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)`.

Answer

`2 xx 10^(5) dm^(3)`

Explanation

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`