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Question

3 gm of CuSO4, 5H2O sample containing inert impurity reacted with KI and the liberated I2 reacted with 10ml of 0.5M Na2S2O3. The % purity of CuSO4.5H2O will be:

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Solution

2CuSO4+2KICu2SO4+K2SO4+I2

I2+2Na2S2O3Na2S4O6+2NaI

10ml of 0.5MNa2S2O3=0.5100 mole= 0.005 mole of Na2S2O3

The number of moles of I2 liberated= 0.0052=0.0025 moles

0.0025 moles of I2 was liberated from 0.0025×2=0.005 moles of CuSO4 i.e. CuSO4.5H2O

Now, 0.005 moles of CuSO4.5H2O=0.005×249.677=1.248gm

The percentage of purity= 1.2483×100=41.6%

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