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Laws of Definite Proportions, Multiple Proportions

A sample of c...

Question

A sample of calcium carbonate $(C a C O_{3})$ has the percentage composition as given: $C a = 40 %, C = 12 %, O = 48 %$ .

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 :

0.016

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0.16

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1.6

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16

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Solution

The correct option is C $1.6$ g
In $100$ g $C a C O_{3}$ , weight of $C a$ is $40$ g.

In $4$ g $C a C O_{3}$ , weight of $C a$ is $= \frac{40}{100} \times 4 = 1.6$ g

Hence, the correct option is $C$

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