Question

What is the number of oxygen atoms present in 126 g of ${HNO}_3$?

• A. $36.132\times {10}^{23}$
• B. $9.033\times {10}^{23}$
• C. $12.044\times {10}^{23}$
• D. $16.132\times {10}^{23}$

Answer 1

The correct option is A $36.132\times {10}^{23}$

Molecular mass of $HN{O}_3$ = atomic mass of hydrogen + atomic mass of nitrogen + 3$\times$ atomic mass of oxygen = 1 + 14 + (3 $\times$ 16) = 63 u.

Number of moles = $\frac{Givenmass}{Grammolecularmass}$

No. of moles in 126g of $HN{O}_3$ = $\frac{126}{63}$
So, 126g of $HN{O}_3$ contain 2 moles.

One mole of $HN{O}_3$ contains 3 atoms of oxygen. So, 2 moles of $HN{O}_3$ contains 6 moles of oxygen atoms.

Since one mole of a substance contains $6.022\times {10}^{23}$ particles, 6 moles of oxygen would contain $6\times 6.022\times {10}^{23}$ = $36.132\times {10}^{23}$ particles of oxygen.

So, 126g of $HN{O}_3$ would contain $36.132\times {10}^{23}$ particles of oxygen.