Question

A 200 mL of solution of 12 is divided into two unequal parts. Part-1 reacts with hypo solution in acidic medium, 300 mL of
0.2 M hypo was consumed, part-II was added with 50 mL of 0.6 M concentrated NaOH solution. What was initial
concnetration of l2?​

Answer 1

Answer:

First part:

Millimoles of Na

2

S

2

O

3

used =8×2×1( n-factor)=16

I

2

+2Na

2

S

2

O

3

→2NaI+Na

2

S

4

O

6

1 mol 2 mol

Millimoles of I

2

used =

2

1

Millimoles of Na

2

S

2

O

3

=

2

16

=18 ....(i)

Second part:

3I

2

+6NaOH→5NaI+NaIO

3

+3H

2

O

Millimoles of H

2

SO

4

= excess NaOH=30×0.1×2=6

Millimoles of total NaOH=300×0.1×1=30

Millimoles of NaOH used =30−6=24

Millimoles of I

2

used =

2

1

millimoles of NaOH used =

2

24

=12 mmol of I

2

used

Total mmol of I

2

used = Part I + Part II =8+12=20 mmol

Molarity of I

2

=

V

m

L

mmol

=

200

20

=0.1 M

20 times the initial M

I

2

=0.1×20=2

Answer 2

Answer:

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