Calcium chlroide when dissolved in water dissociates into its ions according to the following equations.
`CaCl_(2)(aq) to Ca^(2+)(aq) + 2Cl^(-1) (aq)`
Calculate the number of ions obtained from `CaCl_(2)` when 222g of it is dissolved in water.

Explanation

Molar mass of `CaCl_(2)=40+(2xx35.5)=111"g mol"^(-1)`
`111"g of "CaCl_(2)"represent"=1"mol"`
`222"g of "CaCl_(2)"represent"=((222g))/((111g))xx1"mol"=2"mol"`
No. of molecules in 2 moles of `CaCl_(2)=2xxN_(A)=2xx6.022xx10^(23)`
1 molecules of `CaCl_(2)` in solution form ions =3
`2xx6.022xx10^(23)` molecules of `CaCl_(2)` in solution from ions `=3xx2xx6.022xx10^(23)`
`=36.132xx10^(23)`
`=3.6132xx10^(24)`