In an iodometric estimation, the following reactions occur 2Cu2++4I−→Cu2I2+I2;I2+2Na2S2O3→2NaI+Na2S4O6 0.12mole of CuSO4 was added to excess of KI solution and the liberated iodine required 120mL of hypo. The molarity of hypo solution was:
A
2
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B
0.20
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C
0.1
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D
1.0
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Solution
The correct option is D1.0 2Cu+2+4I−⟶Cu2I2+I2
I2+2Na2S2O3⟶2NaI+Na2S4O6
Moles of pure CuSO4=0.12 moles of Na2S2O3 reacted with I2 Moles of Na2S2O3=120×10−3×M=0.12