A sample of CaCO3 has Ca = 40%, C = 12% and O=48%. If the law of constant 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

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) 

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