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. 0.1
B. 0.02
C. 0.15
D. 0.03

Answer

B

Explanation

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