10 cc of H_(2) O_(2)solution when reacted withKIsolution produced 0 . 5 g of iodine . Calculate the percentage purity of H_(2) O_(2)

10 cc of H_(2) O_(2)solution when reacted withKIsolution produced 0 .

1.	10 cc of H_(2) O_(2)solution when reacted withKIsolution produced 0 . 5 g of iodine . Calculate the percentage purity of H_(2) O_(2)
Answer» Solution :`H_(2) O_(2) + KI to KOH + I_(2)` First, applyingPOACfor H atoms to calculate moles of KOH, thenapplying POAC for Katoms to calculate moles of KI, andthenfinallyapplying POACfor Iatoms tocalculate moles ofiodine. POACfor H atoms : molesof H in `H_(2)O_(2) ` = moles of H in KOH `2 xx ` moles of `H_(2)O_(2) = 1 xx ` moles of KOH. . . (i) Applying POAC for K atons, molesof K atoms in KI= moles o F K atoms in KOH `1 xx` molesof KI = ` 1 xx ` moles of KOH `=2 xx ` molesof `H_(2) O_(2)` [from the equ. (i)] Applying POACfor I atoms, molesof I atoms in KI = moles of I atoms in` I_(2)` or moles of `I_(2) = (1)/(2) xx ` moles of KI `= (1)/(2) xx 2 ` molesof `H_(2) O_(2)`[from the eqn. (ii)] Now, `(wt. of I_(2))/( mol. wt. of I_(2)) = (wt. of H_(2) O_(2))/(mol . wt. of H_(2)O_(2))` Suppose x is thewt. of `H_(2)O_(2)` . Then, `(0. 5)/( 254) = (x)/( 34)` `x = (34 xx 0 . 5)/( 254) = 0.0669 G` . `% of H_(2)O_(2) = (0.0669)/(10) xx 100 = 0 . 669 %`