Solution

Osmotic pressure is a colligative property . More the number of particles (or ions) in solution, more will be osmotic pressure. Nacl solution

Osmotic pressure is a colligative property . More the number of particles (or ions) in solution, more will be osmotic pressure. Nacl solution

Given , mass of NaCl =7 g V=1L

∴ Concentration
= mass / mol.mass = 7/58.5 = 0.119 𝑀

NaCl dissociates as follows
NaCl → 𝑁𝑎+𝐶𝑙−(2 ions)

∴ Concentration of ions in solution
=2× 0.119 𝑀
=0.0238 M
MgCl solution

Given, mass of𝑀𝑔𝐶𝑙

_{2}=7g, V=1L∴ Concentration== 𝑚𝑎𝑠𝑠 𝑚𝑜𝑙.𝑚𝑎𝑠𝑠 = 7 95
= 0.0747

𝑀𝑔𝐶𝑙

_{2}dissociates as follows MgC𝑙

_{2}→ 𝑀𝑔_{2}+ 2𝐶𝑙^{-}(3 ions)∴ Concentration of ions in solution =3×0.074 M
=0.222 M

∵ Number of particles in solution B (NaCl)are more than in solution A .

∴ Osmotic pressure of solution B (NaCl) will be more than solution A .

**Q5.**The normality of mixture obtained by mixing 100 mL of 0.2 M 𝐻

_{2}𝑆𝑂

_{4}+ 100 mL of 0.2 M NaOH is

Solution

**Given****H**

_{2}SO_{4}- V=100mL, N=0.2 M**NaOH - V=100mL, N=0.2 M**

**Milliequivalent of 𝐻**

_{2}𝑆𝑂_{4}= 100 × 0.2 × 2 = 40 (∴ It is dibasic acid)**Milliequivalent of NaOH =100 × 0.1 × 2 = 20**

**∴ Moilliequivalent Of 𝐻**

_{2}𝑆𝑂_{4}left =40-20=20**Total volume = 100mL+100mL=200mL**

**Normality of 𝐻**.

_{2}𝑆𝑂_{4}(left0)= 20/200 =0.1 N**Q6.**Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1:1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are

Solution

In Ist case,

In Ist case,

When two liquids X and Y are mixed in the molar ratio 1:1.

Moles of X = 1
Moles of Y =1

Mole fraction of X (𝜘

_{X})= 1 2Mole fraction of Y (𝜘

_{Y})= 1 2We know that

p=𝑝

_{X}^{∘}𝜘_{X}+ 𝑝_{Y}^{∘ }𝜘_{Y}( p =total pressure of mixture) 400 = 1/2 𝑝

_{X}^{∘}+ 1/2 𝑝_{Y}^{∘} 400 × 2 = 𝑝

_{X}^{∘}+ 𝑝_{Y}^{∘}…(i) For case II

^{nd}, When liquids are mixed in the molar ratio of 1:2,

Mole fraction of X =1

Mole fraction of Y = 2

Mole fraction of X (𝜘

_{X}) = 1/3Mole fraction of Y (𝜘

_{Y}) = 2/3𝑃 = 𝑝

_{X}^{∘}𝜘_{X}+ 𝑝_{Y}^{∘}𝜘_{Y}350 = 1/3 𝑝

_{X}^{∘}2/3 𝑝𝑌^{∘}^{ }350 × 3 = 𝑝

_{X}

^{∘ }+ 2𝑝

_{Y}

^{∘ }…(ii)

From Eqs (i) and (ii) , we get

𝑝

_{X}^{ ∘}= 550𝑚𝑚 𝑝_{Y}° = 250 𝑚𝑚**Q7.**Iodine was added to a system of water and CS

_{2}. The concentration of I

_{2}in water and CS

_{2}were found to be 𝐶

_{1}/𝐶

_{2}respectively. The ratio of 𝐶

_{1}/𝐶

_{2}will change if :

Solution

𝐾 = 𝑐

𝐾 = 𝑐

_{1}/𝑐_{2}is constant for a particular solute in a given solvent-solvent system at constant temperature.**Q8.**Which of the following solutions will have the highest boiling point ? NaOH solution

Solution

The molal depression constant (𝑘

The molal depression constant (𝑘

_{f}) for camphor is maximum. Hence depression of freezing point (∆𝑇_{f}) will be maximum for camphor =3.3%**Q9.**Pressure cooker reduces cooking time for food because

Solution

Due to higher pressure inside the boiling point elevated

Due to higher pressure inside the boiling point elevated

**Q10.**The molal boiling point constant of water is 0.53 ͦC. When 2 mole of glucose are dissolved in 4000 g of water, the solution will boil at :

Solution

∆𝑇 = (1000 × 𝐾 × 𝑤) /𝑚.𝑊 ;

∆𝑇 = (1000 × 𝐾 × 𝑤) /𝑚.𝑊 ;

∴ ∆𝑇 =
(1000 × 0.53 × 2 )/4000
= 0.265,

∴ 𝑇

_{b}= 100 + 0.265 = 100.265 ͦC