Question

Calculate the number of atoms of oxygen present in 120g of nitric acid.

Answer 1

Here, first we will calculate the number of moles of $HN{O}_3$ that are present in 120 g of $HN{O}_3$. The molar of $HN{O}_3$ is 63g. Therefore
$63gofHN{O}_3=1$ mole of $HN{O}_3$
1g of $HN{O}_3=\left(\frac{1}{63}\right)$ moles of $HN{O}_3$
Therefore 120g of $HN{O}_3=\left(\frac{120}{63}\right)$ moles of $HN{O}_3$
= 1.905 moles of $HN{O}_3$
Now 1 mole of a substance contains $6.023\times {10}^{23}$ molecules of the substance. Therefore
1 mole of $HN{O}_3=6.023\times {10}^{23}$ molecules of $HN{O}_3$
So 1.905 moles of $HN{O}_3=1.905\times 6.023{10}^{23}$ molecules of $HN{O}_3$
$=1.147\times {10}^{24}$ molecules of $HN{O}_3$
Now 1 molecule of $HN{O}_3$ contains 3 atoms of oxygen. Therefore $1.147\times {10}^{24}$ molecules of $HN{O}_3$ will contain
$=3\times 1.147\times {10}^{24}$ atoms of oxygen
$=3.442\times {10}^{24}$ atoms of oxygen