Correct Answer - B
The degree of ionisation `(alpha)` of ammonium hydrox-ide at the same temperature and concentration is given as follows:
`alpha=(Lambda_(m)^(c))/(Lambda_(m)^(oo))`
`= (9.54 ohm^(-1)cm^(2)mol^(-1))/(238 ohm^(-1)cm^(2)mol^(-1))`
`= 0.04008`
Therefore, percentage ionisation will be `= 0.04008 xx 100% = 4.008%`