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Question

In the standardization of Na2S2O3using K2Cr2O7. The equivalent weight of K2Cr2O7is:


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Solution

Step 1: Finding the valence factor in the molecule of K2Cr2O7:

The ionic equation between Na2S2O3 and K2Cr2O7 is:

4Cr2O72-+3S2O32-+26H+6SO42-+8Cr3++13H2O

So, the oxidation number of Cr in Cr2O72- =2x+7(-2)=-2

So, x=+6

The oxidation number of Cr in Cr+3 =+3

The net change in oxidation number : 3

Valencefactor=numberofCratom×Changeinoxidationstate=2×3=6

Step 2: Finding the equivalent weight of K2Cr2O7:

Equivalentweight=MolecularweightvalencefactorEquivalentweightofK2Cr2O7=Molecularweight6EquivalentweightofK2Cr2O7=2(39)+2(52)+7(16)6EquivalentweightofK2Cr2O7=2946EquivalentweightofK2Cr2O7=49

Therefore, the equivalent weight of K2Cr2O7 is 49.


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