SOLUTIONS QUIZ 17

 

SOLUTIONS QUIZ 17

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry. .

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

•   Mole fraction of solute
•   1/(mole fraction of solute
•   Mole fraction of solvent
•   None of the above

Solution
From Raoult’s law : 𝑃₀−𝑃_s/𝑃₀ = 𝑁₁/(𝑁₁+𝑁₂) 

1 − 𝑃_s /𝑃₀ = 𝑁₁/ (𝑁₁ + 𝑁₂) 

or 𝑃_s /𝑃₀ = 𝑁₂/(𝑁₁+𝑁₂)

 𝑖.𝑒., 𝑃_s = 𝑃₀ =𝑁₂ /(𝑁₁ + 𝑁₂)

 𝑃_s = 𝑃₀ × 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

•   Solutions of non-electrolytes only
•   Solutions of electrolytes only
•   All solutions
•   None of the above

Solution
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 𝑐₁ and that in the second liquid is 𝑐₂. If the solute forms a stable trimer in the first liquid, the distribution law suggests that :

•    3𝑐₁ = 𝑐₂
•   𝑐₁/³√𝑐₂ = constant
•   𝑐₁/3 = 𝑐₂
•   𝑐₂/³√𝑐₁ = constant

Solution
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 𝑀𝑔𝐶𝑙₂ and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of

•   Solution A is greater than B
•   Both have same osmotic pressure
•   Solution B is greater than A
•   Cannot be determine

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 of𝑀𝑔𝐶𝑙₂=7g, V=1L 

∴ Concentration== 𝑚𝑎𝑠𝑠 𝑚𝑜𝑙.𝑚𝑎𝑠𝑠 = 7 95 = 0.0747

 𝑀𝑔𝐶𝑙₂ dissociates as follows 

 MgC𝑙₂ → 𝑀𝑔₂ + 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 𝐻₂𝑆𝑂₄ + 100 mL of 0.2 M NaOH is

•   0.2
•   0.01
•   0.1
•   0.3

Solution
  Given 

 H₂SO₄ - V=100mL, N=0.2 M 

NaOH - V=100mL, N=0.2 M 

Milliequivalent of 𝐻₂𝑆𝑂₄ = 100 × 0.2 × 2 = 40 (∴ It is dibasic acid) 

 Milliequivalent of NaOH =100 × 0.1 × 2 = 20 

 ∴ Moilliequivalent Of 𝐻₂𝑆𝑂₄ left =40-20=20 

Total volume = 100mL+100mL=200mL 

Normality of 𝐻₂𝑆𝑂₄ (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

•   250 mm, 550 mm
•   350 mm, 450 mm
• 350 mm, 700 mm
•   550 mm, 250 mm

Solution
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 2 

Mole fraction of Y (𝜘_Y)= 1 2 

We 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) = 1/3 

Mole fraction of Y (𝜘_Y) = 2/3 

𝑃 = 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 CS₂. The concentration of I₂ in water and CS₂were found to be 𝐶₁/𝐶₂ respectively. The ratio of 𝐶₁/𝐶₂ will change if :  

 

•   More I2 is added
•   More CS2 is added
•   More water is added
•   Temperature is changed

Solution

𝐾 = 𝑐₁/𝑐₂ 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

•   Camphor
• Naphthalene
•   Benzene
•   Water

Solution
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

•   Boiling point of water involved in cooking is increased
•   Heat is more evenly distributed in the cooking space
•   The higher pressure inside the cooker crushes the food material
•   Cooking involves chemical changes helped by a rise in temperature

Solution
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 :  

 

•   100.53 ͦC
•   101.06 ͦC
•   100.265 ͦC
•   99.47 ͦC

Solution
∆𝑇 = (1000 × 𝐾 × 𝑤) /𝑚.𝑊 ; 

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

 ∴ 𝑇_b = 100 + 0.265 = 100.265 ͦC