Calculate number of electrons present in 9.5 g of PO43-:

1.  6

2.  5 NA

3.  0.1 NA

4.  4.7 NA

Subtopic:  Moles, Atoms & Electrons |
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The total no. of neutrons present in 54 mL H2Ol are:

1.  3 NA

2.  30 NA

3.  24 NA

4.  none of these

Subtopic:  Moles, Atoms & Electrons |
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The volume of a drop of water is 0.0018 mL then the number of water molecules present in two drop of water at room temperature is:

1.  12.046×1019

2.  1.084×1018

3.  4.84×1017

4.  6.023×1023

Subtopic:  Moles, Atoms & Electrons |
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Which of the following combinations illustrates the law of reciprocal proportions?

1.  N2O3, N2O4, N2O5

2.  NaCl, NaBr, Nal

3.  CS2, CO2, SO2

4.  PH3, P2O3, P2O5

Subtopic:  Introduction |
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Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide in which the ratio of the weights of carbon and oxygen is respectively      12 : 16 and 12 : 32. These figures illustrate the:

1.  Law of multiple proportions

2.  Law of reciprocal proportions

3.  Law of conservation of mass

4.  Law of constant proportions

Subtopic:  Introduction |
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The formula of an acid is HXO2. The mass of 0.0242 moles of the acid is 1.657 g.
The atomic weight of X is:

1. 35.5 2. 28.1
3. 128 4. 19.0
Subtopic:  Moles, Atoms & Electrons |
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A 6.85 g sample of the hydrates SrOH2.xH2O is dried in an oven to give 3.13 g of anhydrous SrOH2. The value of x is - 

(Atomic weights : Sr=87.60, O=16.0, H=1.0)

1.  8

2.  12

3.  10

4.  6

Subtopic:  Moles, Atoms & Electrons | Equation Based Problem |
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The sulphate of the metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is:

1.  40.3

2.  36.3

3.  24.3

4.  11.3

Subtopic:  Equation Based Problem |
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Sulphur burns according to the reaction

 18S8s+O2g  SO2g
What volume of air, at 1 atm and 273 K, containing 21%  oxygen by volume is required to completely burn sulphur S8 present in 200 g of the sample?
(This sample contains 20% inert material which does not burn)

1.  23.52 litre

2.  320 litre

3.  112 litre

4.  533.33 litre

Subtopic:  Equation Based Problem |
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Phosphoric acid H3PO4 prepared in a two step process.

(1) P4+5O2  P4O10
(2) P4O10+6H2O  4H3PO4

We allow 62g of phosphorus to react with react with excess oxygen which form P4O10 in 85% yield. In the step (2) reaction 90% yield of H3PO4 is obtained. Produced mass of H3PO4 is:

1.  37.485 g

2.  149.949 g

3.  125.47 g

4.  564.48 g

Subtopic:  Equation Based Problem |
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