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2.5 g sample of copper is dissolved in ...

`2.5 ` g sample of copper is dissolved in excess of `H_(2)SO_(4)` to prepare 100 mL of `0.02 M CuSO_(4)` (aq) . 10 mL of `0.02` M solution of `CuSO_(4)` (aq) is mixed with excess of Kl to show the following changes `CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2)`
`2Cu l_(2) to Cu_(2)l_(2) + l_(2)`
The liberated iodine is titrated with hypo `(na_(2)S_(2)O_(3))` and requires V mL of `0.1` M hypo solution for its complete reduction .
The volume (V) of hypo required is :

A

2 mL

B

20 mL

C

1 mL

D

10 mL

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • 2.5 g sample of copper is dissolved in excess of H_(2)SO_(4) to prepare 100 mL of 0.02 M CuSO_(4) (aq) . 10 mL of 0.02 M solution of CuSO_(4) (aq) is mixed with excess of Kl to show the following changes CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2) 2Cu l_(2) to Cu_(2)l_(2) + l_(2) The liberated iodine is titrated with hypo (na_(2)S_(2)O_(3)) and requires V mL of 0.1 M hypo solution for its complete reduction . Percentage purity of sample is :

    A
    `10.16`
    B
    `5.08`
    C
    `2.54`
    D
    `1.27`

  • 2.5 g sample of copper is dissolved in excess of H_(2)SO_(4) to prepare 100 mL of 0.02 M CuSO_(4) (aq) . 10 mL of 0.02 M solution of CuSO_(4) (aq) is mixed with excess of Kl to show the following changes CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2) 2Cu l_(2) to Cu_(2)l_(2) + l_(2) The liberated iodine is titrated with hypo (na_(2)S_(2)O_(3)) and requires V mL of 0.1 M hypo solution for its complete reduction . The amount of l_(2) liberated in the reaction of 10 mL of 0.02 M solution with Kl (excess ) is :

    A
    `0.051` g
    B
    `0.0254` g
    C
    `0.102` g
    D
    `0.204` g

  • The normality of mixture obtained by mixing 100 mL of 0.2 M H_(2)SO_(4) and 200 mL of 0.2 M HCl is

    A
    0.0267
    B
    `0.2670`
    C
    `1.0267`
    D
    `1.1670`

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