If the values of `wedge_(infty)` of `NH_(4)Cl, NaOH` and NaCl are 130, 217 and 109 `ohm^(-1) cm^(2) equiv^(-1)` respectively, the ` wedge_(infty)` of `NH_(4)OH in ohm^(-1) cm^(2) equiv^(-1)` is

Options

A. 238
B. 196
C. 22
D. 456

Answer

A

Explanation

`wedge_(infty) NH_(4)Cl = wedge_(infty) NH_(4)^(+) + wedge_(infty) Cl^(-)= 130` ....... (1)
`wedge_(infty) NaOH = wedge_(infty) Na +wedge_(infty) OH^(-)=217` .....(2)
`wedge_(infty) NaCl = wedge_(infty) Na^(+) + wedge_(infty) Cl^(-) = 109` ........(3)
From the equations (1), (2) and (3), we get `wedge_(infty) NH_(4)OH = 130+217-109 = 238 ohm^(-1) cm^(2) equiv^(-1)`