Question

The number of atoms present in $0.5g$ of nitrogen is same as the atom in:
A.$12g$ of $C$
B.$64g$ of $S$
C.$8g$ of $O$
D.$48g$ of $Mg$

Answer 1

Hint: One mole of anything, including molecules, is (Avogadro's number) of them. Generally you will have a given mass of a component. There are two essential strides to get from the offered mass to the quantity of molecules.

Complete step by step answer:
We know that;
Equivalent moles of an iota contain the equivalent number of particles.
We also know that;
one mole contain ${N_A}$ number of particles inside it and the value of ${N_A}$ is:
${N_A} = 6.023 \times {10^{23}}$.
Now, let us find out the value of $0.5g$ of ${N_2}$:
$0.5g$ of ${N_2} \to \dfrac{{{N_A}}}{2}$ ….(1)
Now, let us check one by one option,
$12g\:C$ = $1mole\:C$ = ${N_A}$ particle
$64g\:S$ = $2mole\:C$ = $2 \times {N_A}$ particle
$8g\:O$ = $0.5mole\:O$ = $\dfrac{{{N_A}}}{2}$ particle
$48g\:Mg$ = $2mole\:Mg$ = $2 \times {N_A}$ particle
Thus, from the above calculations we get to know that the correct expression that matches our required statement in equation (1) is Oxygen.

Hence, the number of atoms present in $0.5g$ of nitrogen is the same as the atom in $8g\:O$; that is, Option C.

Additional information:
What is Avogadro's number?
The quantity of particles present in one mole of any substance is fixed with an estimation of $6.023 \times {10^{23}}$. This is known as Avogadro number or constant represented by ${N_0}$.

Note:
As per mole concept one mole of particles equal to molecular mass in grams. It makes a scaffold between the plainly visible world and the infinitesimal world by relating the measure of substance to the quantity of particles. It likewise gives the connection between other actual constants and properties. In the nuclear level substances are estimated according to the nuclear mass unit.