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A sample of water from a large swimming pool has a resistance of `10000Omega` at `25^(@)C` when placed in a certain conductace cell. When filled with `0.02M KCI` solution, the cell has a resistance of `100 Omega` at `25^(@)C, 585 gm` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `8000 Omega`. Given: Molar conductance of `NaCI` at that concentration is `125 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02 M` is `200W^(-1) cm^(2) mol^(-1)`. Volume (in Litres) of water in the pool is:A. `1.25 xx 10^(-5)`B. `1250`C. `12500`D. None of these |
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Answer» Correct Answer - A `k(NaCI) = 10^(-5)` `V = (125)/(100 xx 10^(-5)) = (125)/(10^(-3)) = 1.25 xx 10^(5)L` |
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