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

Question

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

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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
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%`

5 views Answered 2023-02-05 15:29