Calculate the number of `Cl^(-)` and `Ca^(2+)` ions in `222 g` anhydrous `CaCl_(2)`.

Explanation

`1 mol of CaCl_(2) = 40 + 35.5 xx 2 = 111 g`
`(CaCl_(2) rarr Ca^(2+) + 2Cl^(ɵ))`
` 1 mol 1 mol 2 mol`
a. `111g of CaCl_(2) implies 1 mol of Ca^(2+)`
`222 g of CaCl_(2) implies 2 mol of of Ca^(2+)`
b. `111 g of CaCl_(2) implies 2 mol of Cl^(ɵ)`
`222 g of CaCl_(2) = (2)/(111) xx 222 = 4 mol of Cl^(ɵ)`
Therefore, number of `Ca^(2+)` ions `= 2 xx N`
`= 2 N` or `2 xx 6 xx 6.023 xx 10^(23)`
Number of `Cl^(ɵ)` ions `=4 xx N`
`= 4 N` or `4 xx 6.023 xx 10^(23)`