Equivalent conductances of `NaCl, HCl` and `CH_(3)COONa` at infinite dilution are 126.45, 426.16 and 91 `ohm^(-1) cm^(2) eq^(-1)` respectively. The equivalent conductance of `CH_(3)COOH` at infinite dilution would be :

Options

A. `201.28 Omega^(-1)cm^(2)`
B. `390.71Omega^(-1)cm^(2)`
C. `698.28Omega^(-1)cm^(2)`
D. `540.48Omega^(-1)cm^(-1)`

Answer

B

Explanation

(b) By Kohlrausch's law
`lambda_(oo) " for " NaCl= lambda_(Na^(+))+lambda_(Cl^(-))`…..(i)
`lambda_(oo) " for " HCl= lambda_(H^(+))+lambda_(Cl^(-))`…..(ii)
`lambda_(oo) " for " C_(2)H_(5)COONa= lambda_(Na^(+))+lambda_(C_(2)H_(5)COO^(-))`…..(iii)
So, `lambda_(oo) " for " C_(2)H_(5)COOH` can be obtained by adding
Eqs. (ii) and (iii) and then subtracting Eq. (i)
` =lambda_(oo) "of " C_(2)H_(5)COONa + lambda_(oo) " of " HCl- lambda_(oo) " for " NaCl`
`=(91+426.16-126.45)Omega^(-1)cm^(2)`
`=390.71 Omega^(-1) cm^(2)`