Question

A sample of $CaC{O}_3$ has $Ca=40\%$, $C=12\%$ and $O=48\%$. if the law of constant proportions is true, then the weight of Calcium in $4g$ of a sample of calcium carbonate from another source will be:

• A. $16g$
• B. $1.6g$
• C. $0.016g$
• D. $0.16g$

Answer 1

The correct option is B $1.6g$

$Ca=40\%$, $C=12\%$ and $O=48\%$

The law of constant proportions holds true.

In $100gCaC{O}_3$, $40gCa$ is present.

In $4gCaC{O}_3$, Let $xgCa$ be present.

$\Rightarrow \frac{x}{4}=\frac{40}{100}$

$\Rightarrow x=\frac{40\times 4}{100}$

$\Rightarrow x=\frac{16}{10}$

We get,

$x=1.6g$

Therefore the correct option is (c)