Question

How many oxygen atoms will be present in $88g$ of $C{O}_2$

• A. $24.08\times 10^{23}$
• B. $6.023\times 10^{23}$
• C. $44\times 10^{23}$
• D. $22\times 10^{24}$

Answer 1

Answer: (A)

$1$ mole of $C{O}_2=44g$
$88g$ of $C{O}_2=2$ moles
No. of oxygen atoms in $2$ moles $=2\times 2\times N_A$
$=2\times 2\times 6.023\times 10^{23}$
$=24.08\times 10^{23}$