CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A solution of 0.2 g of a compound containing Cu2+ and C2O24 ions on titration with 0.02 M KMnO4 in the presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dilute acetic acid and treated with excess of KI. The liberated iodine requires 11.3 mL of 0.05 N Na2SO4 solution for complete reduction . Find out the molar ratio of Cu2+ to C2O24 in the compound.

A
1 : 2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2 : 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1 : 4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4 : 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1 : 2
The mixture of Cu2+ and C2O24 are reacting seperately first with KMnO4 solution and then with KI to liberate I2. It can be seen that Cu2+ cannot be oxidised and C2O24 cannot be reduced. So, reactions involved are:
2MnO4+5C2O24+16H+2Mn2++10CO2+8H2O
2Cu2++4I Cu2I2+I2
I2+2S2O23 2I+S4O26

Equivalents of KMnO4 solution,
= 0.02×5×22.61000=2.26×103

moles of C2O24,
= 2.26×1032=1.13×103

Equivalents of Na2S2O3 solution
= 0.05×11.31000=5.65×104

moles of Cu2+
= 5.65×1041=5.65×104

molar ratio of Cu2+ to C2O24
= 5.65×1041.13×103=12

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image
similar_icon
Similar questions
View More