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Stoichiometric Calculations

2.5 g sample ...

Question

$2.5$ 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:

A

10.16

%

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B

5.08

%

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C

2.54

%

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D

1.27

%

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Solution

The correct option is B $5.08$ %
The molar mass of $C u$ is $63.5$ g/mol.

$2$ mmol of $C u S O_{4}$ corresponds to $\frac{2}{1000} \times 63.5 = 0.127$ g of $C u$ .

The percentage purity of $C u$ is $\frac{0.127 \times 100}{2.5} = 5 %$ .

Hence, the correct option is $(B)$

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Similar questions

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

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

10

0.02 M

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

0.1 M

hypo solution for its complete reduction.

V

) of hypo required is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

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

10

0.02 M

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

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

0.02 M

solution with excess

K I

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

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

10

0.02 M

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

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}

638.0

C u S O_{4}

solution is titrated with excess of

0.2 M

K I

solution. The liberated

I_{2}

400

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:

K C l O_{3}

are dissolved in conc. HCl and the solution was boiled. Chlorine gas evolved in the reactions was in then passed through a solution of KI and liberated iodine was titrated with 100 ml of hypo. 12.3 ml of same hypo solution required 24.6 ml of 0.5 iodine for complete neutralization. Calculate % purity of

K C l O_{3}

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