Question

What is the weight of CO having the same number of oxygen atoms as much oxygen present in 88 g of ${\text{CO}}_2$?

• A. 112 g
• B. 56 g
• C. 84 g
• D. 140 g

Answer 1

The correct option is A

112 g

Molecular mass of ${\text{CO}}_2$ = 44 g
88 g of ${\text{CO}}_2$ = 2 moles of ${\text{CO}}_2$

Also, 1 mole of ${\text{CO}}_2$ $\to$ $6.022\times {10}^{23}$ molecules of ${\text{CO}}_2$.
1 molecule of ${\text{CO}}_2$ have 2 oxygen atoms.
So, 88 g (2 moles) of $C{O}_2$ $\to$ $2\times 2\times 6.022\times {10}^{23}$ atoms of oxygen.

1 mole of CO $\to$ 28 g
1 molecule of CO have 1 oxygen atom.
$\Rightarrow$ $6.022\times {10}^{23}$ oxygen atoms present in 28 g CO.
$\therefore$ $4\times 6.022\times {10}^{23}$ oxygen atoms present in 28 $\times$ 4 = 112 g of CO
So, 112 g of CO will have the same number of oxygen present as in 88 g of ${\text{CO}}_2$.