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# 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.

 Definition: Colligative properties are those that depend on the number of solute particles, regardless of their nature, in relation to the total number of particles in the solution.

## 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

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 (Kb(H2O) = 0.52 K kg/mol)

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

Explanation:

Given: Molality, m = 1m = 1mol/kg , Degree of dissociation,πͺ= 60% = 0.6

For dissociation,

Van’t Hoff factor, i = 1+(n-1)πͺ

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

A2B3 →2A3+ +3B2-

This makes n= 5

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

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

On solving,

ΔTb= 1.768 K

ΔTb= Tb-Tbo

Where, Tbo is boiling point of water = 373 K

Tb is the boiling point of solution at 1 atm

Substituting the values,

1.768 K= Tb-373K

Tb = 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

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

Solute

Van’t Hoff factor(i)

Na2SO3

3

C12H22O11

1

Al(NO3)3

4

MgCl2

3

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

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.

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?(Kf 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, ΔTf= Kf× 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

Explanation:

Cryoscopic constant Kf = ΔTf/ m

Unit of ΔTf = K

Unit of m(molality) is mol/Kg

This makes unit of Kf = 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. Kb
2. 0.2 Kb
3. 0.02Kb
4. 0.98Kb

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, ΔTb= Kb× m = 0.98Kb

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 Kb and the factors that influence it.

Answer: The molal elevation constant is denoted by Kb. 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

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

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

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