A conductance cell was calibrated by filling it with a 0.02 M solution of potassium chloride (specific conductance `= 0.2768 ohm^(-1) m^(-1)`) and measuring the resistance at 298 K which was found to be 457.3 ohm. The cell was then filled with a calcium chloride solution containing 0.555 g of `CaCl_(2)` per litre. The measured resistance was 1050 ohm. Calculate the molar conductivity of `CaCl_(2)` solution.

Answer

`0.0241` mho `m^(2) mol^(-1)`

Explanation

Determine cell constant with KCl data. Its value is `126.6 m^(-1)`.
Sp. Cond. Of `CaCl_(2)` soln. = Cell const. `xx 1/("Resistance")`
`=126.6xx1/1050=0.1206 ohm^(-1) m^(-1)`
Molar conductance `=("Sp. Cond.)/("Conc.)`
Conc. `=0.555/111.0=0.005" mol dm"^(-3) =5" mol m"^(-3)`