Unit 5: Solutions — Online MCQ Test
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Practice Unit 5: Solutions with a free chapter-wise online MCQ test.
This chapter covers: Different methods for expressing the concentration of solution - molality molarity mole fraction percentage (by volume and mass both) the vapour pressure of solutions and Raoult's....
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Unit 5: Solutions — Important Questions & Answers
The mixture which shows positive deviation from Raoult’s law is:
- A. Benzene + Toluene
- B. Acetone + Chloroform
- C. Chloroethane + Bromoethane
- D. Ethanol + Acetone
Answer: D. Ethanol + Acetone
Ethanol + Acetone shows positive deviation because the interaction between the components is weaker than the pure component interactions, leading to increased vapor pressure.
Ethanol + Acetone shows positive deviation because the interaction between the components is weaker than the pure component interactions, leading to increased vapor pressure.
The mixture which shows positive deviation from Raoult’s law is:
- A. Benzene + Toluene
- B. Acetone + Chloroform
- C. Chloroethane + Bromoethane
- D. Ethanol + Acetone
Answer: D. Ethanol + Acetone
Ethanol + Acetone is an example of a solution showing positive deviation from Raoult's law as the hydrogen bonds between ethanol molecules are weakened by the addition of acetone.
Ethanol + Acetone is an example of a solution showing positive deviation from Raoult's law as the hydrogen bonds between ethanol molecules are weakened by the addition of acetone.
Match the List-I (Example) with List-II (Type of Solution). List-I: (A) Humidity, (B) Alloys, (C) Amalgams, (D) Smoke. List-II: (I) Solid in solid, (II) Liquid in gas, (III) Solid in gas, (IV) Liquid in solid.
- A. A-II, B-IV, C-I, D-III
- B. A-II, B-I, C-IV, D-III
- C. A-III, B-I, C-IV, D-II
- D. A-III, B-II, C-I, D-IV
Answer: B. A-II, B-I, C-IV, D-III
Humidity is liquid in gas (A-II). Alloys are solid in solid (B-I). Amalgams are liquid in solid (C-IV). Smoke is solid in gas (D-III).
Humidity is liquid in gas (A-II). Alloys are solid in solid (B-I). Amalgams are liquid in solid (C-IV). Smoke is solid in gas (D-III).
Which of the following aqueous solution will exhibit highest boiling point?
- A. 0.01 M Urea
- B. 0.01 M KNO3
- C. 0.01 M Na2SO4
- D. 0.015 M C6H12O6
Answer: C. 0.01 M Na2SO4
Boiling point elevation depends on the Van't Hoff factor (i) and concentration. i*c: (1*0.01)=0.01, (2*0.01)=0.02, (3*0.01)=0.03, (1*0.015)=0.015. 0.01M Na2SO4 gives the highest value.
Boiling point elevation depends on the Van't Hoff factor (i) and concentration. i*c: (1*0.01)=0.01, (2*0.01)=0.02, (3*0.01)=0.03, (1*0.015)=0.015. 0.01M Na2SO4 gives the highest value.
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
- A. The solution shows positive deviation.
- B. The solution shows negative deviation.
- C. The solution is ideal.
- D. The solution has volume greater than the sum of individual volumes.
Answer: B. The solution shows negative deviation.
Mole fractions: X = 5/(5+10) = 1/3, Y = 10/(5+10) = 2/3. Ideal P = 63(1/3) + 78(2/3) = 21 + 52 = 73 torr. Since observed 70 torr < 73 torr, it shows negative deviation.
Mole fractions: X = 5/(5+10) = 1/3, Y = 10/(5+10) = 2/3. Ideal P = 63(1/3) + 78(2/3) = 21 + 52 = 73 torr. Since observed 70 torr < 73 torr, it shows negative deviation.
At 100ºC the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:
- A. 100ºC
- B. 102ºC
- C. 103ºC
- D. 101ºC
Answer: D. 101ºC
Using Raoult's Law (P0 - Ps)/P0 = n2/(n1 + n2) ≈ n2/n1, molality m is calculated as 2.125. Then ΔTb = Kb * m = 0.52 * 2.125 ≈ 1.10. Boiling point = 100 + 1.10 = 101.1ºC, closest to 101ºC.
Using Raoult's Law (P0 - Ps)/P0 = n2/(n1 + n2) ≈ n2/n1, molality m is calculated as 2.125. Then ΔTb = Kb * m = 0.52 * 2.125 ≈ 1.10. Boiling point = 100 + 1.10 = 101.1ºC, closest to 101ºC.
Fog is a colloidal solution of:
- A. Gas in liquid
- B. Solid in gas
- C. Gas in gas
- D. Liquid in gas
Answer: D. Liquid in gas
Fog is an aerosol consisting of liquid droplets of water dispersed in a gas (air).
Fog is an aerosol consisting of liquid droplets of water dispersed in a gas (air).
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25ºC. (Given, Vapour Pressure Data at 25ºC, benzene = 12.8 kPa, toluene = 3.85 kPa)
- A. The vapour will contain a higher percentage of toluene
- B. The vapour will contain equal amounts of benzene and toluene
- C. Not enough information is given to make a prediction
- D. The vapour will contain a higher percentage of benzene
Answer: D. The vapour will contain a higher percentage of benzene
Since the vapour pressure of benzene is higher than that of toluene at 25ºC, the vapour phase will be richer in the more volatile component (benzene) for an ideal mixture.
Since the vapour pressure of benzene is higher than that of toluene at 25ºC, the vapour phase will be richer in the more volatile component (benzene) for an ideal mixture.
The Henry’s law constant (K_H) values of three gases A, B and C in water are 145, 2 × 10^-5 and 35 kbar, respectively. The solubility of these gases in water follow the order:
- A. A > C > B
- B. A > B > C
- C. B > A > C
- D. B > C > A
Answer: B. A > B > C
By Henry’s law, solubility is inversely proportional to K_H. Thus the gas with the smallest K_H is most soluble: B > C > A.
By Henry’s law, solubility is inversely proportional to K_H. Thus the gas with the smallest K_H is most soluble: B > C > A.
The freezing point depression constant (Kf) of benzene is 5.12 K kg mol⁻¹. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places):
- A. 0.40 K
- B. 0.60 K
- C. 0.20 K
- D. 0.80 K
Answer: A. 0.40 K
ΔTf = Kf × m = 5.12 × 0.078 = 0.3994 K ≈ 0.40 K.
ΔTf = Kf × m = 5.12 × 0.078 = 0.3994 K ≈ 0.40 K.