MHT CET · Chemistry · Some Basic Concepts of Chemistry
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. \(\mathrm{Ca}=40, \mathrm{C}=40, \mathrm{C}=12,0=16\) )
- A \(1.6 \times 10^{-2} \mathrm{~g}\)
- B \(1.6 \mathrm{~g}\)
- C \(0.1 \mathrm{~g}\)
- D \(0.2 \mathrm{~g}\)
Answer & Solution
Correct Answer
(B) \(1.6 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
(A)
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.
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.
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