Question

A sample of calcium carbonate has the following percentage composition: 
\(\mathrm{Ca}=40%,\)\(\mathrm{C}=12%,\) \(\mathrm{O}=48%. A\).According to the law of definite proportion, the weight of calcium in \(4\text{g}\) of a sample of calcium carbonate from another source will be \(\text{ (at. no. Ca = 40, C = 12, O = 16)}\). 

• A. \(1.6 \times 10^{-2}\,\text{g}\)
• B. \(1.6\,\text{g}\)
• C. \(0.1\,\text{g}\)
• D. \(0.2\,\text{g}\)

Hint:

In numerical problems based on definite proportion, always use the given percentage composition directly.

Answer 1

The Correct Option is B

Step 1: Apply the law of definite proportion.
The law states that a pure compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source.
Step 2: Use the given percentage of calcium.
Calcium constitutes \(40%\) of calcium carbonate by mass.
Step 3: Calculate the mass of calcium in 4 g of sample.
\[ \text{Mass of Ca} = \frac{40}{100} \times 4 = 1.6\,\text{g} \]
Step 4: Conclusion.
Hence, the mass of calcium present is \(1.6\,\text{g}\).