The molar conductivities ΛNaOAc∞ and ΛHCl∞ at infinite dilution in water at 25 °Care 910 and 426.2 S cm2 mol-1′ respectively. To calculate, ΛNOAc∞ the additional value required is

The molar conductivities ${\mathrm{\Lambda }}_{\mathrm{NaOAc}}^{\mathrm{\infty }}$ and ${\mathrm{\Lambda }}_{\mathrm{HCl}}^{\mathrm{\infty }}$ at infinite dilution in water at 25 °Care 910 and 426.2 S ' respectively. To calculate, ${\mathrm{\Lambda }}_{\mathrm{NOAc}}^{\mathrm{\infty }}$ the additional value required is

1. A

${\mathrm{\Lambda }}_{\mathrm{NaCl}}^{\mathrm{\infty }}$

2. B

${\mathrm{\Lambda }}_{{\mathrm{H}}_{2}\mathrm{O}}^{\alpha }$

3. C

${\mathrm{\Lambda }}_{\mathrm{KCl}}^{\mathrm{\infty }}$

4. D

${A}_{\mathrm{NaOH}}^{\mathrm{\infty }}$

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Solution:

We have ${\mathrm{\Lambda }}_{\mathrm{HOAc}}^{\mathrm{\infty }}={\mathrm{\Lambda }}_{\mathrm{NaOAc}}^{\mathrm{\infty }}+{\mathrm{\Lambda }}_{\mathrm{HCl}}^{\mathrm{\infty }}+{\mathrm{\Lambda }}_{\mathrm{NaCl}}^{\mathrm{\infty }}$ Hence, in addition to ${\mathrm{\Lambda }}^{\mathrm{\infty }}\left(\mathrm{NaOAc}\right)$ and  we must have the value of ${\mathrm{\Lambda }}^{\mathrm{\infty }}\left(\mathrm{NaCl}\right)$

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