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The resistance of 0.15 M solution of an electrolyte is 50Omega. The specific conductance of the solution is 2.4 Sm^(-1). The resistance of 0.5 N solution of the same electrolyte measured using the same conductivity cell is 480Omega. Find the equivalent conductivity of 0.5 N solution of the electrolyte. |
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Answer» Solution :Given that `R_(1)=50 Omega""R_(2)=480Omega` `kappa_(1)=2.4Sm^(-1) ""kappa_(2)=?` `N_(1)=0.15N""N_(2)=0.5N` `wedge=(kappa_(2)(SM^(_1)) times 10^(-3)("gram EQUIVALENT")^(-1)m^(3))/(N)` `""=(0.25 times 10^(-3)S("gram equivalent")^(-1)m^(2))/(0.5)` `wedge=5 times 10^(-4)Sm^(2)" gram equivalent"^(-1)` We know that `KAPPA="Cell constant"/R` `therefore kappa_(2)/kappa_(1)=R_(1)/R_(2)` `kappa_(2)=kappa_(1) times R_(1)/R_(2)` `""=2.4Sm^(-1) times (50 Omega)/(480 Omega)` `""=0.25Sm^(-1)` |
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