Question

The values ${\wedge }_{eq}^{\infty }$ for $N{H}_{4}Cl,NaOH$ and $NaCl$ are 149.74, 248.1 and 126.4 $oh{m}^{-1}$ $c{m}^2$ $eq{v}^{-1}$ . The value of ${\wedge }_{eq}^{\infty }$ of $N{H}_{4}OH$ :

• A. 371.44
• B. 271.44
• C. 71.44
• D. It cannot be calculated from the data given

Answer 1

The correct option is B 271.44

According to kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivites of its ions.
Hence,
${\Lambda }_{eq}^{\infty }\left({NH}_{4}OH\right)={\Lambda }_{eq}^{\infty }\left({NH}_{4}Cl\right)+{\Lambda }_{eq}^{\infty }\left(NaOH\right)-{\Lambda }_{eq}^{\infty }\left(NaCl\right)$
Substitute values in the above equation
${\Lambda }_{eq}^{\infty }\left({NH}_{4}OH\right)=149.74+248.1-126.4=271.44{ohm}^{-1}{cm}^2{eq}^{-1}$

Hence, option B is correct.