Question

Ionic conductance at infinite dilution of $Al^{3 +}$ and $SO_{4}^{2 -}$ ions are $189ohm^{- 1}cm^{2}eq^{- 1}$ and $160ohm^{- 1}cm^{2}eq^{- 1}$ respectively. Calculate the $\lambda_{ eq }^{\infty}$ of $Al _2\left( SO _4\right)_3:-$

• A. $349ohm^{- 1}cm^{2}eq^{- 1}$
• B. $143ohm^{- 1}cm^{2}eq^{- 1}$
• C. $720ohm^{- 1}cm^{2}mole^{- 1}$
• D. $120ohm^{- 1}cm^{2}eq^{- 1}$

Answer 1

Answer: (A)

$\Lambda_{ eq }^{\infty} Al _2\left( SO _4\right)_3=\wedge_{ eq }^{\infty}\left( Al ^{+3}\right)+\lambda_{ eq }^{\infty}\left( SO _4^{-2}\right)$
$189+160$
$=349ohm^{- 1}cm^{2}eq^{- 1}$