Given
Ca = 40%, C = 12% and O = 48%
The law of constant proportions holds true
In 100 g CaCO3, 40 g Ca is present
In 4 g CaCO3, let x g Ca is present
⇒ x/4 = 40/100
x = (40 x 4)/100
x = 16/10
We get,
x = 1.6 g
Therefore, the correct option is (c)