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Chemistry

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 $0.20 M - C u S O_{4}$
12.5 mL
25 mL
50 mL
2.5 mL

25 mL

50 mL

12.5 mL

2.5 mL

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Solution

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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

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}

View Solution

(a)

C u S O_{4}

reacts with

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

and

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

g of

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}

liberated from

0.7245

g of

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

? (Molar mass of

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

g/mol and of

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

g/mol)

View Solution

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:

View Solution

To a 25 ml

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}

solution is:

View Solution

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)

to react with

I_{2}

liberated by adding

0.04

mole of

K I

50 mL

0.20 M C u S O_{4}

solution?

View Solution

What volume of 0.40M−Na2S2O3 would be required to react with the I2 liberated by adding the excess KI to 50 mL of 0.20M−CuSO412.5 mL25 mL50 mL2.5 mL

GivenMolarity of Na2S2O3 = 0.4MMolarity of CuSO4 = 0.2MVolume of CuSO4= 50mLSolutionUsing Volumetric principleM1×V1=M2×V20.4×V1=0.2×50V1=25mlThe correct option is B

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 $0.20 M - C u S O_{4}$
12.5 mL
25 mL
50 mL
2.5 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