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 mL of `H_(2)O_(2)`,
Meq of Kl = Meq. Of `H_(2)O_(2)` in 50 mL = Meq. Of `Na_(2)S_(2)O_(3)`
`(wxx1000)/(34//2)=20xx0.1`
`therefore W_(H_(2)O_(2))" in 50 mL"=(20xx0.1xx34)/(2000)=0.034`
`therefore W_(H_(2)O_(2))" in 1000 mL"=(0.034xx1000)/(50)=0.68g//L`