The vapour pressure of 1 molal solution of a non-volatile solute in water at 300 K would be:

(The vapour pressure of water at 300 K = 12.3 kPa)

1. 21.08 kPa 2. 12.08 kPa
3. 33.08 kPa 4. 4.08 kPa

Subtopic:  Relative Lowering of Vapour Pressure |
 77%
Level 2: 60%+
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A solution containing 30 g of non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K.

The molar mass of the solute will be :

1.  23 g mol−1

2.  34 g mol−1

3.  15 g mol−1

4.  46 g mol−1

Subtopic:  Relative Lowering of Vapour Pressure |
Level 3: 35%-60%
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Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The atomic masses of A and B are respectively:

(Kf for benzene is 5.1 K kg mol-1)

1. 15.59 u and 52.64 u

2. 25.59 u and 42.64 u

3. 13.59 u and 52.64 u

4. 23.59 u and 32.64 u

Subtopic:  Depression of Freezing Point |
 64%
Level 2: 60%+
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The type of inter-molecular interactions present in:

(a) n-Hexane and n-octane (i) Van der Waal’s forces of attraction
(b) NaClO4 and water (ii) Ion-dipole interaction
(iii) Dipole-dipole interaction
 
(a) (b)
1. (i) (ii)
2. (ii) (ii)
3. (i) (iii)
4. (iii) (iii)
Subtopic:  Azeotrope |
 66%
Level 2: 60%+
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The insoluble compound in water is /are:

1. Phenol

2. Formic acid and toluene

3. Phenol and toluene 

4. Toluene and chloroform

Subtopic:  Introduction & Colligative properties |
 61%
Level 2: 60%+
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The mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN will be:
1. 1.424%
2. 4.424%
3. 5.124%
4. 2.124%

Subtopic:  Concentration Terms & Henry's Law |
 85%
Level 1: 80%+
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Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphene generally given is 1.5 mg.

Calculate the mass of 1.5 × 10−3m aqueous solution required for the above dose.
1. 13.22 g
2. 3.22 g
3. 11.22 g
4. 9.22 g

Subtopic:  Concentration Terms & Henry's Law |
 77%
Level 2: 60%+
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The amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol is:

1. 4.57 g 2. 3.57 g
3. 1.57 g 4. 12.57 g
Subtopic:  Concentration Terms & Henry's Law |
 81%
Level 1: 80%+
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Find the correct order of depression in freezing point (ΔTf) of water for equal amounts of acetic acid, trichloroacetic acid, and trifluoroacetic acid.

1. Acetic acid < Trichloroacetic acid < Trifluoroacetic acid

2. Acetic acid > Trichloroacetic acid > Trifluoroacetic acid

3.  Acetic acid < Trichloroacetic acid > Trifluoroacetic acid

4. Acetic acid > Trichloroacetic acid < Trifluoroacetic acid

Subtopic:  Depression of Freezing Point |
 60%
Level 2: 60%+
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The depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water will be:

 (Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1)

1. 0.32 K

2. 2.87 K

3. 0.65 K

4. 5.03 K

Subtopic:  Depression of Freezing Point |
 67%
Level 2: 60%+
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