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Molarity

638.0 g of ...

Question

$638.0$ g of $C u S O_{4}$ solution is titrated with excess of $0.2 M$ $K I$ solution. The liberated $I_{2}$ required $400$ mL of $1.0 M$ $N a_{2} S_{2} O_{3}$ for complete reaction. The percentage purity of $C u S O_{4}$ in the sample is:

5 %

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10 %

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15 %

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20 %

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Solution

The correct option is D $10 %$
The overall reaction is as follows:
$2 C u^{2 +} + 2 e^{-} \to C u_{2}^{⊝}$
$2 I^{⊝} \to I_{2} + 2 e^{-}$
________________________________________
$2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
________________________________________
Reduction of $I_{2}$ is given by, $I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{⊝} + S_{4} O_{4}^{2 -}$
$2$ mol $C u S O_{4} \equiv 4$ mol $K I \equiv 1$ mol $I_{2}$ $\equiv 2$ mol $S_{2} O_{3}^{2 -}$

milli moles $C u S O_{4}$ $\equiv$ milli moles $S_{2} O_{3}^{2 -}$ $\equiv 400 \times 1$ mmoles $= 0.4$ mol
Weight of $C u S O_{4} = 400 \times 10^{- 3} \times 159.5$ g
$% C u S O_{4} = \frac{400 \times 10^{- 3} \times 159.5 \times 100}{638} = 10 %$

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Similar questions

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

Percentage of purity of sample is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

ml of

0.02 M

solution with excess

K I

is :

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

The color of the solution developed during the addition of

K I

C u S O_{4}

solution is :

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

The volume (

V

) of hypo required is:

5

g sample of brass was dissolved in one litre dil.

H_{2} S O_{4}

20

mL of this solution were mixed with

K I

, liberating

I_{2}

and

C u^{+}

and the

I_{2}

required

20

mL of

0.0327

N hypo solution for complete titration. Calculate the percentage of

C u

in the alloy.

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638.0 g of CuSO4 solution is titrated with excess of 0.2M KI solution. The liberated I2 required 400 mL of 1.0M Na2S2O3 for complete reaction. The percentage purity of CuSO4 in the sample is:

$638.0$ g of $C u S O_{4}$ solution is titrated with excess of $0.2 M$ $K I$ solution. The liberated $I_{2}$ required $400$ mL of $1.0 M$ $N a_{2} S_{2} O_{3}$ for complete reaction. The percentage purity of $C u S O_{4}$ in the sample is: