The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with an equimolar solution of benzene and toluene is -

1. 0.50 2. 0.6
3. 0.27 4. 0.73

Subtopic:  Dalton’s Law of Partial Pressure |
 64%
Level 2: 60%+
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The vapour pressure of a solvent decreases by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be if the decrease in vapour pressure is to be 20 mm of mercury then the mole fraction of the solvent is [CBSE PMT 1998]

(1) 0.8

(2) 0.6

(3) 0.4

(4) None

Subtopic:  Concentration Terms & Henry's Law |
 54%
Level 3: 35%-60%
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Find the molar concentration (in mol L⁻¹) of a solution whose osmotic pressure is 0.0821 atm at 300 K.

(1) 0.033

(2) 0.066

(3) 0.33 × 10–2

(4) 3.0

Subtopic:  Osmosis & Osmotic Pressure |
 85%
Level 1: 80%+
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The osmotic pressure of 5 % (mass/volume) solution of cane sugar at 150 °C (mol. mass of sugar = 342 g/mole) is:

1. 4 atm 2. 5.07 atm
3. 3.55 atm 4. 2.45 atm
Subtopic:  Osmosis & Osmotic Pressure |
 79%
Level 2: 60%+
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A solution containing 3.3 g of a substance in 125 g of benzene (b.p. 80°C) boils at 80.66°C. If Kb for one litre of benzene is 3.28°C, the molecular weight of the substance shall be 

1. 127.20

2. 131.20

3. 137.12

4. 142.72

Subtopic:  Elevation of Boiling Point |
 77%
Level 2: 60%+
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Calculate the boiling point of a solution prepared by dissolving 0.1 mole of sugar in 200 g of water at 1 atm pressure.
(Given: Molal boiling point elevation constant of water, Kb = 0.513 °C kg mol⁻¹.)

1.  100.513°C

2. 100.0513°C

3. 100.256°C

4.  101.025°C

Subtopic:  Elevation of Boiling Point |
 78%
Level 2: 60%+
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An aqueous solution containing 1 g of urea boils at 100.25 °C. The aqueous solution containing 3 g of glucose in the same volume will boil at:
1. 100.75 °C
2. 100.5 °C
3. 100 °C
4. 100.25 °C

Subtopic:  Osmosis & Osmotic Pressure |
 75%
Level 2: 60%+
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Solution of sucrose (Mol. Mass = 342) is prepared by dissolving 34.2 gm of it in 1000 gm of water. Freezing point of the solution is : (Kf for water is 1.86 K kg mol–1

1.  272.814 K

2.  278.1 K

3.  273.15 K

4.  270 K

Subtopic:  Depression of Freezing Point |
 76%
Level 2: 60%+
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Find the molecular mass of a weak monobasic acid when 0.1 g of the acid is dissolved in 21.7 g of water, and the solution freezes at 272.817 K.
Given: Kf for water = 1.86 K kg mol⁻¹.

1.  45.0

2.  40.6

3.  49.8

4. 46.8

Subtopic:  Depression of Freezing Point |
Level 3: 35%-60%
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1. 0.01

2. 1.00

3. 0.001

4. 100

Subtopic:  Depression of Freezing Point |
 89%
Level 1: 80%+
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