20g of naphthoic acid (C11H8O2) dissolved in 50g of benzene (Kf=1.72 K Kg mol-1) shows a depression in freezing point of 2K. The Vant Hoff factor is?

1. 0.5

2. 0.1

3. 2

4. 3

Subtopic:  Van’t Hoff Factor |
 76%
Level 2: 60%+
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A 0.002 molar solution of NaCl having degree of dissociation of 90% at 270C, has osmotic pressure equal to:-

1. 0.94 bar

2. 9.4 bar

3. 0.094 bar

4. 9.4 x 10-4 bar

Subtopic:  Osmosis & Osmotic Pressure |
 59%
Level 3: 35%-60%
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Determining the normality of a solution with a density of 0.6 g/ml and composed of 35% NH4OH by mass.

1. 4.8 N

2. 10 N

3. 0.5 N

4. 6 N

Subtopic:  Concentration Terms & Henry's Law |
 61%
Level 2: 60%+
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The relative lowering of vapour caused by dissolving 71.3 g of a substance in 1000 g of water is 7.13 x 10-3. The molecular mass of the substance is-

1. 180 g mol-1

2. 18 g mol-1

3. 1.8 g mol-1

4. 360 g mol-1

Subtopic:  Relative Lowering of Vapour Pressure |
 72%
Level 2: 60%+
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Total vapour pressure of a mixture of 1 mol A (pA°  = 150 torr) and 2 mol B (pB° = 240 torr) is 200 torr. In this case:-

1. There is a positive deviation from Raoult's law
2. There is a negative deviation from Raoult's law
3. There is no deviation from Raoult's law
4. None of these 

Subtopic:  Introduction & Colligative properties | Relative Lowering of Vapour Pressure |
 67%
Level 2: 60%+
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Which of the following aqueous solution should have the highest osmotic pressure?
(1) 0.011 M AICI
at 50°C
(2) 0.03 M NaCI at 25°C
(3) 0.012 M (NH
4)2SOat 25°C
(4) 0.03 M NaCI at 50°C

Subtopic:  Osmosis & Osmotic Pressure |
 57%
Level 3: 35%-60%
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Calculate the mass of ice that separates when 50 g of ethylene glycol is added to 200 g of water
 and the solution is cooled to −9.3°C.
Given:
Kf of water = 1.86 K kg mol⁻¹
Molar mass of ethylene glycol, C₂H₆O₂ = 62 g mol⁻¹


1. 42 mg
2. 42 g
3. 38.71 g
4. 38.71 mg

Subtopic:  Depression of Freezing Point |
 65%
Level 2: 60%+
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Calculate the freezing point of a pure solvent if 40 g of NaOH (i = 2) is dissolved in 1 L of the solvent and the vapour pressure of the resulting solution becomes equal to that of the solid solvent at 300 K.
Given:
Density of solvent = 1 g mL⁻¹
Kf = 1.8 K kg mol⁻¹
Molar mass of NaOH = 40 g mol⁻¹

1. 296.4 K

2. 303.6 K 

3. 297.5 K

4. 302.4 K

Subtopic:  Depression of Freezing Point |
 63%
Level 2: 60%+
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P1 and P2 are vapour pressure of solvent and solution at temperature T. The solution is prepared by dissolving non-volatile solute in the same solvent. P1 =760 torr, P2 =684 mm Hg, Mole fraction of solvent in the solution will be:-

1. 0.1

2. 0.2

3. 0.9

4. Can't be predicted

Subtopic:  Relative Lowering of Vapour Pressure |
 62%
Level 2: 60%+
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100 g solute is dissolved in 1400 g of solvent. Density of resultant solution is 1.5 g/mL. The ratio of its molarity and molality will be :-
(1) 1.5

(2) 1.3

(3) 1.4

(4) 1.2
 

Subtopic:  Concentration Terms & Henry's Law |
 66%
Level 2: 60%+
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