A sample of calcium carbonate $(C a C O_{3})$ has the following percentage composition:
Ca = 40%; C = 12%; O = 48%.
If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be ?

0.016 g

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0.16 g

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1.6 g

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16 g

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Solution

The correct option is C 1.6 g
Calcium carbonate from any source contains 40% calcium.
So each 1 g of calcium carbonate contains 0.4g of calcium.
Thus, for 4 g of calcium carbonate, the amount of calcium $= 4 \times 0.4 = 1.6 g$

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