At 298 K molar conductivities at infinite dilution `( wedge_(m)^@)` of `NH_(4)Cl`, KOH and KCl are 152.8, 272.6 and 149.8 S `cm^(2)mol^(-1)` respectively. The ` wedge_(m)^@` of `NH_(4)OH` in `Scm^(2) mol^(-1)` and % dissociation of 0.01 M `NH_(4)OH` with `wedge_(m) =25.1 Scm^(2) mol^(-1)` at the same temperature are,

Options

A. 275.6,0.91
B. 275.6,9.1
C. 266.6,9.6
D. 30,84

Answer

B

Explanation

`wedge_(m)^@ NH_(4)OH = wedge_(m)^@ (NH_(4)Cl + KOH) - wedge_(m)^@ (KCl)``therefore wedge_(m)^@ NH_(4)OH = 275.6 Scm_(2) mol^(-1)`
Degree of dissociation `(alpha) = wedge_(m)/ wedge_(m)^@ =25.1/275.6=0.091`
`therefore` degree of dissociation = 9.1%