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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:A. `4.8xx10^(22)`B. `4.8xx10^(24)`C. `9.6xx10^(24)`D. `9.6xx10^(22)` |
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Answer» Correct Answer - B Mole of `K_(4)Fe(CN)_(6)=2xx0.8=1.6` Also `1 "mole of" K_(4)Fe(CN)_(6)` gives `4K^(+)` and `1 Fe(CN)_(6)^(4-) "ion"`. Thus total ions in `1.6 "mole"` `=1.6xx5xxN=48.184xx10^(23)` |
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