InterviewSolution
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InterviewSolution
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`500ml` of `0.1MKCI, 200ml` of `0.01M NaNO_(3)` and `500ml` of `0.1M AgNO_(3)` was mixed. The molarity of `K^(+), Ag^(+),CI^(-),Na^(+),NO_(3)^(-)` in the solution would beA. `[K^(+)] = 0.04, [Ag^(+)] = 0.04, [Na^(+)] = 0.002`B. `[K^(+)] = 0.04, [Na^(+)] = 0.00166, [NO_(3)^(-)] = 0.04333`C. `[K^(+)] = 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025`D. `[K^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO_(3)^(-)] = 0.0525` |
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Answer» Correct Answer - B `[K^(+)] = (0.1 xx 500)/(500 +200 +500) = (50)/(1200) = 0.04167M` `[Na^(+)] = (0.01 xx 200)/(500 +200 +500) = (2)/(!200) = 0.00167M` `[NO_(3)^(-)] = (200 xx 0.01 xx 500 xx 0.1)/(500 +200 +500) = (52)/(1200) = 0.0433M` Since 50 milli moles of `Cl^(-)` are mixed with 50 milli moles of `Ag^(+)` ions, total `Ag^(+)` & `Cl^(-)` get precipitated as `AgCl`. Hence the solution does not contain `Ag^(+)` & `Cl^(-)` ions |
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