Question

The total number of atoms present in $1$ mole of potash alum, in terms of Avogadro number $N_A$, is:

[ Potash Alum $=K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O$]

• A. $3N_A$
• B. $5N_A$
• C. $8N_A$
• D. $32N_A$

Answer 1

The correct option is D $32N_A$

The molecular formula of potash alum is $K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O$.
One molecule of potash alum contains $32$ atoms.
$\therefore 1$ mole of potash alum contains $N$ molecules.
Thus, the number of atoms present in $1$ mole of potash alum will be $32N$.
Here, $N$ is the Avogadro number.