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A sample of calcuium carbonate (CaCO(3))...

A sample of calcuium carbonate `(CaCO_(3))` 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

A

0.016g

B

0.16g

C

1.6g

D

16g

Text Solution

Verified by Experts

The correct Answer is:
C
In `CaCO_(3)` % of `Ca = 40%`
According to law of constant proportion % composition will remain same.
`therefore ` Weight of `Ca = (4 xx 40)/(100) = 1.6 g`
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