wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Potassium acid oxalate K2C2O4.3H2C2O4.4H2O can be oxidized by MnO4 in acidic medium. Calculate the volume of 0.3 M MnO4 reacting in acid solution with 10.16 g of the acid oxalate.
(Molar mass of potassium acid oxalate is 508 g mol1.)

A
106.7 mL
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
160.0 mL
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.06 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.16 L
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 106.7 mL
K2+3C2O4.3H2+3C2O4.4H2OPotassium acid oxalate++7MnO4H++2Mn2+++4CO2

nf=(|Change in O.S.|)×Number of atoms
nfpotassium acid oxalate=(|+3(+4)|)×8=8
nfMnO4=(|+7(+2)|)×1=5

From the law of equivalence,
Equivalents of potassium acid oxalate = Equivalents of MnO4
Wpotassium acid oxalateMolar mass of potassium acid oxalate×nfpotassium acid oxalate=MMnO4×VMnO4×nfMnO4
10.16 g508 g mol1×8=(0.3 mol L1)×VMnO4×5
VMnO40.1067 L106.7 mL

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon