InterviewSolution
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InterviewSolution
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Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .A. ` 6250 S m^2 "mol"^(-1)`B. ` 6.25 xx 10^(-4) Sm^2 "mol"^(-1)`C. `625 xx 10^(-4) S m^2 "mol"^(-1)`D. ` 62.5 S m^2 "mol"^(-1)` |
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Answer» Correct Answer - B For ` 0.2 M` solution. Specific conductance k = Conductance `(1/R)` `x` Cell constant `(l//a)` `rArr k = 1/(50) xx l/a = 1.3 Sm^(-1)` `rArr 1/a = 13. xx 50 m^(-1)` `1/a = 1.3 xx 50 xx 10^(-2) cm^(-1)` For `0.4 M` solutin, `R= 2600 Omega` `R= rho 1/a` `rArr 1/(rho) = k = 1/R xx 1/a` ` k= 1/(260) x 1.3 xx 50 xx 10^(-2) S cm^(-1)` Molar conductivity of the solution is given by `Lambda_m = (kxx 1000)/M` `1/(260) xx (1.3 xx 50 xx 10^(-2) xx 1000)/4` ` =6. 26 S cm^2 "mol"^(-1) 6, 25 xx 10^(-4) m^2 "mol"^(-1)`. |
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