Colligative Properties Questions

Addition of a non-volatile solute to a volatile solvent, results in the decrease in the vapour pressure of the solution. Many properties of solutions are associated with this decrease in vapour pressure. These are: (1) relative lowering of vapour pressure of the solvent (2) depression of freezing point of the solvent (3) elevation of boiling point of the solvent and (4) osmotic pressure of the solution. Such properties are called colligative properties.

Colligative Properties Chemistry Questions with Solutions

Q-1: Which of the following aqueous solutions has the highest vapour pressure at 300 K?

1. 1 M Na₃PO₄
2. 1 M CaCl₂
3. 1 M KNO₃
4. 1 M C₆H₁₂O₆

Answer: d) 1 M C₆H₁₂O₆

Explanation: The Van’t Hoff Factor(i) quantifies the effect of a solute on various colligative properties of solutions.An electrolytic solute’s van’t Hoff factor is always equal to the number of ions in which it is ionised. It is equal to one for a non-electrolytic solute.

Upon ionisation of electrolytic solutes Na₃PO₄, CaCl₂ and KNO₃, the value of the van’t hoff factor(i) comes out to be 4,3 and 2 respectively. Because C₆H₁₂O is a non-electrolytic solute, it has i=1.

The greater the value of the Van’t Hoff factor(i), the lower the vapour pressure, and vice versa. As a result of the lower van’t hoff factor(i) value, C₆H₁₂O₆ has a higher vapour pressure.

Q-2: 1 molal aqueous solution of an electrolyte A₂B₃ is ionised 60%. The boiling point of the solution at 1 atm is (K_b(H₂O) = 0.52 K kg/mol)

1. 274.76 K
2. 377 K
3. 374.76 K
4. 376.5 K

Answer: c) 374.76 K

Explanation:

Given: Molality, m = 1m = 1mol/kg , Degree of dissociation,𝞪= 60% = 0.6

For dissociation,

Van’t Hoff factor, i = 1+(n-1)𝞪

The value of “n” is equal to the number of ions in which the electrolyte A₂B₃ dissociates.

A₂B₃ →2A³⁺ +3B^2-

This makes n= 5

Van’t Hoff factor, i = 1+(5-1)0.6 = 3.4

Elevation in boiling point, ΔT_b= i × K_b× m = (3.4)× (0.52K kg/mol)×(1mol/kg)

On solving,

ΔT_b= 1.768 K

ΔT_b= T_b-T_b^o

Where, T_b^o is boiling point of water = 373 K

T_b is the boiling point of solution at 1 atm

Substituting the values,

1.768 K= T_b-373K

T_b = 374.76 K

Q-3: Which of the following has equal boiling point?

1. 0.1 M Na₂SO₃
2. 0,1 M C₁₂H₂₂O₁₁
3. 0.1 MgCl₂
4. 0.1 M Al(NO₃)₃

Answer: a,c

Explanation: Boiling point will be the same for those solutes which will have the same value for van’t Hoff factor.

Hence, Na₂SO₃ and MgCl₂ will have the same value for van’t hoff factor(i).

Note: C₁₂H₂₂O₁₁ is a non electrolyte, therefore the value of the van’t hoff factor will be equal to one. In the rest of the cases, it will be equal to the number of ions it produces.

Q-4: In a closed container, vapour pressure of a liquid depends upon

1. Volume of the container
2. Temperature
3. Volume of the liquid
4. All of the above

Answer: b) Temperature

Explanation: The tendency of a material to change into a gaseous state is referred to as vapour pressure. In other words, the pressure exerted by a liquid’s vapour in thermodynamic equilibrium with the condensed phases in a closed system is termed as vapour pressure.

Effect of Temperature on Vapour pressure:

The K.E associated with the liquid increases as the temperature of the liquid rises. And as kinetic energy increases, the escaping tendency of the molecule increases, causing vapour pressure to rise. As a result, we can conclude that vapour pressure is directly proportional to temperature.

The volume of the container and liquid has no effect on the vapour pressure.

Q-5: Which of the following observations demonstrates colligative properties?

a) A 0.1 M NaCl solution has a higher vapour pressure than 0.1 M CaCl₂ solution.

b) A 0.1 M KOH solution freezes at lower temperature than pure water

c) Pure water freezes at a high temperature than pure methanol.

Answer:a) and b)

Explanation:

a) The greater the value of the Van’t Hoff factor(i), the lower the vapour pressure, and vice versa. The van’t Hoff factor value(i) of NaCl is 2, while CaCl₂ has a value of 3. NaCl has a higher vapour pressure than CaCl₂ solution due to its lower “i” value.

This demonstrates the relative lowering in vapour pressure colligative property.

b) Substances in a pure state have a higher freezing point than the solutions with solute present in it.This demonstrates the depression in freezing point colligative property.

c) Colligative properties cannot be demonstrated in pure substances.

Q-6: Examine the diagram below and select the appropriate options.

1. There will be no substance movement across the membrane.
2. CaCl₂ will flow towards the KCl solution
3. The CaCl₂ solution will be approached by KCl.
4. The osmotic pressure of 0.1 M KCl solution is greater than that of 0.05 M CaCl₂ solution.(assuming complete dissociation of electrolyte)

Answer: d) The osmotic pressure of 0.1 M KCl solution is greater than that of 0.05 M CaCl₂ solution.

Explanation: The diagram depicts the osmosis phenomenon. Osmosis is the flow of solvent molecules from its high concentration to its low concentration.

Only solvent molecules, not solute particles, move across the membrane. Because CaCl₂ and KCl are solutes, they will not move or flow.

Osmotic pressure,𝝅 = i×c×R×T

Osmotic pressure is directly proportional to the magnitude of (i×c) provided temperature and R are constant.

