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

User Excray
by
6.6k points

2 Answers

3 votes

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).

User Pietro Marchesi
by
6.9k points
3 votes
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
User Julien N
by
7.2k points