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. 0.20
C. 0.15
D. 0.03

Answer

B

Explanation

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