Value of “i” is 2 and 3 for KCl and CaCl₂ respectively.

For KCl, 𝝅 = 2×0.1= 0.2

For CaCl₂,𝝅 = 3 ×0.05=0.15

We can clearly see, 0.1 M KCl has higher osmotic pressure than 0.05 M CaCl₂ .

Q-7: When 0.643 g of a compound is added to 50 mL of benzene (density:0.879g/mL),the freezing point is reduced from 5.51 to 5.03 degrees Celsius. What is the compound’s molecular mass?(K_f for benzene = 5.12)

Answer: Let molecular mass of solute(compound) be y

Density of benzene = Mass of benzene/Volume

0.879 g/mL= Mass of benzene/ 50mL

Mass of benzene = 43.95 g = 0.04395 Kg

Moles of solute( compound) = Weight of solute/ Molecular mass of solute

= 0.643 g/ y

Molality(m) = Moles of solute/ Mass of benzene( Kg)

= 0.643 g/(y × 0.04395 Kg)

Depression in freezing point, ΔT_f= K_f× m

(5.51-5.03) K = 5.12 K Kg/mol ×[ 0.643 g/(y × 0.04395 Kg)]

On solving, y = 156 g/mol

Q–8: The unit of cryoscopic constant is

1. K g/mol
2. Kg/mol K
3. K Kg/mol
4. mol/ Kg K

Answer: c) K Kg/mol

Explanation:

Cryoscopic constant K_f = ΔT_f/ m

Unit of ΔT_f = K

Unit of m(molality) is mol/Kg

This makes unit of K_f = K Kg/mol

Q-9: Determine the elevation of the boiling point of a aqueous solution with 1 mol of solute(glucose).( density=1.2g/mL)

1. K_b
2. 0.2 K_b
3. 0.02K_b
4. 0.98K_b

Answer: d) 0.98K_b

Explanation:

Density of solution = Mass of solution/Volume of solution

1.2 g/mL= Mass of solution/ 1000 mL

Mass of solution = 1200g = 1.2Kg

Now, glucose acts as a solute in the solution, and its molar mass is 180g.

Mass of solvent = Mass of solution-Mass of solute = 1200 – 180 = 1020g = 1.02 Kg

Molality of solution, m = Moles of solute/Weight of solvent(Kg)

= 1/1.02 = 0.98

Elevation in boiling point, ΔT_b= K_b× m = 0.98K_b

Q-10: Why is osmotic pressure preferable to other colligative properties?

Answer: The osmotic pressure method has an advantage over other methods because pressure is measured at room temperature and molarity of the solution is used instead of molality.Its magnitude is large in comparison to other colligative properties, even in very dilute solutions.

Q-11: Define K_b and the factors that influence it.

Answer: The molal elevation constant is denoted by K_b. The molal elevation constant is the increase in boiling point caused by dissolving one mole of solute in 1000g (1Kg) of solvent. Ebullioscopic constant is another name for the molal elevation constant.

Molal elevation constant is characteristic for the type of solvent and not on the solute dissolved in it.

Q-12: When SPM is placed between two isotonic solutions, then

1. Osmosis occur more efficiently
2. Osmosis do not occur
3. Reverse osmosis occur
4. Flow of solute takes place

Answer: b) Osmosis do not occur

Explanation: Isotonic solutions are two solutions that have the same osmotic pressure at the same temperature. Osmosis does not occur when such solutions are separated by a semipermeable membrane.

Q-13: Define

1. Hypertonic solutions
2. Hypotonic solutions

Answer:

a) If the water flows out of the cells when placed in a solution, it shrinks and the solution is said to be hypertonic.

b) If water flows into the cells and they swell when placed in a solution, then the solution is said to be hypotonic.

Q-14: Which colligative property is used for the determination of molar masses of biomolecules?

1. Lowering in vapour pressure
2. Depression in freezing point
3. Osmotic pressure
4. Boiling point

Answer: c) Osmotic pressure

Q-15: In 1L of glacial CH₃COOH, 0.1 M acetamide is dissolved. When the solution is cooled, the molecule contains the first crystal that forms at the freezing point is

1. Glacial CH₃COOH
2. Acetamide
3. Both acetamide and glacial CH₃COOH
4. None of the above

Answer: a) Glacial CH₃COOH

Explanation: Only solvent molecules solidify at the freezing point in the depression of the freezing point experiment. Because the solvent molecule in this case is glacial CH₃COOH, the first crystal of it appears.

Practise Questions on Colligative Properties

Q-1: Which of the following statements are correct for depression in freezing point?

I) The solution has a lower vapour pressure than pure solvent.

II) The solution has a higher vapour pressure than pure solvent.

III) At the freezing point, only solvent molecules solidify.

IV) At the freezing point, only solute molecules solidify.

Q-2: Which of the following 0.1M aqueous solution will have the lowest freezing point?

1. Urea
2. Barium sulphate
3. Potassium chloride
4. Fructose

.

Q-3: Which of the following liquid pairs is most likely to separate into two layers when 10 mL of each is mixed and allowed to stand?

1. CCl₄ and C₆H₆
2. C₂H₅OH and CH₃OH
3. CCl₄ and CH₃OH
4. C₆H₁₄ and C₅H₁₂

Q-4: What is the molecule mass of a compound with a concentration ,W= 1.2g/L, an osmotic pressure equal to 0.20 atm, and a temperature of T=300 K?

Q-5: Which law states that “the mole fraction of solute equals the relative lowering in vapour pressure” for a dilute solution?

1. Dalton’s law
2. Raoult’s law
3. Henry’s law
4. Avogados law

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