Question

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

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

Answer 1

The correct option is C

$1.6\text{g}$

Explanation of the correct option

(C)$1.6\text{g}$

Law of constant proportions

• The Law of constant proportion is given by chemist Joseph Proust.
• In this law, elements combine in simple ratios of natural numbers.
• "In a chemical substance, the elements are always present in definite proportions by mass”.
• The proportional law still holds valid.
• $40\mathrm{g}\mathrm{Ca}$ is found in $100\mathrm{g}{\mathrm{CaCO}}_3.$
• Let $\mathrm{x}\mathrm{g}\mathrm{Ca}$ be present in $4\mathrm{g}{\mathrm{CaCO}}_{3.}$

$$\begin{gathered} \frac{\mathrm{x}}{4\mathrm{g}}=\frac{40\mathrm{g}}{100\mathrm{g}} \\ \mathrm{x}=\frac{40}{100}\times 4 \\ =1.6\mathrm{g} \end{gathered}$$

So, the weight of Calcium is $1.6\text{g}$.

Explanation for incorrect options

The weight of Calcium is $1.6\text{g}$. Hence, the options (A), (B), and (D) are incorrect.

Therefore, the correct option is (C). The weight of Calcium in $4\mathrm{g}$ of a sample of calcium carbonate from another source will be $1.6\text{g}$.