Question

A sample of $CaC{O}_3$ has Ca =40%, C = 12% and O = 48% by mass. If the law of constant proportion is true then the weight of Ca in 16g of $CaC{O}_3$ made from different processes will be -

• A. 64 g
• B. 0.64g
• C. 6.4g
• D. 4.6 g

Answer 1

Answer: (C)

6.4g

Molecular weight of $CaC{O}_3$ = 100g

Percentage of Ca = 40 %

Let, law of constant proportion is true.

So, mass of Ca in 100 g of $CaC{O}_3$ = $\frac{40}{100}\times 100=40g$

So, weight of Ca in 16g of $CaC{O}_3$ = $\frac{40}{100}\times 16g=6.4g$