Question

A sample of calcium carbonate (CaCO₃) has the following percentage composition : 

Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be ……………..

(A) 0.016 g 

(B) 0.16 g 

(C) 1.6 g 

(D) 16 g

Answer 1

(C) 1.6 g

Law of constant proportional, state that "the individual elements that constitute a chemical compound are always present in a fixed ratio (by mass)". This ratio does not depend on the source of the chemical compound.

\(\therefore \) weight of calcium in 4g of sample = \(4\times\frac{40}{100}\)

= \(1.6\) g

Answer 2

Option : (C) 1.6 g