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The molar conductivity (Lambda_(m)) of KCl solutions at different concentrations at 298 K is plotted as shown in the figure below: Determine the value of Lambda_(m)^(@) and A for KCl. |
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Answer» Solution :On extrapolating the STRAIGHT line to y-axis, it MEETS at `Lambda_(m) = 150 S cm^(2) MOL^(-1)` `Lambda_(m)^(@) = 150 S cm^(2) mol^(-1)` A=- SLOPE, Taking points (1) and (2) `A = (-(148.2-149.1) S cm^(2) mol^(-1))/((0.02 - 0.01)(mol L^(-1))^(1//2)) = 90 S cm^(3) mol^(-1)//(mol L^(-1))^(2)` |
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