InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The ionization constant of a weak acid is `1.6xx10^(-5)` and the molar conducitvity at infinite dilutionis `380xx10^(-4)" S " m^(2) mol^(-1)`. If the cell constant is `0.01 m^(-1)`, then the conductance of 0.01 M solution is :A. `1.52xx10^(-5)" S "`B. `1.52" S "`C. `1.52xx10^(-3)" S "`D. `1.52xx10^(-4)" S "` |
|
Answer» Correct Answer - D (d) For a weak acid `K_(a)=C alpha^(2)` `K_(a)=C((lambda_(m)^(c ))/(lambda_(m)^(alpha)))^(2)` `1.6xx10^(-5)=0.01((lambda_(m)^(c ))/(380xx10^(-4)))^(2)` `16xx10^(-6)=10^(-2)((lambda_(m)^(c ))/(380xx10^(-4)))^(2)` `(lambda_(m)^(c ))/(380xx10^(-4))=(16xx10^(-4))^(1//2)=4xx10^(-2)` `lambda_(m)^(c )=(380xx10^(-4))xx(4xx10^(-2))` `=152xx10^(-5)=1.52xx10^(-3)` |
|