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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 obtained from another source will be

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0.016 g

0.16 g

1.6 g

16 g

answer is C.

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Detailed Solution

Law of constant proportions states that “ A sample of pure compound always contains same elements combined together in the same proportions by mass ”

Given 100g of ${CaCO}_{3}$ contains 40g of ‘Ca' or

$\frac{100}{25} =$ 4g of Calcium carbonate obtained from other source contains $\frac{40}{25} = 1.6$ g of Calcium

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