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An excess KI solution is mixed in a solution of K_(2)Cr_(2)O_(7) and liberated iodine required 72 mL of 0.05 N Na_(2)S_(2)O_(3) for complete reaction. How many grams of K_(2)Cr_(2)O_(7) were present in the solution of K_(2)Cr_(2)O_(7) ? The reaction occurs as : Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O |
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Answer» SOLUTION :The REACTION involved may be given as : `Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` `3[I_(2)+2Na_(2)S_(2)O_(3)]to 3[2NaI+Na_(2)S_(3)O_(6)]` 1 mole `K_(2)Cr_(2)O_(7)-=6 " moles of" Na_(2)S_(2)O_(3)` No. of moles of hypo`=("Mass")/(M.w. (158))=(ExxNxxV)/(1000xx158)` `N_(Na_(2)S_(2)O_(3))=(158xx0.05xx72)/(1000xx158)=3.6xx10^(-3)` No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)` [No. of moles of `Na_(2)S_(2)O_(3)`] `=(1)/(6)[3.6xx10^(-3)]=6xx10^(-4)` mole Mass of `K_(2)Cr_(2)O_(7)` in the given solution =No. of moles `xx` Molecular WEIGHT `=6xx10^(-4)xx294=0.1764` |
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