wiz-icon
MyQuestionIcon
MyQuestionIcon
6
You visited us 6 times! Enjoying our articles? Unlock Full Access!
Question

A 10 g sample of CuS and Cu2S was treated with 100 mL of 1.25 M K2Cr2O7 to produce Cr3+, Cu2+ and SO2. The excess oxidant was reacted with 50 mL of Fe2+ solution. 25 mL of the same Fe2+ solution required 0.875 M KMnO4 under acidic condition, the volume of KMnO4 used was 20 mL. Find the percentage of Cu2S in the sample.
Given: molar mass of Cu is 63.5 g mol1

A
28.7 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
71.3 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
57.4 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
42.6 %
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 42.6 %
Redox changes are,
+6Cr2O27+3Cr3+ (n-factor = 6)
+7MnO4+2Mn2+ (n-factor = 5)
+2Cu2S+2Cu2+++4SO2 (n-factor = 6)
+1Cu22S+2Cu2+++4SO2 (n-factor = 8)

Equivalents of dichromate present initially=1.25×6×1001000=0.75
Equivalents of Fe2+ in 25 mL=0.875×5×201000=0.0875
Equivalents of Fe2+ in 50 mL=0.0875×2=0.175
Equivalents of excess dichromate=0.175
Equivalents of dichormate consumed by CuS and Cu2S=0.750.175=0.575
Equivalents of CuS+Equivalents of Cu2S=0.575

If x g is the mass of CuS, the mass of Cu2S is (10x) g.

x63.5+32×6+(10x)2(63.5)+32×8=0.575
x=5.74 g
% CuS=5.7410×100=57.4 %
% Cu2S=(105.74)10×100=42.6 %

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon