Question

Which of the following contains the maximum number of atoms?
(A) $6.023 \times {10^{21}}$ molecules of $C{O_2}$
(B) $22.4L$ of $C{O_2}$at STP
(C) $0.44g$ of $C{O_2}$
(D) None of these

Answer 1

As we know that atoms are small elements which are the smallest units of matter. All forms of matter are constituted with and around matter. It is thus important to know an approximate number of atoms present which can be calculated with the formula N = no. of moles x no. of atoms in one molecule $xN_A$.

Complete step by step answer:
Let number of atoms = N
Avogadro number = $6.022 \times {10^{23}} = {N_A}$
Now, N = no. of moles x no. of atoms in one molecule $xN_A$
Considering, (A) $6.023 \times {10^{21}}$ molecules of $C{O_2}$:
We know that,
N = no. of moles x no. of atoms in one molecule x NA
No. of moles = given no. of molecules / Avogadro number (NA)
Let no. of moles = n
$\Rightarrow n = \dfrac{{6.023 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}}}mol$
Given no. of atoms in one molecule = $3$
$\Rightarrow N = \dfrac{{6.023 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}}} \times 3 \times {N_A}$
$\Rightarrow N = \dfrac{{6.023 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}}} \times 3 \times 6.022 \times {10^{23}}$
$\Rightarrow N = 6.023 \times {10^{21}} \times 3$
$\Rightarrow N = 18.069 \times {10^{21}}$
(B) $22.4L$ of $C{O_2}$at STP:
We know that
N = no. of moles x no. of atoms in one molecule $xN_A$
No. of moles = given volume in L / 22.4
Let no. of moles = n
$\Rightarrow n = \dfrac{{22.4}}{ 22.4 \\\ \\\ }mol$
$\Rightarrow n = 1mol$
Given no. of atoms in one molecule = $3$
$\Rightarrow N = 1 \times 3 \times {N_A}$
$\Rightarrow N = 1 \times 3 \times 6.022 \times {10^{23}}$
$\Rightarrow N = 18.066 \times {10^{23}}$
So, number of atoms = $18.066 \times {10^{23}}$

(C) $0.44g$ of $C{O_2}$:
We know that,
N = no. of moles x no. of atoms in one molecule $xN_A$
No. of moles = given weight in g / molecular mass of compound
Let no. of moles = n
$\Rightarrow n = \dfrac{{wg(g)}}{{{M_O}}}$
$\Rightarrow n = \dfrac{{0.44}}{{44}}mol$
$\Rightarrow n = 0.01mol$
Given no. of atoms in one molecule = 3
$\Rightarrow N = 0.01 \times 3 \times {N_A}$
$\Rightarrow N = 0.01 \times 3 \times 6.022 \times {10^{23}}$
$\Rightarrow N = 0.18 \times {10^{23}}$
So, number of atoms = $0.18 \times {10^{23}}$
Thus, we conclude that after calculation Option (B) has the maximum number of atoms $22.4L$ of $C{O_2}$at STP = $18.066 \times {10^{23}}$ atoms.

Note: Carbon dioxide $(C{O_2})$ is a colourless gas which contains carbon atoms double bonded with two oxygen atoms. It is used as a refrigerant and in fire extinguishers and is also used in carbonated drinks and for inflating rafts, jackets, rubber, plastic and other products. It is a harmful gas as it promotes growth of greenhouse gases.