Question

How many atoms of oxygen are present in 300 grams of ${CaCO}_3$?

• A. $54.207\times {10}^{23}$
• B. $18.06\times {10}^{23}$
• C. $36.12\times {10}^{23}$
• D. $12.04\times {10}^{23}$

Answer 1

Answer: (A)

$54.207\times {10}^{23}$

Mass of ${CaCO}_3$ = 40+12+48 = 100g
Moles of ${CaCO}_3$ = $\frac{300}{100}$ = 3
Number of molecules of ${CaCO}_3$ = $3\times 6.02\times {10}^{23}$
1 molecule of ${CaCO}_3$ contain 3 atoms of oxygen
$3\times 6.02\times {10}^{23}moleculewillcontain3\times 3\times 6.023\times {10}^{23}=54.027\times {10}^{23}$ atoms