InterviewSolution
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InterviewSolution
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The reaction between gaseous `NH_(3)` and `HBr` produces a white solide `NH_(4)Br`. Suppose a small quantity of gaseous `NH_(3)` and gaseous `HBr` are introduced simultaneously into opposite ends of an open tube which is one metre long. Calculate the distance of white solid formed from the end which was used to introduce `NH_(3)` |
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Answer» Let the distance of white solid from `NH_(3) " end" x = cm` The distance of which solid from `HBr " end " = (100 - x) cm`. Rates of diffusion shall be proportional to these distances. `(r_(1))/(r_(2)) = (x)/((100 - x)) = sqrt((M_(HBr))/(M_(NH_(3))))` Mol. mass of `HBr = 1 + 80 = 81` Mol. mass of `NH_(3) = 14 + 3 = 17` So, `(x)/((100 - x)) = sqrt((81)/(17))` or, `(x)/((100 - x)) = 2.18` So, `x = 100 xx 2.18 - 2.18 x` So, `x = (100 xx 2.18)/(3.18) = 68.55 cm` |
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