$500 ml$ of $N{H}_3$ contains $6.00 x {10}^{23}$ molecules at $S.T.P.$ How many molecules are present in $100 ml$ of $C{O}_2$ at $S.T.P.$

 

$\frac{P{V}_1}{n_{1}RT}=\frac{P{V}_2}{n_{2}RT}$ or, $\frac{V_1}{n_1}=\frac{V_2}{n_2}$

or, $\frac{500}{1}=\frac{100}{n_2}\text{ or, }{n}_2=\frac{1}{5}$

[$6.0 x {10}^{23}$ molecules of $N{H}_3=1$mole] 

$1$ mole of $C{O}_2$ contains $6.0 x {10}^{23}$ molecules

$\therefore \frac{1}{5}$ mole of $C{O}_2$ contains $1.2 x {10}^{23}$ molecules.