The resistance of 0.15M solution of an electrolyte is 50 Omega. 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 480 Omega. Find the equivalent conductivity of 0.5 N solution of the electrolyte.

The resistance of 0.15M solution of an electrolyte is 50 Omega. The sp

1.	The resistance of 0.15M solution of an electrolyte is 50 Omega. 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 480 Omega. Find the equivalent conductivity of 0.5 N solution of the electrolyte.
Answer» Solution :GIVEN that `R_1 = 50 Omega , R_2 = 480 Omega` `k_1 = 2.4 Sm^(-1) k_2 = ?` `N_1 = 0.15 N N_2 = 0.5 N` `LAMBDA = (k_2(Sm^(-1)) xx 10^(-3)("GRAM equivalent")^(-1)m^(3))/(N) = (0.25 xx 10^(-3)S("gram equivalent")^(-1)m^2)/(0.5)` `= 5 xx 10^(-4) Sm^(2) "gram equivalent"^(-1)` We KNOW that `k = ("Cell constant")/R "" :. (k_2)/(k_1) = (R_1)/(R_2)` `k_2 = k_1 xx (R_1)/(R_2) = 2.4 Sm^(-1) xx (50 Omega)/(480 Omega) = 0.25 Sm^(-1)`.