Question

A sample of $CaC{O}_3$ has Ca - 40%, C = 12% and O = 48%. If the law of constant proportions is true, then the mass of Ca in 5 g of $CaC{O}_3$ from another source will be:

• A. 2.0 g
• B. 0.2 g
• C. 0.02 g
• D. 20.0 g

Answer 1

The correct option is A 2.0 g

Since, mass percentage of Ca in $CaC{O}_3=40\text{\%}$ and law of constant proportion is true.
So, the mass percentage of $Ca$ in $5gm$ $CaC{O}_3$ will also be $40\text{\%}.$
Hence,$40\text{\% of 5gm =}$$5\times$$\frac{40}{100}$ $=2gm$
Hence, answer is option A.