To a 25 mL `H_(2)O_(2)` solution excess of an acidified solution of potassium iodide was added. The iodine liberated required 20 " mL of " 0.3 N sodium thiosulphate solution Calculate the volume strength of `H_(2)O_(2)` solution.

Options

A. (a)`1.344 g//L`
B. (b)`3.244 g//L`
C. (c )`5.4 g//L`
D. (d)`4.08 g//L`

Answer

d

Explanation

`2KI+H_(2)SO_(4)+H_(2)O_(2) rarr K_(2)SO_(4)+2H_(2)O+I_(2)`
`2Na_(2)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+2Nal`
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`
`=6/25xx1000=240`
Equivalent per litre`=240/1000`
Gram per litre of `H_(2)O_(2)=240/1000xx17=4.08 g//L`
`("Equivalent weight" of H_(2)O_(2)=34/2=17)`