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Three solutions eaach of 100 mL containing 0.4 M `As_(2)S_(3),5M` NaOH and `6MH_(2)O_(2)`, respectively were mixed to form `AsO_(4)^(3-)` and `SO_(4)^(2-)` as products. Q. When the above solution is allowed to stand for some time what volume of `O_(2)` will be obtained `STP`?A. 0.112 LB. 0.224 LC. 0.224LD. 0.448L |
Answer» Correct Answer - C If a `H_(2)O_(2)` solution is allowed to stand, it decomposes to `O_(2)` and `H_(2)O` `underset(40)(H_(2)O_(2)tounderset(0)(H_(2)O)+(1)/(2)underset(0)(O_(2))` `{:(mmol es),(of H_(2)O_(2)),(l eft)]:}-(1)/(2)xx40=20m" mol of "O_(2)` Thus, mmol es of O_(2)formed`=40xx(1)/(2)=20` Volume of `O_(2)` at `STP=20xx10^(-3)xx22.4L` `(because1 " mol of "O_(2) at STP=22.4L)` `=0.448L` |
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