Question

Assertion: $1\;{\rm{g}}\;{O_2}$ and $1\;{\rm{g}}\;{O_3}$ have equal numbers of atoms.
Reasons: Mass of $1\;{\rm{mole}}$ atom is equal to its gram-atomic mass.
i) If both (A) and (R) are correct and (R) is the correct explanation of (A).
ii) If both (A) and (R) are correct and (R) is not the correct explanation of (A).
iii) If (A) is correct but (R) is wrong.
iv) If (A) is wrong but (R) is correct.
v) If both (A) and (R) are wrong.

Answer 1

Hint: We can determine the number of particles, amount or mass of a given substance by using their relationships with each other. In quantitative chemistry, we have various relationships between different quantities that are quite useful. One such relationship is between the amount of a substance and the number of particles of that substance in it.

Complete step by step answer:
It has been defined as a mole which contains $6.02 \times {10^{23}}$ particles. So ${\rm{1}}\;{\rm{mol}}$ of every substance would have an equal number of particles. For example, ${\rm{1}}\;{\rm{mol}}$ of H would have $6.02 \times {10^{23}}$ hydrogen atoms whereas , ${\rm{1}}\;{\rm{mol}}$ of ${H_2}$ would have $6.02 \times {10^{23}}$ hydrogen gas molecules. Similarly, ${\rm{1}}\;{\rm{mol}}$ of NaCl would have $6.02 \times {10^{23}}$ NaCl formula units.
We can determine the amount of a substance in a given mass by using its molar mass which is the mass of ${\rm{1}}\;{\rm{mol}}$ of it. The relationship can be written as:
${\rm{Amount}}\left( {mol} \right) = \dfrac{{{\rm{Given\; mass }}\left( m \right)}}{{{\rm{Molar \;mass }}\left( M \right)}}$
Similarly, we can define the gram-atomic mass of a substance as the mass of ${\rm{1}}\;{\rm{mol}}$ of its atoms.
Now, we will consider the given statements by using the above explanations. For assertion, let’s calculate the number of atoms in $1\;{\rm{g}}\;{O_2}$ as follows:
${\rm{Amount\;of\;oxygen}}\left( {{O_2}} \right) = \dfrac{{{\rm{1 g}}\;}}{{{\rm{32 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\;}}\\
{\rm{Number\;of\; }}{O_2}{\rm{\; molecules}} = 2\times\left( {\dfrac{{{\rm{1 g}}\;}}{{{\rm{32 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\;}}} \right)6.02 \times {10^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\
= {\rm{0}}{\rm{.37625}} \times {10^{23}}{\rm{ }}O\;{\rm{ atoms}}$
We can do the same for $1\;{\rm{g}}\;{O_3}$ as follows:
${\rm{Amount\;of\;ozone}}\left( {{O_3}} \right) = \dfrac{{{\rm{1 g}}\;}}{{{\rm{48 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\;}}\\
{\rm{Number \;of \;}}{O_3}{\rm{\; molecules}} = 3\times\left( {\dfrac{{{\rm{1 g}}\;}}{{{\rm{48 g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}\;}}} \right)6.02 \times {10^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\\
= {\rm{0}}{\rm{.37625}} \times {10^{23}}{\rm{ }}O{\rm{ \;atoms}}$
So, we have established that the assertion is correct. Now let’s have a look at the reason. We have already defined the gram-atomic mass as the mass of ${\rm{1}}\;{\rm{mol}}$ of atoms of a given substance which makes the statement of the reason to be correct as well. However, the reason is the correct explanation of the assertion.

Hence, the correct option is A.

Note: We cannot determine the amount or number of atoms simply by looking at the mass or formula and need to calculate properly by keeping in mind the atomicity of the molecules.