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Question

In the decomposition of 10 g of MgCO3, 0.1 mole CO2 and 4.0 g MgO are obtained. Hence, percentage purity of MgCO3 is:

A
50%
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B
60%
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C
40%
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D
84%
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Solution

The correct option is B 84%
Solution:- (D) 84%
Molecular weight of MgCO3=84g
Molecular weight of MgO=40g
Decomposition of MgCO3-
MgCO3MgO+CO2
Now, from the above reaction-
Weight of pure MgCO3 required to produce 40g of MgO=84.3g
Weight of pure MgCO3 required to produce 4g of MgO=84.340×4=8.43g
Given weight of MgCO3=10g
Now,
Amount of pure MgCO3 in 10g of given MgCO3=8.43g
Thus,
Amount of pure MgCO3 in 100g of given MgCO3=8.4310×100=84.3g
Therefore,
The percentage purity of given MgCO3=84.3%84%

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