1.

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)

Answer»

`31.75%`
`63.5%`
`53%`
`37%`

SOLUTION :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%`


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