Question

Equivalent conductance of $\text{NaCl, HCl}$ and $\text{C}{\text{H}}_{\text{3}}\text{COONa}$ at infinite dilution are 126.45, 425.16 and 91 ohm${}^{-1}$ cm${}^2$ respectively. The equivalent condcutance of $\text{C}{\text{H}}_{\text{3}}\text{COOH}$ at infinite dilution would be:

• A. $101.38\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}$
• B. $253.62\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}$
• C. $389.71\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}$.
• D. $678.90\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}$

Answer 1

The correct option is C $389.71\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}$.

$C{H}_{3}COONa+HCl⟶C{H}_{3}COOH+NaCl$

${\lambda }_{eq\left(C{H}_{3}COOH\right)}^{\infty }={\lambda }_{eq\left(C{H}_{3}COONa\right)}^{\infty }+{\lambda }_{eq\left(HCl\right)}^{\infty }-{\lambda }_{eq\left(NaCl\right)}^{\infty }$

${\lambda }_{eq\left(C{H}_{3}COOH\right)}^{\infty }=91+425.16-126.8=389.71oh{m}^{-1}c{m}^{2}eq$

where ${\lambda }_{eq}^{\infty }=$ Equivalence conductivity at infinitie dilution.

Hence, the correct answer is $\left(C\right)$