Question

$4.4$ gm of $C{O}_2$ and $2.24$ litres of $H_2$ at STP (1 atm, 273 K)are mixed in a container. The total number of molecules present in the container will be:

• A. $6.023\times {10}^{23}$
• B. $1.2046\times {10}^{23}$
• C. $6.023\times {10}^{22}$
• D. $6.023\times {10}^{24}$

Answer 1

The correct option is B $1.2046\times {10}^{23}$

Total number of molecules = No. of molecules of $C{O}_2$ + No. of molecules of $H_2$.
Let us assume 'n' represent moles and 'N' represent molecules.
$n_{C{O}_2}$ = $\frac{4.4}{44}$ = $0.1$ mol
$n_{H_2}$ = $\frac{2.24}{22.4}$ = $0.1$ mol

$N_{C{O}_2}$ = $n_{C{O}_2}\times N_A$ = $0.1\times N_A$ molecules
$N_{H_2}$ = $n_{H_2}\times N_A$ = $0.1\times N_A$ molecules

$\therefore$ Total no of molecules = $0.1\times N_A+0.1\times N_A$ = $0.2\times N_A$ where $N_A$ is Avogadro's No. = $6.023\times {10}^{23}$
Total no of molecules = $1.2046\times {10}^{23}$ molecules