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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)` ?A. 0.034B. 0.68C. 6.8D. 5.8 |
Answer» Correct Answer - B 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` |
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