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20.0 g 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 ?

(A)

60

(B)

84

(C)

75

(D)

96

NEET - Medical UG-2015 Medium

Answer & Solution:

Answer: 84 %

The chemical reaction is \(MgCO_{3}\overset{heat}{\rightarrow}MgO+CO_{2}\)  

molecular weight of Magnesium carbonate = 84 g/mol

molecular weight of Magnesium Oxide =40 g/mol

Now  1 mol carbonate produces 1 mol of oxide ,

So , 20 g or 0.238 mol of Magnesium Carbonate must produce 0.238 mol of Oxide if it is 100 % pure 

Now , moles of Oxide produced = 0.2 mol 

So , percentage purity of Magnesium Carbonate is =\((0.2\div 0.238)\times 100\)  = 84 %

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