1.

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

Answer»

`4.8xx10^(23)`
`4.8xx10^(24)`
`9.6xx10^(24)`
`9.6xx10^(22)`

Solution :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)`


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