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Q.

A sample of calcium carbonate (CaCO3) has the following percentage composition : Ca = 40%, C : 12%, O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be :

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a

0.016 g

b

0.16 g

c

1.6 g

d

16 g

answer is C.

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Detailed Solution

Molecular weight of CaCO3  = 100 gm 100 gm of sample contains = 40 gm of Ca4 gm of sample contains = ?= 4×40100 = 1.6 gm
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