Question

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

Answer 1

Step-by-step explanation:

According to the mole concept, 1 mole of an atom contains
`6.022 * 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 * 6.022 * 10^(23)` atoms

=
`0.6022 * 10^(23)`

=
`6.022 * 10^(22)`

Therefore, we can conclude that there are
`6.022 * 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