Excess of Kl and dill. `H_(2)SO_(4)` were mixed in 50 mL `H_(2)O_(2)`. Thus, `l_(2)` liberated requires 20 mL of 0.1 N `Na_(2)S_(2)O_(3)`. What will be the strenght of `H_(2)O_(2)` in g `L^(-1)` ?

Options

A. 0.034
B. 0.68
C. 6.8
D. 5.8

Answer

B

Explanation

For 20 m L of `H_(2)O_(2)`
`Meq. Kof kI = Meq. of H_(2)O_(2) in 50 mL = Meq. Of Na_(2) S_(2) O_(3)`
`(Wxx 100)/(34//2)= 20 xx 0.1`
`therefore W_(H_(2)O_(2))" in" 50 mL` = `(20 xx 0.1 xx 34)/(2000)= 0.034`
`therefore W_(H_(2)O_(2))" in " 1000 mL` = `(0.034 xx 1000)/(50)= 0.68g//L`