Test: Solutions (October 22) - NEET MCQ

Test: Solutions (October 22) - NEET MCQ

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15 Questions MCQ Test Daily Test for NEET Preparation - Test: Solutions (October 22)

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Test: Solutions (October 22) - Question 1

An ideal solution is formed when its components [1988]

Detailed Solution for Test: Solutions (October 22) - Question 1

For ideal solution,

Test: Solutions (October 22) - Question 2

The relative lowering of the vapour pressure is equal to the ratio between the number of [1991]

Detailed Solution for Test: Solutions (October 22) - Question 2

According to Raoult's law, the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the solution.

mole fraction of solute

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Test: Solutions (October 22) - Question 3

Which of the following modes of expressing concentration is independent of temperature ? [1992,1995]

Detailed Solution for Test: Solutions (October 22) - Question 3

The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature.

Test: Solutions (October 22) - Question 4

If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that [1992]

Detailed Solution for Test: Solutions (October 22) - Question 4

As both the solutions are isotonic hence there is no net movement of the solvent through the semipermeable membrane between two solutions.

Test: Solutions (October 22) - Question 5

Which one of the following salts will have the same value of van’t Hoff factor (i) as that of K4[Fe (CN)6]. [1994]

Detailed Solution for Test: Solutions (October 22) - Question 5

K4 [Fe(CN)6] and Al2(SO4)3 both dissociates to give 5 ions or i = 5

and

Test: Solutions (October 22) - Question 6

The number of moles of oxygen in one litre of air containing 21% oxygen by volume, in standard conditions, is [1995]

Detailed Solution for Test: Solutions (October 22) - Question 6

Percentage volume of oxygen = 21%. (Given)
∵ 100 ml of air contains = 21ml of O2
∴ Volume of oxygen in one litre of air

Therefore no. of moles

(∵ volume of 1 litre of gas at S.T.P. is 22400 ml)

Test: Solutions (October 22) - Question 7

According to Raoult's law, relative lowering of vapour pressure for a solution is equal to  [1995

Detailed Solution for Test: Solutions (October 22) - Question 7

Relative lowering of vapour pressure depends upon the mole fraction of solute.

mole fraction of solute

Test: Solutions (October 22) - Question 8

Which of the following 0.10 m aqueous solutions will have the lowest freezing point ? [1997]

Detailed Solution for Test: Solutions (October 22) - Question 8

Depression in F.P. ∝ No. of particles.
Al2(SO4)3 provides five ions on ionisation

while KCl provides two ions
KCl     K    Cl
C6H12O6 and C12H22O11 are not ionised so they have single particle.
Hence, Al2(SO4)3 have maximum value of depression in F.P or lowest F.P

Test: Solutions (October 22) - Question 9

A 5% solution of cane sugar (mol. wt. =342) is isotonic with 1% solution of a substance X. The molecular weight of X is [1998]

Detailed Solution for Test: Solutions (October 22) - Question 9

Osmotic pressure of 5% cane sugar solution 5% cane sugar solution means 100ml of solution contain cane sugar = 5g

∴    1000 of solution contain cane sugar

Similarily 1% is solution

Osmotic pressure of 1% solution of susbtance  (π2)

As both are isotonic, So
π1 = π2

∴  M (mol. wt. of X)

Test: Solutions (October 22) - Question 10

Which of the following statements, regarding the mole fraction (x) of a component in solution, is incorrect? [1999]

Detailed Solution for Test: Solutions (October 22) - Question 10

Mole fraction of any component A

As total no. of moles > no. of moles of A thus x can never be equal to one on zero.

Test: Solutions (October 22) - Question 11

The beans are cooked earlier in pressure cooker, because [2001]

Detailed Solution for Test: Solutions (October 22) - Question 11

The beans are cooked earlier in pressure cooker because boiling point increases with increasing pressure.

Test: Solutions (October 22) - Question 12

Molarity of liquid HCl will be, if density of solution is 1.17 gm/cc [2001]

Detailed Solution for Test: Solutions (October 22) - Question 12

Test: Solutions (October 22) - Question 13

A solution con tains non -volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure ? [2002]

Detailed Solution for Test: Solutions (October 22) - Question 13

Test: Solutions (October 22) - Question 14

Camphoris often used in molecular mass determination because [2004]

Detailed Solution for Test: Solutions (October 22) - Question 14

Solvent having high cryoscopic constant can be used in determination of molecular mass by cryoscopic method.

Test: Solutions (October 22) - Question 15

A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]

Detailed Solution for Test: Solutions (October 22) - Question 15

As ΔTf = Kf. m
ΔTb = Kb. m

Hence, we have

= [ΔTb = 100.18 - 100 = 0.18°C]

As the Freezing Point of pure water is 0°C,
ΔTf = 0 –Tf
0.654 = 0 – Tf
∴  Tf = – 0.654 thus the freezing point of solution will be – 0.654°C.

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