InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Resistance of ` 0.2 M` solution of an electrolue is ` 50 Omega`. The specific conductance of the solution is ` 1.4 S m^^(-1)`. The resistance of `0.5` M solution of the same electrolyte is `280. Omega`. The molar conducitivity of ` 0.5 M` solution of the electrolyte is ` S m^2 "mol"^(-1)` is.A. ` 5 xx10^(-4)`B. `5 xx10^(-3)`C. `5 xx10^3`D. `5 xx10^2` |
|
Answer» Correct Answer - A For `0.2 M` solution , ` R= 50 Omega` ` k= 14 S m^(-1) = 1.4 xx 10^(-2) S cm^(-1)` The resistivity of the solution is related to specific cnductance by ` rho = 1/k = 1/(1.4 xx 10^(-2)) Omega cm` It is known that , `R= rho 1/a` It is kinown tht , `R= rho = 1/a` `rArr l/a = R xx 1/(rho) = 50 xx 1.4 xx 10^(-2) cm` for `0.5 M` solution, `R= 280 Omega` ` k=? ` `l/a = 50 xx 1.4 xx 10^(-2) xm` The specific conductance of the solution is given by ` k= 1/(rho) = 1/R xx l/a` ` k= 1/(280) xx 50 xx 1.4 xx 10^(-2)` `2.5 xx 10^(-3) S cm^(-1)` Now , the molar conductivity of the solution is given by `Lambda_m = (K xx 1000)/M = (2.5 xx 10^(-3) xx 1000)/(0.5)` ` 5.5 cm^2 "mol"^(-1)` ` = 5 xx 10^(-4) S m^2 "mol"^(-1)` Hence, the molar conductivity of `0.5 M` solution of the electrolyte is `5xx 10^(-4) Sm^2 "mol"^(-1)`. |
|