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InterviewSolution
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A sample of water from a large swimming pool has a resistance of `9200 Omega` at `25^(@)C` when placed in a certain conductance cell. When filled with `0.02M KCI` solution the cell has a resistance of `85 Omega` at `25^(@)C. 500g` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `7600 Omega`. calculate the volume of water in the pool. Given: Molar conductance of `NaCI` at that concentration is `126.5 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02M` is `138 Omega^(-1) cm^(2) mol^(-1)`. |
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Answer» Correct Answer - `2 xx 10^(5) dm^(3)` For `KCI` `lambda_(M) = (k xx 1000)/(M) rArr 138 (k xx 1000)/(0.02)` `k = 2.76 xx 10^(-3) = (1)/(R) ((l)/(a)) = (1)/(85) ((l)/(a))` `(l//a) = 0.2346` For `H_(2)O: k_(H_(2)O) = (1)/(9200) (l//a)` For NaCI: `k_(NaCI) = (1)/(7600) (l//a)` `lambda_(M) = ((k_(NaCI-k_(H_(2)O)))xx1000)/(M)` `126.5=(((1)/(7600)-(1)/(9200))xx0.2346xx1000)/(M)` ` = 4.2438 xx 10^(-5)` `M = (500)/(58.5xxV) =4.2438 xx 10^(-5)` `V = 201400 L` `V = 2.014 xx 10^(-5)L` |
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