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The conductivity of sodium chloride at 298K has been determined at different concentrations and the results are given below: Calculate Lamda_(m) for all concentration and draw a plot between Lamda_(m) and c^((1)/(2)). Find the value of Lamda_(m)^(@). |
Answer» Solution :* Here, the value of given concentration and concentration and conductivity and the values of `c^((1)/(2))` and `Lamda_(m)` re given in the following table.  * Plotting the graph, from the calculation of above table `c^((1)/(2))` is on X-axis and its respective value of `Lamda_(m)` on the Y-axis, we will get following graph.  * This graph having negative slopping linear line. Value of `Lamda_(m)^(0)" S "m^(2)MOL^(-1)`: Value of `c^((1)/(2))=0` which is the value of `Lamda_(m)^(0)` `k=1,237xx10^(-1)S " "m^(-1),c=0.001M` Then, `k=1.237xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.0316M^((1)/(2))` `therefore Lamda_(m)=(k)/(c)=(1.237xx10^(-2)" S "cm^(-1))/(0.001" mol "L^(-1))XX(1000cm^(3))/(L)` `=123.7" S "cm^(2)mol^(-1)` Given, `k=11.85xx10^(-2)" S "m^(-1),c=0.010M` Then, `k=11.85xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.01M^((1)/(2))` `=118.5" S "cm^(2)mol^(-1)` Given, `k=23.15xx10^(-2)" S "m^(-1),c=0.020M` Then, `k=23.15xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.1414M^((1)/(2))` `therefore Lamda_(m)=(k)/(c)=(23.15xx10^(-4)" S "cm^(-1))/(0.020" mol "L^(-1))xx(1000cm^(3))/(L)` `=115.8" S "cm^(2)mol^(-1)` Given, `k=55.53xx10^(-2)" S "m^(-1),c=0.050M` Then, `k=55.53xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.2236M^((1)/(2))` `therefore k=(k)/(c)=(55.53xx10^(-4)" S "cm^(-1))/(0.050" mol "L^(-1))xx(1000cm^(3))/(L)` `=111.1" S "cm^(2)mol^(-1)` Given, `k=106.74xx10^(-4)" S "m^(-1),c=0.3162M` `therefore Lamda_(m)=(k)/(c)=(106.74xx10^(-4)" S "cm^(-1))/(0.100" mol "L^(-1))xx(1000cm^(3))/(L)` `=106.74" S "cm^(2)mol^(-1)` Now, we have the following data:  Since the line interrupts `Lamda_(m)` at `124.0" S "cm^(2)` `mol^(-2),Lamda_(m)^(@)=124.5" S "cm^(2)mol^(-1)`. |
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