0 votes
in CBSE by (90.1k points)
State and explain Kohlrausch's law.

3 Answers

0 votes
by (90.1k points)
According to this law, "molar conductivity of an electrolyte at infinite dilution is expressed as sum of the contributions from its individual ions." These contributions are called ionic conductances of cations and anion.
If \(\lambda^{2}{ }_{+}\)and \(\lambda^{2}\)-are ionic conductances of cation and anion, then
$$
\Lambda_{m}^{\circ}=v_{+} \lambda_{+}^{\circ}+v_{-} \lambda_{-}^{\circ} \ldots(1)
$$
Where \(V_{+}\)and \(V\), represents number of cations and number of anions.
(b) (i) Calculation of molar conductivity of weak electrolyte
Suppose we have to determine \(\wedge=\mathrm{m}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\). For this, limiting molar conductivities of strong electrolytes like \(\mathrm{CH}_{3} \mathrm{COONa}, \mathrm{HCl}\) and \(\mathrm{HCl}\) are determined.
\(\wedge_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)=\lambda^{\circ} \mathrm{CH}_{3} \mathrm{COO}^{-}\)
\(+\lambda_{ }^{\circ} \mathrm{Na}^{+} \ldots(1)\)
\({ }_{\Lambda_{m}{ }^{a}}(\mathrm{HCl})=\lambda{ }^{\circ} \mathrm{H}^{+}+\lambda \mathrm{Cl}^{-} \ldots(2)\)
\({ }^{\wedge}_{m}(\mathrm{NaCl})=\lambda_{\mathrm{Na}^{+}}+\lambda \mathrm{Cl}^{-} \ldots(3)\)
Applying \((1)+(2)-(3)\), we get,
$$
\begin{aligned}
\lambda_{ }^{\circ} \mathrm{CH}_{3} \mathrm{COO}^{-}+\lambda_{\mathrm{Na}^{+}}^{\circ}+\lambda_{ }^{\circ} \mathrm{H}^{+}+\lambda_{\mathrm{Cl}^{-}} & \\
&-\lambda_{\mathrm{Na}}+-\lambda_{\mathrm{Cl}}
\end{aligned}
$$

$$
\begin{gathered}
=\wedge_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)+\wedge_{m}^{\circ}(\mathrm{HCl}) \\
-\wedge_{m}^{\circ}(\mathrm{NaCl}) \\
\text { or, } \lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\circ}+\lambda_{\mathrm{H}^{+}}^{\circ}=\wedge_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right) \\
+\wedge_{m}^{\circ}(\mathrm{HCl})-\wedge_{m}^{\circ}(\mathrm{NaCl}) \\
\text { or, } \wedge_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=\wedge_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right) \\
+\wedge_{m}^{\circ}(\mathrm{HCl})-\wedge_{m}^{\circ}(\mathrm{NaCl})
\end{gathered}
$$
(ii) Calculation of degree of dissociation of weak electrolyte
By the knowledge of limiting molar conductivity of electrolyte, degree of dissociation of weak electrolyte is calculated by the formula-
$$
\propto=\frac{\wedge_{m}^{c}}{\wedge_{m}^{\circ}}
$$
Where \(\Lambda_{\mathrm{m}}^{c}\) is the molar conductivity of solution at any concentration.
\(\wedge_{m}^{0}\) is the limiting molar conductivity.
0 votes
by
cialis 10mg oral <a href="https://ordergnonline.com/">cialis price costco</a> buy ed pills no prescription
0 votes
by
buy cialis online safely <a href="https://ordergnonline.com/">price of cialis</a> best place to buy ed pills online

Related questions

0 votes
3 answers
0 votes
8 answers
asked Dec 19, 2021 in CBSE by kritika (90.1k points)
0 votes
2 answers
0 votes
3 answers
asked Dec 25, 2021 in CBSE by kritika (90.1k points)
0 votes
3 answers
Welcome to Admisure, where you can ask questions and receive answers from other members of the community.
...