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The conductance of a salt solution (AB) ...

The conductance of a salt solution (AB) measured by two parallel electrodes of area `100 cm^(2)` separated by 10 cm was found to be `0.0001 Omega^(-1)`. If volume enclosed between two electrode contains 0.1 mole of salt, and the molar conductivity `(S cm^(2) mol^(-1))` of salt at same concentration is `1.0 xx 10^(-x)`, x is

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The correct Answer is:
1
`C = 10^(-4)Omega , a = 100 cm^(2) , l =10 cm , R = P. (1)/(a) implies (1)/(C) = (1)/(K) xx (1)/(a) implies K = C xx (1)/(a) = 10^(-4) xx (10)/(100)= 10^(-5)`
` M = (n)/(V) = (0.1)/( 100 xx 100) xx 1000 = 0.1M , ^^_M = (K xx 1000)/(M) = (10^(-5) xx 1000)/(0.1) = 0.1 = 1 xx 10^(-x) = 1 xx 10^(-1)`
x=1
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