3 gm of CuSO4, 5H2O sample containing inert impurity reacted with KI and the liberated I2 reacted with 10ml of 0.5MNa2S2O3. The % purity of CuSO4.5H2O will be:
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Solution
2CuSO4+2KI⟶Cu2SO4+K2SO4+I2
I2+2Na2S2O3⟶Na2S4O6+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