A 0.0200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced to copper (I) by iodine, `2Cu^(2+)+4I^(-)to2 CuI+I_2`
If 20.0 mL of 0.10 M `Na_2S_2O_3` is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu=63.5 g/mole)

Options

A. `31.75%`
B. `63.5%`
C. `53%`
D. `37%`

Answer

B

Explanation

2 moles of `Cu^(2+)`=1 moles of `I_2` =2 moles of hypo.
so moles of hypo used =`20xx10^(-3)xx0.1=2` milli moles = milli moles of copper
hence percentage of copper =`(2xx10^(-3)xx63.5)/0.2xx10%=63.5%`