The number of `Cl^(-)` ions present in 222 g anhydrous
`CaCl_(2)` is

Options

A. `2.5N_A`
B. `2N_A`
C. `3N_A`
D. `4N_A`

Answer

D

Explanation

`underset(1 " mol")(CaCl_(2)) rarrunderset(1 " mol") (Ca^(2+)) + underset(2" mol")(2Cl^(Θ))`
1 mole of `CaCl_(2)= 40 + 35. 5 xx 2 = 111` g
111 g mole of `CaCl_(2) = 2 ` mol of `Cl^(Θ)`
222 g of `CaCl_(2) = 2/111 xx 222 = 4 ` mol of `Cl^(Θ)`
Therefore, number of `Ca^(2+)` ions `= 2 xx N `
`= 2 N or 2 xxx 6. 023 xx 10^(23)`
NUmber of `Cl^(Θ) ` ions `= 4 xx N`
`= 4 xx 6.023 xx 10 ^(23)`
`= 2.409 xx 10^(24)`