Question

A sample of calcium carbonate $\left(CaC{O}_3\right)$ 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 $4\text{g}$ of a sample of calcium carbonate from another source will be

• A. $0.016\text{g}$
• B. $0.16\text{g}$
• C. $1.6\text{g}$
• D. $16\text{g}$

Answer 1

The correct option is C $1.6\text{g}$

$\Rightarrow 100\text{g}$ of $CaC{O}_3$ produces $40\text{g}$ of $Ca$
$\Rightarrow 1\text{g}$ of $CaC{O}_3$ will produce $\frac{40}{100}\text{g}$ of $Ca$
$\Rightarrow$Weight of $Ca$ in $4\text{g}$ of a sample of calcium carbonate from another source will be $=\frac{40}{100}\times 4=1.6\text{g}$