Question

Which of the following has the minimum number of oxygen atoms?
a.) $10ml {H_2}O(l)$ [density of the water = $1gm{l^{ - 1}}$]
b.) $0.1mol {V_2}{O_5}$
c.) $12g {O_3}(g)$
d.) $12.044 \times {10^{22}}$ molecules of $C{O_2}$

Answer 1

Hint: In order to solve this question we will use the basic concepts of chemistry as the number of oxygen atoms is directly proportional to the product of number of moles and number of Oxygen atoms in the molecule and 1 mole of ${H_2}O$ contains $22400ml$ of ${H_2}O$ molecules. Use these concepts to reach the answer.

Complete step by step answer:
By taking option A
Number of Oxygen atoms in 10ml of water molecules:
1mole of ${H_2}O$=22400ml of molecules ${H_2}O$
1ml of ${H_2}O$ molecules = $\dfrac{1}{{22400}}$moles of ${H_2}O$ molecules.
1ml of ${H_2}O$ molecules = $\dfrac{1}{{22400}} \times 10$moles of ${H_2}O$ molecules
1ml of ${H_2}O$ molecules= $4.46 \times {10^{ - 4}}$ moles of the ${H_2}O$ molecules.
Number of${H_2}O$ molecules= ($4.46 \times {10^{ - 4}} \times $Avogadro’s number) molecules.
Number of${H_2}O$ molecules= $\left( {4.46 \times {{10}^{ - 4}} \times 6.022 \times {{10}^{23}}} \right)$molecules.
Number of${H_2}O$ molecules= $26.8 \times {10^{19}}$molecules.
Since, each ${H_2}O$ molecule has only 1 Oxygen atom.
So, we can say that,
Number of ${H_2}O$ molecules = Number of oxygen atoms
=$26.8 \times {10^{19}}$ atoms of oxygen

By considering option B.
Number of Oxygen atoms in 0.1 mole of ${V_2}{O_5}$ .
The number of Oxygen atoms is directly proportional to the product of number of moles and number of Oxygen atoms in the molecule.
Now,
Number of ${V_2}{O_5}$ molecules= ($0.1 \times $Avogadro’s number) molecules.
=$\left( {0.1 \times 6.022 \times {{10}^{23}}} \right)$molecules
= $\left( {6.022 \times {{10}^{22}}} \right)$molecules
Number of Oxygen atoms= $5 \times $(number of ${V_2}{O_5}$molecules)
(Since, one molecule of ${V_2}{O_5}$ contains 5 atoms)
=$5 \times 6.022 \times {10^{22}}$atoms
= $3.011 \times {10^{23}}$ atoms
Therefore, Number of Oxygen atoms is $3.011 \times {10^{23}}$ atoms

Taking option C
Number of Oxygen atoms in12g of ${O_3}$ -
The molar weight of ${O_3}$is 48g
So, number of moles of${O_3}$ = $\dfrac{{12g}}{{48g}} = 0.25$ moles
Number of ${O_3}$ molecules = ($0.25 \times $Avogadro’s number) molecules
=$(0.25 \times 6.022 \times {10^{23}})$molecules
=$1.5 \times {10^{23}}$molecules.
Number of Oxygen atoms = $3 \times $Number of ${O_3}$ molecules (Since, one molecule of ${O_3}$contains 3 atoms)
$ = 3 \times 1.5 \times {10^{23}}$ atoms
= $4.5 \times {10^{23}}$ atoms

Considering option D.
Number of Oxygen atoms in $12.044 \times {10^{22}}$molecules of $C{O_2}$-
Number of Molecules of $C{O_2}$ = $12.044 \times {10^{22}}$ molecules
Atoms of Oxygen $ = 2 \times $ Number of $C{O_2}$molecules (Since, one molecule of $C{O_2}$ contains 2 atoms)
$ = 3 \times 12.044 \times {10^{22}}$ atoms
$ = 2.4 \times {10^{23}}$ atoms
Atoms of Oxygen $ = 2.4 \times {10^{23}}$ atoms of Oxygen
Hence, the minimum number of oxygen atoms are present in 10ml of ${H_2}O$ molecules.
So, the correct answer is “Option A”.

Note: The Avogadro constant is the proportionality constant element that compares the number of sampled constituent particles to the volume of material in that sample. It takes its name from Amedeo Avogadro, an Italian scientist. The SI unit is the inverse mole, which is marked as $Na = 6.022 \times {10^{23}}$${(mol)^{ - 1}}$. Hence, the number of the elementary entity that one mole of the any sample will have, it will be equal to the numerical value of the Avogadro constant, $Na = 6.022 \times {10^{23}}$. The elementary entities may be molecules, atoms or the ions in a given sample of the compound or element.
Firstly, the value of the Avogadro’s number was obtained by dividing the charge of a mole of electrons by the charge of a single electron which is equal to the $6.022 \times {10^{23}}$ particles per mole. The numerical value of the Avogadro’s constant was used by the French physicist Jean Baptiste Perfin for the first time to explain the Brownian motion.