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Resistance of a conductivity cell filled with 0.1 mol L^(-1)KCl solution is 100 ohms. If the resistance of the same cell filled with 0.02 mol L^(-1)KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 mol L^(-1) KCl solution. The conductivity of 0.1 mol L^(-1)KCl solution is 1.29 xx 10^(-2) S cm^(-1) |
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Answer» SOLUTION :Cell constant, `G^(@)` = Conductivity `XX` Resonance `=1.29 xx 10^(-2) S cm^(-1) xx 100` ohm `=1.29 cm^(-1)` Conductivity of 0.02 mol `L^(-1)` KCl `=G^(@)/("Resistance") = (1.29 cm^(-1))/(520 ohm) = 0.00248 S cm^(-1)` `Lambda_(m) = (K xx 1000)/("concentration") = (2.48 xx 10^(-3) xx 1000)/0.02 = 124 S cm^(2) mol^(-1)` |
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