The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectiley 425, 90 and 125 mho `cm^(2) " mol"^(-1)` and `25^(@)C`. The molar conductivity of 0.1M `CH_(3)COCH` solution is 7.8 mho `cm^(2) "mol"^(+1)` at the same temperature then degree of dissociation is

Options

A. `0.10`
B. `0.02`
C. `0.15`
D. `0.03`

Answer

B

Explanation

`Lambda^(@) (CH_(3) COOH)`
` = (lambda_(CH_(3) COO^(-))^(@) + lambda_(Na^(+))^(@) ) + (lambda_(H^(+))^(@) + lambda_(Cl^(-))^(@)) - (lambda_(Na^(+)) + lambda_(Cl^(-)))`
`= 90 + 425 - 125 = 390 mho cm^(2) mol^(-1)`
Degree of dissociation= `(Lambda^(c))/(Lambda^(@)) = (7.8)/(390) = 0.02`