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

The correct option is A $24.08\times {10}^{23}$

1 mol of $C{O}_2=12+32=44g$

Number of oxygen in 1 molecule of $C{O}_2=2$

$\therefore$ 44 g $C{O}_2$ contains 2 mol oxygen

$88gC{O}_2$ contains $=\frac{2}{44}\times 88=4$ mol oxygen

as, $1mol=6.022\times {10}^{23}$ atoms

$4moloxygen=4\times 6.022\times {10}^{23}$ oxygen atoms

$=24.08\times {10}^{23}$ oxtgen atoms

Hence, option $A$ is correct.