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`

Explanation

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`