Question

The number of atoms present in 78 g of potassium is equal to the number of potassium ions present in .......... g of potassium chloride.

• A. 74.5
• B. 37.25
• C. 223.5
• D. 149

Answer 1

The correct option is D 149

No. of atoms present in 39 g of K = $N_A$
$\therefore$ No. of atoms present in 78 g of K = 2$N_A$
Now, $N_A$ ions of potassium are present in 74.5 g of $KCl$.

2$N_A$ ions of potassium are present in 149 g of $KCl$.