1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL of acidic solution.

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

## The correct option is B 10 NNormality is the number of gram or mole equivalents of solute present in one litre of a solution. Normality (N) =W×1000E×V (in mL) where, W = Weight of the solute E = Equivalent mass of solute V = Volume of solution in mL Equivalent mass of solute=Molecular massValence factor Valence factor is the number of electron released or absorbed. Reaction of KMnO4 in acidic solution, MnO−4+8H++5e−→Mn2++4H2O Here, Mn+7 is reduced to Mn+2. Hence, it gains 5 e− so valence factor is 5 Molecular mass of KMnO4 is 158 g mol−1 E=Molar mass of KMnO4Valence factor=158/5=31.6 Normality (N) =15.8×100031.6× 50 So, Normality of the solution = 10 N

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program