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Ideal Gas Equation

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.The amount of $I_{2}$ liberated in the reaction of $10$ ml of $0.02 M$ solution with excess $K I$ is :

A

0.051

g

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B

0.0254

g

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C

0.102

g

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D

0.204

g

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Solution

The correct option is B $0.0254$ g
100 ml of 0.02 M $C u S O_{4}$ solution corresponds to $100 \times 0.02 = 2$ mmol.
Out of this, 10 ml (i.e 0.2 mmol) are used .
$2 C u S O_{4} \equiv 2 C u I_{2} \equiv I_{2}$
2 moles of $C u S O_{4}$ will liberate 1 mol of $I_{2}$ .
0.2 mmoles of $C u S O_{4}$ will liberate 0.1 mmol of $I_{2}$ .
The molar mass of $I_{2}$ is 253.8 g/mol.
Hence, 0.1 mmol of $I_{2}$ corresponds to $\frac{0.1 m o l \times 253.8 g / m o l}{1000} = 0.0254 g$ of $I_{2}$ .

Hence option (B) is correct.

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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.

Percentage of purity of sample 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 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:

Q. 5 g of a sample of bleaching powder is treated with excess acetic acid and KI solution. The liberated

I_{2}

required 50 ml of N/10 hypo solution. The % available chlorine in sample is

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