Question

Which of the following formulas is/are valid for potash alum? (${\lambda }_{eq}^{\infty }$ is equivalent conductance at infinite dilution)

• A. ${\lambda }_{eq}^{\infty }\left(alum\right)=\frac{1}{4}{\lambda }_{eq}^{\infty }\left(K^{+}\right)+\frac{3}{4}{\lambda }_{eq}^{\infty }\left({Al}^{3+}\right)+{\lambda }_{eq}^{\infty }\left({SO}_4^{2-}\right)$
• B. ${\lambda }_{eq}^{\infty }\left(alum\right)=8{\lambda }_m^{\infty }\left(alum\right)$
• C. ${\lambda }_{eq}^{\infty }\left(alum\right)=\frac{1}{4}{\lambda }_{eq}^{\infty }\left(K^{+}\right)+\frac{1}{3}{\lambda }_{eq}^{\infty }\left({Al}^{3+}\right)+{\lambda }_{eq}^{\infty }\left({SO}_4^{2-}\right)$/
• D. ${\lambda }_{eq}^{\infty }\left(alum\right)=\frac{1}{4}{\lambda }_{eq}^{\infty }\left(K^{+}\right)+\frac{3}{2}{\lambda }_{eq}^{\infty }\left({Al}^{3+}\right)+{\lambda }_{eq}^{\infty }\left({SO}_4^{2-}\right)$

Answer 1

Answer: (A), (B)

The correct options are
A ${\lambda }_{eq}^{\infty }\left(alum\right)=\frac{1}{4}{\lambda }_{eq}^{\infty }\left(K^{+}\right)+\frac{3}{4}{\lambda }_{eq}^{\infty }\left({Al}^{3+}\right)+{\lambda }_{eq}^{\infty }\left({SO}_4^{2-}\right)$
B ${\lambda }_{eq}^{\infty }\left(alum\right)=8{\lambda }_m^{\infty }\left(alum\right)$