Question

The values of $\Lambda_{e q}^{\infty}$ for $NH_{4}Cl$ , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 $ohm^{- 1}cm^{2}eq^{- 1}$ . The value of $\Lambda_{e q}^{\in fty}NH_{4}OH$ is
$\left(\right.$ in ohm $\left.{ }^{-1} cm ^2 eq ^{-1}\right)$

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

Answer 1

Answer: (B)

$\Lambda_{\text {eq }}^{\infty}$ for $NH _4 Cl$ can be find out by the relation given below
$\Lambda_{ eq }^0$ or $\Lambda^{\infty}\left( NH _4 OH \right)=\Lambda^{\infty} eq \left( NH _4 Cl \right)+\Lambda^{\infty} eq ( NaOH )-\Lambda^{\infty} eq ( NaCl )$
$=(149.74+248.1-126.4)$
$=271.44 ohm ^{-1} cm ^2 eq ^{-1}$