Question

A sample of $CaC{O}_3$ has $Ca=40\%,C=12\%$ and $O=48\%$. 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. $64g$
• B. $0.64g$
• C. $6.4g$
• D. $4.6g$

Answer 1

Answer: (C)

$6.4g$

Gram Molecular Mass of $CaC{O}_3$
$=40+12+16\times 3=40+12+48=100gm$
$100gmCaC{O}_3$ contain $=40gCa$
$\therefore 16gCaC{O}_3$ will contain $=\frac{40}{100}\times 16=6.4gCa$