Question

Which has the minimum number of oxygen atoms?
A.10mL ${{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)\left[ {{\text{density of water}} = 1{\text{g m}}{{\text{L}}^{ - 1}}} \right]$
B.$0.1{\text{mol }}{{\text{V}}_2}{{\text{O}}_5}$
C.$12{\text{g }}{{\text{O}}_3}\left( {\text{g}} \right)$
D.$12.044 \times {10^{23}}{\text{molecules of C}}{{\text{O}}_2}$

Answer 1

First calculate the number of moles of the given substance, then as we known one mole of any substance contains ${{\text{N}}_{\text{A}}}$ entity of that substance; however atoms of different elements can be determine using atomicity.

Formula: The mole is a concept of quantity in terms of number, mass and volume. It can be mathematical represented as:
${\text{mole}} = \dfrac{{{\text{given mass}}}}{{{\text{mol}}.{\text{wt}}./{\text{At}}.{\text{wt}}.}}$ , ${\text{mole}} = \dfrac{{{\text{number of particle given}}}}{{6.022 \times {{10}^{23}}}}$ , ${\text{mole}} = \dfrac{{{\text{volume at NTP}}\left( {{\text{in Litre}}} \right)}}{{22.4{\text{L}}}}$ .

Complete step-by-step answer: Now in case of 10mL ${{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)\left[ {{\text{density of water}} = 1{\text{g m}}{{\text{L}}^{ - 1}}} \right]$ , mass of water can be given by: ${\text{mass}} = {\text{volume}} \times {\text{density}} = 10{\text{mL}} \times 1{\text{g m}}{{\text{L}}^{ - 1}} = 10{\text{g}}$ .
Therefore number of mole in 10g of water can be given as:
${\text{mole}} = \dfrac{{10}}{{18}} = 0.55{\text{mol}}$ .
Therefore number of oxygen atom of water in $0.55{\text{mol}}$ can be given by: ${\text{number of oxygen atom}} = {\text{mole}} \times {{\text{N}}_{\text{A}}} \times {\text{atomicity of oxygen}}$ .
We get ${\text{number of oxygen atom}} = 0.55 \times \left( {6.022 \times {{10}^{23}}} \right) \times 1 = 3.35 \times {10^{23}}{\text{atom}}$ .
Similarly in case of $0.1{\text{mol }}{{\text{V}}_2}{{\text{O}}_5}$ , we get number of oxygen atom: ${\text{number of oxygen atom}} = 0.1 \times \left( {6.022 \times {{10}^{23}}} \right) \times 5 = 3 \times {10^{23}}{\text{atom}}$ , by using the equation: ${\text{number of oxygen atom}} = {\text{mole}} \times {{\text{N}}_{\text{A}}} \times {\text{atomicity of oxygen}}$ .
In case of $12{\text{g }}{{\text{O}}_3}\left( {\text{g}} \right)$ , moles of ozone can be given as: ${\text{mole}} = \dfrac{{12}}{{48}} = 0.25{\text{mol}}$ . Therefore, number of oxygen atom in $0.25{\text{mol}}$ of ozone can be given by: ${\text{number of oxygen atom}} = 0.25 \times \left( {6.022 \times {{10}^{23}}} \right) \times 3 = 4.5 \times {10^{23}}{\text{atom}}$ , by using the equation: ${\text{number of oxygen atom}} = {\text{mole}} \times {{\text{N}}_{\text{A}}} \times {\text{atomicity of oxygen}}$ .
In case of $12.044 \times {10^{23}}{\text{molecules of C}}{{\text{O}}_2}$ , moles of carbon dioxide can be given as: ${\text{mole}} = \dfrac{{12.044 \times {{10}^{22}}}}{{6.022 \times {{10}^{23}}}} = 2{\text{mol}}$ .
Therefore, number of oxygen atom in $2{\text{mol}}$ of carbon dioxide can be given by:
${\text{number of oxygen atom}} = 2 \times \left( {6.022 \times {{10}^{23}}} \right) \times 2 = 24.088 \times {10^{23}}{\text{atom}}$ , by using equation: ${\text{number of oxygen atom}} = {\text{mole}} \times {{\text{N}}_{\text{A}}} \times {\text{atomicity of oxygen}}$ .
Thus, a minimum number of oxygen atoms are present in $0.1{\text{mol }}{{\text{V}}_2}{{\text{O}}_5}$ .

Therefore, the correct option is B.

Note: A mole is the SI unit for the amount of substance. It is define as the amount of substance that contains as many entities as there are atoms in exactly 12g of ${{\text{C}}^{12}}$ isotope. 1 mole is equivalent to ${{\text{N}}_{\text{A}}}$ atoms, molecules, ions, electron or any entity. 1 mole is equivalent to molecular weight or atomic weight of substance. 1 mole is equivalent to the volume of $22.4{\text{L}}$ of a gas occupied at NTP.