Saved Bookmarks
| 1. |
Conducitivity of 2.5 xx 10^(-4) M methonic acid is 5.25 xx 10^(-5) S cm^(-1). Calculate the molar conductivity and degree of dissociation. Given: lambda^(@) (H^(+)) = 349.5 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 50.5 S cm^(2) mol^(-1) |
|
Answer» SOLUTION :`Lambda_(HCOOH)^(@)= Lambda_(H^(+))^(@) + Lambda_(HCOO^(-))^(@) =349.5 + 50.5 = 400 S cm^(2) mol^(-1)` `Lambda_(HCOOH) = K xx 1000 cm^(3) L^(-1) "MOLARITY"^(-1)` `=(5.25 xx 10^(-5) S cm^(-1) xx 1000 cm^(3) L^(-1))/(2.5 xx 10^(-4)"mol" L^(-1))= 210 S cm^(2) mol^(-1)` Degree of DISSOCIATION `(ALPHA) = Lambda_(m)/Lambda_(,)^(@) = 210/400 = 0.525` or `52.5%` |
|