0.5 g of fuming H2SO4 (oleum) is diluted  with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:

1. 30.6% 

2. 40.6%

3. 20.6%

4. 50%

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One g of a mixture of Na2CO3 and NaHCO3 consumes y equivalent of HCl for complete neutralisation. One g of the mixture is strongly heated, then cooled and the residue treated with HCl. How many equivalent of HCl would be required for complete neutralisation?

1. 2y equivalent

2. y equivalent

3.  3y/4 equivalent

4. 3y/2 equivalent

Subtopic:  Equivalent Weight |
 54%
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The chloride of a metal contains 71% chlorine by mass and the vapour density of it is 50. The atomic mass of the metal will be:

1. 29

2. 58

3. 35.5

4. 71

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 53%
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The equivalent mass of Zn(OH)2 in the following reaction is equal to its,

Zn(OH)2 + HNO3 Zn(OH)(NO3) + H2O :

1. Formula mass/1

2. Formula mass/2

3. 2 x formula mass

4. 3 x formula mass

Subtopic:  Equivalent Weight |
 60%
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What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 2:1 by volume?

1. 0.4 N

2. 0.5 N

3. 1.05 N

4. 0.15 N

Subtopic:  Equivalent Weight |
 74%
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0.7 g of Na2CO3.xH2O were dissolved in water and the volume was made to 100mL, 20mL of this solution required 19.8 mL of N/10 HCl for complete neutralization. The value of x is:

1. 7

2. 3

3. 2

4. 5

Subtopic:  Equivalent Weight |
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A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 3.15 g of the oxide have yielded 1.05g of the metal. We may deduce that:

1. the atomic mass of the metal is 8

2. the atomic mass of the metal is 4

3. the equivalent mass of the metal is 4

4. the equivalent mass of the metal is 8

Subtopic:  Equivalent Weight |
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An oxide of metal has 20% oxygen, the equivalent mass of oxide is:

1. 32 

2. 40

3. 48

4. 52

Subtopic:  Equivalent Weight |
 56%
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How much water is to be added to dilute 10 mL of 10N HCl to make it decinormal?

1. 990mL

2. 1010 mL

3. 100mL

4. 1000mL

Subtopic:  Equivalent Weight |
 57%
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W1 g of an element combines with oxygen forming W2 g of its oxide. The equivalent mass of the element is:

1. [W1 / W2]x8

2. [W1 / W2-W1]x8

3. [W2-W1/W1 ]x8

4. [W1 / W1-W2]x8

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 65%
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