At S.T.P. the density of CCl4 vapour in g/L will be nearest to [CBSE PMT 1988]

(1) 6.84

(2) 3.42

(3) 10.26

(4) 4.57

Subtopic:  Moles, Atoms & Electrons |
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4.4 g of an unknown gas occupies 2.24 litres of volume at NTP. The gas may be: 

1. Carbon dioxide 2. Carbon monoxide
3. Oxygen 4. Sulphur dioxide
Subtopic:  Moles, Atoms & Electrons |
 87%
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The number of gram molecules of oxygen in 6.02 × 1024 CO molecules are:

1. 10 g molecules

2. 5 g molecules

3. 1 g molecules

4. 0.5 g molecules

Subtopic:  Moles, Atoms & Electrons |
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The mass of carbon present in 0.5 mole of K4[Fe(CN)6] is:

1. 1.8 g

2. 18 g

3.  3.6 g

4.  36 g

Subtopic:  Moles, Atoms & Electrons |
 72%
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The number of moles of BaCO3 which contains 1.5 moles of oxygen atoms is:

1. 0.5

2. 1

3. 3

4. 6.02 ×1023

Subtopic:  Equation Based Problem |
 74%
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The oxide of metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal is

1. 84.7

2. 74.7

3. 64.7

4. 44.7

Subtopic:  Millimole/Equivalent Concept |
 55%
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The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre of this organic gas at STP  is exactly equal to that of one litre of N2 at STP. Therefore, the molecular formula of the organic gas is:

1. C2H4

2. C3H6

3. C6H12

4. C4H8

Subtopic:  Moles, Atoms & Electrons | Empirical & Molecular Formula |
 77%
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A sample of pure compound is found to have Na = 0.0887 mole, O = 0.132 mole, C = 2.65 × 1022 atoms.

The empirical formula of the compound is 

(1) Na2CO3

(2) Na3O2C5

(3) Na0.088700.132C2.65×1022

(4) NaCO

Subtopic:  Empirical & Molecular Formula |
 62%
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An organic compound containing C, H, and N gave the following on analysis: C = 40%, H = 13.3% and N = 46.67%. Its empirical formula would be:

1. CHN
2. C2H2N
3. CH4N
4. C2H7N

Subtopic:  Empirical & Molecular Formula |
 81%
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An organic substance containing C, H, and O gave the following percentage composition :

C = 40.687%, H = 5.085% and O = 54.228%. The vapour density of this organic substance is 59.

The molecular formula of the compound will be:

1. C4H6O4

2. C4H6O2

3. C4H4O2

4. None of the above

Subtopic:  Moles, Atoms & Electrons |
 72%
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