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Question

A mineral consists of an equimolar mixture of carbonates of two bivalent metals(MCO3). One metal is present to the extent of 13.2 % by weight. 2.58 g of mineral on heating lost 1.233 g of CO2. Calculate the percentage by weight of the other metal.

A
65.12 %
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B
21.68 %
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C
13.20 %
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D
42.80 %
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Solution

The correct option is B 21.68 %
Let us consider MCO3 and MCO3 be two carbonates with respect to M and M bivalent metals respectively.

On heating
(a) MCO3MO+CO2 (g).........(i)
(b) MCO3MO+CO2 (g) .......(ii)

1 mole of CO2 (g) i.e. 44 g is produced from 1 mole of carbonate (CO23) ion i.e. 60 g (using equation (i) and (ii))

1.233 g of CO2 is produced from 60×1.23344=1.68 g of carbonate (CO23) ion.

Percentage of (CO23) ion = Mass of CO23ionTotal mass×100

= 1.68 g2.58 g×100

= 65.12%

Percentage of one of the metals = 13.2
Percentage of the other metal
= 100 - percentage of CO23 ion - percentage of one metal
= 100 - 65.12 - 13.2
= 21.68%

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