Question

How many atoms are present in 0.10mol of ptcl2 (nh3)2?

Answer 1

Explanation:

According to the mole concept, 1 mole of an atom contains $6.022 \times 10^{23}$ atoms or molecules.

Therefore, we can calculate the number of atoms present in 0.10 mol of $PtCl_{2}(NH_{3})_{2}$ as follows.

               $0.10 mol \times 6.022 \times 10^{23}$ atoms

                   = $0.6022 \times 10^{23}$

                   = $6.022 \times 10^{22}$

Therefore, we can conclude that there are $6.022 \times 10^{22}$ atoms present in 0.10 mol of $PtCl_{2}(NH_{3})_{2}$.

Answer 2

The number of atoms  present in 0.10 mol of PtCl2(NH3) is calculated using Avogadro law  formula

that is

1 mole = 6.02 x10^23 atoms
what about 0.10 moles = ? atoms

= 6..02 x10 ^23 x 0.1 = 6.02 x10^22  atoms