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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. $\mathrm{Ca}=40, \mathrm{C}=40, \mathrm{C}=12,0=16$ )
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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. $\mathrm{Ca}=40, \mathrm{C}=40, \mathrm{C}=12,0=16$ )
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The correct answer is:
$1.6 \mathrm{~g}$
(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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