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Identify A to E |
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Answer» Solution :(C ) is colourlesssoluble solution (C ) gives blackppt , with `H_(2)SHCI` `RARR` (C ) isof group II (C ) gives yellow ppt , with `KI` as WELL as with `K_(2)CrO_(4) rArr (c )` has `Pb^(2+)` THUS (C) is `Pb(NO_(3))_(2)` obtained from (A),(B) on HEATINGWITH `Mn^(2+)` and `HNO_(3)` gives purple colour whichis dueto oxidationof `Mn^(2+)`in `HMaO_(4)`(purple) `rArr`(B)is oxidising agent `rArr` (B) is `PbO_(2)` (blackish brown ) `rArr`(A) is doubled oxide of lead (A ) is `Pb_(2)O_(4)` (red lead )`(2PbO.PbO_(2))` `underset((A))(Pb_(3)O_(4)) + 4HNO_(3) rarr underset((B))(PbO_(2))+ underset((C))(2Pb(NO_(3))_(2)) + 2H_(2)O` `underset((C))(Pb(NO_(3))_(2)) + 2KI rarr underset("Yellow")(PbI_(2)darr)` `Pb (NO_(3))_(2) +H_(2)S rarr underset((D)black)(PbCrO_(4)darr)` `PbO_(2) +Mn^(2+) +HNO_(3) rarr underset((E)"pink")(HMnO_(4)) +Pb^(2+)` |
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