Number of ions present in `2.0 "litre"` of a solution of `0.8 M K_(4)Fe(CN)_(6)` is:

Options

A. `4.8xx10^(23)`
B. `4.8xx10^(24)`
C. `9.6xx10^(24)`
D. `9.6xx10^(22)`

Answer

b

Explanation

Moles of `K_(4)[Fe(CN)_(6)]` present in 2 L of solution
`=(2 L) xx (0.8 mol L^(-1))=1.6` mol
`K_(4)[Fe(CN)_(6)]hArr4K^(+)+[Fe(CN)_(6)]^(-)`
No. of ions = 5
Total ions present = `1.6xx5xxN_(0)`
`=1.6 mol xx 5xx 6.022 xx 10^(23) mol^(-1)`
`=4.8xx10^(24)`