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  • The limiting molar conductivities of <img alt="\mathrm{NaI}, \mathrm{NaNO}_{3}$ and $\mathrm{AgNO}_{3}" src="/latex-image/?%5Cmathrm%7BNaI%7D%2C%20%5Cmathrm%7BNaNO%7D_%7B3%7D%24%20and%20%24%5Cmathr

The limiting molar conductivities of \mathrm{NaI}, \mathrm{NaNO}_{3}$ and $\mathrm{AgNO}_{3} are 12.7,12.0$ and $13.3 \mathrm{mS} \mathrm{m}^{2} \mathrm{~mol}^{-1}, respectively (all at 25^{\circ} \mathrm{C} ). The limiting molar conductivity of \mathrm{AgI} at this temperature is_________\mathrm{mS} \mathrm{m}^{2} \mathrm{~mol}^{-1}.

Option: 1

14


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\Lambda_{m}^{\infty}(\operatorname{Ag} I)} =?

From Kohlrausch's law of infinite dilution, we have 

\mathrm{\Lambda_{m}^{\infty}\left(\mathrm{Ag\; I}\right)=\Lambda_{m}^{\infty}\left(\mathrm{AgNO}_{3}\right)+\Lambda_{m}^{\infty}(\mathrm{NaI})-\Lambda_{m}^{\infty}\left(\mathrm{NaNO}_{3}\right)}

                       \begin{aligned} &=13.3+12.7-12.0 \\ &=26.0-12.0 \\ &=14\; \mathrm{mS} \mathrm{m}^{2} \mathrm{~mol}^{-1} \end{aligned}

Hence, the answer is 14.

Posted by

Rakesh

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