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Relation between Cp and Cv

What volume o...

Question

What volume of $0.40 M - N a_{2} S_{2} O_{3}$ would be required to react with the $I_{2}$ liberated by adding the excess KI to 50 mL of $0.20 M - C u S O_{4}$

A

12.5 mL

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B

25 mL

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C

50 mL

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D

2.5 mL

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Solution

The correct option is B 25 mL
Given
Molarity of $N a_{2} S_{2} O_{3}$ = 0.4M
Molarity of $C u S O_{4}$ = 0.2M
Volume of $C u S O_{4}$ = 50mL
Solution
Using Volumetric principle

$M_{1} \times V_{1} = M_{2} \times V_{2}$
$0.4 \times V_{1} = 0.2 \times 50$
$V_{1} = 25 m l$
The correct option is B

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Similar questions

(a)

C u S O_{4}

K I

in acidic medium to liberate

I_{2}

.

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

(b)

Mercuric perisdate,

H g_{5} (I O_{6})_{2}

reacts with a mixture of

K I

H C l

according to the following the equation:

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

The liberated iodine is titrated against

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

solution, one mL of which is equivalent to

0.0499

C u S O_{4} .5 H_{2} O

What volume in mL of

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

solution will be required to react with

I_{2}

0.7245

H g_{5} (I O_{6})_{2}

? (Molar mass of

C u S O_{4} .5 H_{2} O = 249.5

H g_{5} (I O_{6})_{2} = 1448.5

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:

H_{2} O_{2}

solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N

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

solution. The volume strength of

H_{2} O_{2}

Q. In iodometric titrations, an oxidizing agent such as

K M n O_{4}, K_{2} C r_{2} O_{7}, C u S O_{4}, H_{2} O_{2}

is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine

(I_{2})

.
Free iodine is titrated against standard reducing agent usually with sodium thiosulphate. The reactions are as follows:

K_{2} C r_{2} O_{7} + 6 K I + 7 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 4 K_{2} S O_{4} + 7 H_{2} O + I_{2}

2 C u C O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

I_{2} + N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

In iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, blue or voilet colour disappears when iodine is completely changed to iodide.

What volume of

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

would be required

(in mL)

I_{2}

liberated by adding

0.04

K I

50 mL

0.20 M C u S O_{4}

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

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