Question

How many oxygen atoms will be present in $88\, g$ of $CO_2$

• A. $24.08\times10^{23}$
• B. $6.023\times10^{23}$
• C. $44\times10^{23}$
• D. $22\times10^{24}$

Answer 1

Answer: (A)

$24.08\times10^{23}$

Answer 2

$1\space mol\space of\space CO_2 = 12+32 = 44 g$
Number of oxygen in $1\space molecule\space of\space CO_2 = 2$
This means than 44g of $CO_2$ contains 2 mol oxygen
Now, 88 g of $CO_2$ will contain $= \frac{2}{44} \times 88 =$ 4 mol of oxygen
Now, 1 molecule $= 6.033 \times 10^{23}\space atoms.$
$4\space molecules = 4 \times 6.033 \times 10^{23}\space atoms$
$= 24.08 \times 10^{23}\space atoms$