1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq cm in area was found to offer a resistance of 50 ohm. Calculate t
1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq cm in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution.
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Given, `l=2.1` cm, a=4.2 sq cm, R=50 ohm
Specific conductance, `kappa=l/a. 1/R`
or `kappa=2.1/4.2xx1/50=0.01 ohm^(-1) cm^(-1)`
Equivalent conductivity `= kappa xx V`
V = the volume containing 1 g-equivalent =1000 mL
So, Equivalent conductivity `=0.01 xx 1000`
`=10 ohm^(-1) cm^(2) eq^(-1)`
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