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

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Number of ions present in `2.0 "litre"` of a solution of `0.8 M K_(4)Fe(CN)_(6)` is:
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`K_4[Fe(CN)_6]to4K^(o+)+[Fe(CN)_6]^(4-)`
5 ions (mol)`=5xx6xx10^(23)` ions
Number of ions `=5xx6xx10^(23)xx0.8xx2=48xx10^(24)` ions
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