Search any question & find its solution
Question:
Answered & Verified by Expert
The limiting molar conductivities $\wedge^0$ for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126,152 and $150 \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ respectively. The $\wedge^0$ for $N a B r$ is
Options:
Solution:
2357 Upvotes
Verified Answer
The correct answer is:
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\left(152 \mathrm{scm}^2\right) \wedge_{K B r}^0=\wedge_{X^{+}}^0+\wedge_{\zeta \zeta^{-}}^0$ ...(i)
$\left(126 \mathrm{scm}^2\right) \wedge_{\mathrm{NGCl}}^0=\wedge_{\mathrm{Nb}^{+}}^0+\wedge_{\mathrm{cr}}^0$ ...(ii)
$\left(150 \mathrm{scm}^2\right) \wedge_{\mathrm{KCl}}^0=\wedge_{\mathrm{K}^{+}}^0+\wedge_{\mathrm{Cl}}^0$ ...(iii)
By equation (i)+(ii)-(iii)
$=126+152-150=128 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
\(\begin{array}{|l|l|l|l|l|l|} \hline \text { Electrolyte: } & \mathrm{KCl} & \mathrm{KNO}_3 & \mathrm{HCl} & \mathrm{NaOAc} & \mathrm{NaCl} \\ \hline \Lambda^*\left(\mathrm{Scm}^2 \mathrm{~mol}^{-1}\right): & 149.9 & 145.0 & 426.2 & 91.0 & 126.5 \\ \hline \end{array}\)
$\left(126 \mathrm{scm}^2\right) \wedge_{\mathrm{NGCl}}^0=\wedge_{\mathrm{Nb}^{+}}^0+\wedge_{\mathrm{cr}}^0$ ...(ii)
$\left(150 \mathrm{scm}^2\right) \wedge_{\mathrm{KCl}}^0=\wedge_{\mathrm{K}^{+}}^0+\wedge_{\mathrm{Cl}}^0$ ...(iii)
By equation (i)+(ii)-(iii)
$=126+152-150=128 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
\(\begin{array}{|l|l|l|l|l|l|} \hline \text { Electrolyte: } & \mathrm{KCl} & \mathrm{KNO}_3 & \mathrm{HCl} & \mathrm{NaOAc} & \mathrm{NaCl} \\ \hline \Lambda^*\left(\mathrm{Scm}^2 \mathrm{~mol}^{-1}\right): & 149.9 & 145.0 & 426.2 & 91.0 & 126.5 \\ \hline \end{array}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.