A 1.10 g sample of copper ore is dissolved and the Cu^(2+)is treated with excess KI. The liberated I_2requires 12.12 mL of 0.10M Na_(2) S_(2)O_3solution for titration. Find the % of copper by mass in ore.

A 1.10 g sample of copper ore is dissolved and the Cu^(2+)is treated w

1.	A 1.10 g sample of copper ore is dissolved and the Cu^(2+)is treated with excess KI. The liberated I_2requires 12.12 mL of 0.10M Na_(2) S_(2)O_3solution for titration. Find the % of copper by mass in ore.
Answer» Solution :`Cu^(2+) +erarrCu^(+), 2I^(-) rarrI_(2)+2E` Meq.of `Cu^(+2)` = Meq.of liberated `I_2` = Meq.of `Na_(2)S_2O_3` `= 12.12 xx0.1 XX1 =1.212` `:.(W_(cu^(2+)))/(63.6//1)xx1000=1.212` ` :. W_(cu) =W_(cu^(2+))=0.077` `( :. Cuoverset(H_2SO_4)rarrCuSO_(4))` `:. %Cu=(0.077)/(1.10)XX100 =7%`