A sample of calcium carbonate (CaCO3) 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

Question

A sample of calcium carbonate (CaCO3) 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 (A) 0.016 g (B) 0.16 g (C) 1.6 g (D) 16 g

Tardigrade Admin · Accepted Answer

In 100 g CaCO 3, 40 g Ca is present. In 4 g CaCO 3, Let x g Ca be present ( x /4)=(40/100) ⇒ x =1.6 g

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

$0.016 g$

$0.16 g$

$1.6 g$

$16 g$

In $100 g C a C O_{3}, 40 g C a$ is present.
In $4 g C a C O_{3}$ , Let $x g C a$ be present
$\frac{x}{4} = \frac{40}{100}$
$\Rightarrow x = 1.6 g$

Q. A sample of calcium carbonate (CaCO3​) 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 4g of a sample of calcium carbonate obtained from another source will be

0.016 g

0.16 g

1.6 g

16 g