Question

Number of atoms of oxygen present in $10.6g$ of $N{a}_{2}C{O}_3$ will be

• A. $6.02\times 10^{23}$
• B. $12.04\times 10^{22}$
• C. $1.806\times 10^{23}$
• D. $31.80\times 10^2$

Answer 1

Answer: (C)

Molecular mass of
$N{a}_{2}C{O}_3=2\times 23+12+3\times 16=106$
$\because 106gN{a}_{2}C{O}_3$ contains
$=3\times 6.023\times 10^{23}$ oxygen atoms
$\therefore 10.6g$ of $N{a}_{2}C{O}_3$ will contain
$=\frac{3\times 6.023\times 10^{23}}{106}\times 10.6$
$=18.069\times 10^{24}$
$=1.806\times 10^{23}$ oxygen atoms