Question

$4.4g$ of $C{O_2}$ and $2.24$ litre of ${H_2}$ at STP are mixed in a container .The total number of molecules present in the container will be :
A . $6.022 \times {10^{23}}$
B . $1.2044 \times {10^{23}}$
C . $6.022 \times {10^{26}}$
D . $6.022 \times {10^{24}}$

Answer 1

Hint: We know the formula of number of moles that is Number of moles of carbon dioxide=weight of carbon dioxide in gram/molecular weight of carbon dioxide .When volume of gas is given at STP then we can calculate number of moles of molecule with the help of formula ; Number of moles of molecule =Volume of gas at STP (in litre)/$22.4$ (in litre).

Complete answer:

It is given in the problem that weight of $C{O_2}$ in gram $= 4.4g$,molecular weight of $C{O_2}$=atomic weight of carbon+2×atomic weight of oxygen $= 12 + 2 \times 16 = 44$. So here in the given problem the number of moles of carbon dioxide $= \dfrac{{4.4}}{{44}} = 0.1$ mole. Now we will calculate the number of moles of the Hydrogen molecule.
$2.24$ litre volume of ${H_2}$ is present at STP ,we know that molar volume of ${H_2}$at STP is $22.4$litre.
Hence the number of moles of hydrogen molecule $= \dfrac{{2.24}}{{22.4}} = 0.1$ mole.
Total number of moles present in the mixture $= 0.1 + 0.1 = 0.2$ mole.
We know that one mole occupies ${N_A}$ (Avogadro’s Number) molecules .So here the total number of molecules present in the container $= 0.2 \times {N_A} = 0.2 \times 6.022 \times {10^{23}}$
$= 1.2044 \times {10^{23}}$
Hence the option B is correct, that is $1.2044 \times {10^{23}}$ .

Note : We have approached this problem by calculating the total number of moles in the container. Once we get total number of moles present in the mixture we have multiplied it with Avogadro’s number which is equal to $6.022 \times {10^{23}}$ .As we know that one mole is equal to Avogadro’s number at STP. When $4.4g$ of $C{O_2}$ and $2.24$ litre of ${H_2}$ at STP are mixed in a container then the total number of molecules present in the container will be $1.2044 \times {10^{23}}$ .