1.

The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at 25^(@)C using a conductance cell with a cell constant 0.88 cm^(-1). Calculate the specific conductance and equivalent conductance of the solution.

Answer»

Solution :`R="210 ohm, "(L)/(a)=0.88cm^(-1)`
`"Specific conductance "KAPPA=(l)/(a)xx(1)/(R )`
`=("0.88cm"^(-1))/("210 ohm")=4.19xx10^(-3)" mho.cm"^(-1)`
`=4.19xx10^(-3)"mho.m"^(-1)`
Equivalent conductance, `lambda=kappa xxV`
V has 1 GRAM equivalent dissolved given is 0.01 N in 1000 ml.
`therefore V=(1000)/(0.01)=1,00,000ml`
`lambda=4.19xx10^(-3)xx1,00,000`
`lambda=419.05"mho. cm"^(2)."gm. equiv"^(-1)`.
`=4.190xx10^(-2)" mho m"^(2)("gm. equiv")^(-1)`


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