Question

ACTIVITY
Find the molecular weight of:
* CaCO3
4. BaSO4
2. KMnO4
5.FeCl3
3. NaOH
6.00
​

Answer 1

KMnO4 is a strong oxidising agent . In acidic medium, itself reduced to Mn2+ while in basic or neutral medium it reduced to MnO2, logic behind this is very simple , as acidic medium is also called as electron deficient medium . So there is no support from medium to reaction that why it compeleted its all deficiency of electron from other source (substrate) . While in basic medium (electron rich medium ) there is good support from the medium to reaction thatswhy it take only 3e from other source remaning electron provided by solution.

MnO4-  H+  Mn2+ 

from this reaction we can deduce that Mn7+ reduced to Mn2+ so  for this reaction is 5

Eq.wt = molecular weight / n factor

So for KMnO4 molecular weight is 158.g

So Eq.wt = 158/5 = 31.6 g per equivalent.



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Answer 2

CaCO3=40+12+16*3=100u

BaSO4=137+22+16*4=223u

KMnO4=26+54+16*4=144u

FeCl3=55+35*3=160u

NaOH=22+16+1=39u

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