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To a 25 mL of H(2)O(2) solution, excess ...

To a `25 mL` of `H_(2)O_(2)` solution, excess of acidified solution of `KI` was added. The iodine liberated required `20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution. Calculate the volume strength of `H_(2)O_2` solution.

A

`1.344g//L`

B

`3.244g//L`

C

`5.4g//L`

D

`4.08g//L`

Text Solution

Verified by Experts

The correct Answer is:
D
`2KI+H_(2)SO_(4)+H_(2)O_(2)rarrK_(2)SO_(4)+2H_(2)O+I_(2)`
`2Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+"2NaI"`
`"milli eq. of "H_(2)O_(2)" in 50 ml = milli eq. of "I_(2)`
`="milli eq. of "Na_(2)S_(2)O_(3)`
`"milli eq. of "H_(2)O_(2)" in 25 ml "=20xx0.3=6`
`"milli eq. of "H_(2)O_(2)" in 1000 ml "=(6)/(25)xx1000=240`
Equivalent per litre `=(240)/(1000)`
Gram per litre of `H_(2)O_(2)=(240)/(1000)xx17=4.08g//L`
(Equivalent weight of `H_(2)O_(2)=(34)/(2)=17`).
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