A sample of calcium carbonate has the following percentage composition. $\mathrm{Ca}=40 \%, \mathrm{C}=12…
A sample of calcium carbonate has the following percentage composition. $\mathrm{Ca}=40 \%, \mathrm{C}=12 \%$ and $0=48 \%$
According to law of definite proportion the weight of calcium in $4 \mathrm{~g}$ of a sample of calcium carbonate from another source will be (at. no. $\mathrm{Ca}=40, \mathrm{C}=40, \mathrm{C}=12,0=16$ )
$1.6 \times 10^{-2} \mathrm{~g}$
$1.6 \mathrm{~g}$
$0.1 \mathrm{~g}$
$0.2 \mathrm{~g}$
Solution
In $100 \mathrm{~g}$ of $\mathrm{CaCO}_{3}, 40 \mathrm{~g} \mathrm{Ca}$ is present
$\therefore$ In $4 \mathrm{~g}$ of $\mathrm{CaCO}_{3}=\frac{4 \times 40}{100}=1.6 \mathrm{~g}$ of $\mathrm{Ca}$ is present.