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Question

How much amount of CuSO4.5H2O required for liberation of 2.54g I2 when titrated with KI?

A
2.5 gm
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B
4.99 gm
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C
2.4 gm
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D
1.2 gm
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Solution

The correct option is A 4.99 gm
Molecular weight of CuSO4.5H2O=249.5gmol1

The reaction for liberation of I2;
2CuSO4.5H2O+KI=I2+Cu2I2+2K2SO4+5H2O

2.54g of I2=2.54254=0.01 moles of I2

Now, 1 mole of I2 is liberated from 2 moles of CuSO4.5H2O.

0.01 mole of I2 will be liberated from 2×0.01 mole i.e. 0.02 mole of CuSO4.5H2O.

Amount of CuSO4.5H2O=0.02 mole ×249.5gmol1=4.99gm

Hence, option B is correct.

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