Answer :
Answer: There are [tex]5.42 \times 10^{23}[/tex] [tex]Ca^{2+}[/tex] ions present in 0.9 mol of [tex]CaCl_{2}[/tex].
Explanation:
According to the mole concept, there are [tex]6.022 \times 10^{23}[/tex] atoms present in one mole of a substance.
As here, 0.9 mol of [tex]CaCl_{2}[/tex] are present. Also, there is only one atom of Ca is present in a molecule of [tex]CaCl_{2}[/tex].
Hence, calcium ions present in 0.9 mol of [tex]CaCl_{2}[/tex] are calculated as follows.
[tex]No. of ions = 1 \times 0.9 \times 6.022 \times 10^{23}\\= 5.42 \times 10^{23}[/tex]
Thus, we can conclude that there are [tex]5.42 \times 10^{23}[/tex] [tex]Ca^{2+}[/tex] ions present in 0.9 mol of [tex]CaCl_{2}[/tex].