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20.0g of a magnesium carbonate sample de...

20.0g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample (Mg=24)-

A

75

B

96

C

60

D

84

Text Solution

Verified by Experts

The correct Answer is:
D
`MgCO_(3)(s) to MgO(s)+CO_(2)(g)` . . [1]
Moles of `MgCO_(3)=("weight in gram")/("molecular mass")=(20)/(84)=0.238` mol
From equation [1], 1 mol of `MgCO_(3)=1` mol MgO
`therefore0.238` ol `MgCO_(3)=0.238` mol MgO
`=0.238xx40g=9.52g" "MgO`
Now,, practical yield of MgO=8g
% purity`=(8)/(9.52)xx100=84%`
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