Question

If the values of $\mathop \wedge \nolimits_m$ of $\mathop {NH}\nolimits_4 \mathop {Cl}\nolimits_{}$ , $\mathop {NaOH}\nolimits_{}$ and are 130 , 217 and 109 $\mathop {ohm}\nolimits^{ - 1} \mathop {cm}\nolimits^2$.$equiv^{-1}$ respectively , the $\mathop \wedge \nolimits_m$ $\mathop {NH}\nolimits_4 \mathop {OH}\nolimits_{}$ in $\mathop {ohm}\nolimits^{ - 1} .\mathop {cm}\nolimits^2 \mathop {.equiv}\nolimits^{ - 1}$is
A.238
B.198
C.22
D.456

Answer 1

The molar conductivity of an electrolyte at infinite dilution or as concentration approaches zero acquires a unique value with each ion making a definite contribution towards it. The interionic reactions are negligibly small. Kohlrausch made these observations and proposed the law.
At infinite dilution the molar conductivity of an electrolyte can be expressed as the sum of contributions from its individual ions.

Complete answer:
Let us use this method to calculate of a weak electrolyte $\mathop \wedge \nolimits_m$ $\mathop {NH}\nolimits_4 \mathop {OH}\nolimits_{}$. For this purpose we take three strong electrolytes $\mathop {NH}\nolimits_4 \mathop {Cl}\nolimits_{}$ , $\mathop {NaOH}\nolimits_{}$ and $\mathop {NaCl}\nolimits_{}$ and determine $\mathop \wedge \nolimits_m$ values separately by extrapolation method. Then according to Kohlrausch law.
$\mathop \wedge \nolimits_m$ : $\mathop {NH}\nolimits^4 \mathop {Cl}\nolimits_{}$ = $\mathop \wedge \nolimits_m$ $\mathop {NH}\nolimits_4^ + + \mathop \wedge \nolimits_m \mathop {Cl}\nolimits^ -$
$\mathop \wedge \nolimits_m$ $\mathop {NaOH}\nolimits_{}$= $\mathop \wedge \nolimits_m$ $\mathop {Na}\nolimits^ + + \mathop \wedge \nolimits_m \mathop {OH}\nolimits^ -$
$\mathop \wedge \nolimits_m$ $\mathop {NaCl}\nolimits_{}$= ∧∞$\mathop {Na}\nolimits^ + + \mathop \wedge \nolimits_m \mathop {Cl}\nolimits^ -$

$\mathop \wedge \nolimits_m \left( {N{H_4}OH} \right) = \mathop \wedge \nolimits_m \left( {N{H_4}Cl} \right) + \mathop \wedge \nolimits_m \left( {NaOH} \right) - \mathop \wedge \nolimits_m \left( {NaCl} \right)$

$\mathop \wedge \nolimits_m \left( {N{H_4}OH} \right) = 130 + 217 - 109$

$\mathop \wedge \nolimits_m \left( {N{H_4}OH} \right) = 238\mathop {ohm}\nolimits^{ - 1} .\mathop {cm}\nolimits^2 \mathop {.equi}\nolimits^{ - 1}$

238 - by using Kohlrausch law when we found the molar conductivity of ammonium hydroxide it comes out to be 238 . Thus this option is correct.
198 - by using Kohlrausch law when we found the molar conductivity of ammonium hydroxide it does not come out to be 198 . Hence this option is not correct.
22 - this value is not in accordance with Kohlrausch law. Hence this option is not correct.
456 - it is the double of the value coming hence this option is not correct.
Our required and is a that is 238

So, the correct answer is Option A.

Note:
Sometimes, the molar conductivities of the ions are not available. In such cases following procedure is adopted
Selecting three strong electrolytes such that sum/difference of molar conductivities of their ions gives molar conductivities of the ions of weak electrolyte.
m values of these strong electrolytes are measured separately by extrapolation method.
Add and subtract these values to get m of the weak electrolyte.