The following figure shows the dependence of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Electrolyte $B$ is a strong electrolyte,and electrolyte $A$ is a weak electrolyte.Statement $(A)$ is incorrect because $\Lambda_m^0$ for a weak el

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Electrolyte $B$ is a strong electrolyte,and electrolyte $A$ is a weak electrolyte.Statement $(A)$ is incorrect because $\Lambda_m^0$ for a weak electrolyte $(A)$ cannot be obtained by extrapolation; it is calculated using Kohlrausch's law.Statement $(B)$ is correct because strong electrolytes follow the Debye-$H$ückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$,which is a straight line.Statement $(C)$ is incorrect because at infinite dilution,the degree of dissociation $(\alpha)$ for any electrolyte approaches $1$ (or $100\%$),not zero.Statement $(D)$ is correct because Kohlrausch's law of independent migration of ions states that $\Lambda_m^0$ can be calculated for any electrolyte using the limiting molar conductivities of its individual ions.Therefore,statements $(A)$ and $(C)$ are incorrect. The total number of incorrect statements is $2$.

Similar Questions

The values of $\lambda^{\infty}_m$ for $NH_4Cl$,$NaOH$,and $NaCl$ are $129.8$,$248.1$,and $126.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively. Calculate $\lambda^{\infty}_m$ for $NH_4OH$ solution (in $\Omega^{-1} \ cm^2 \ mol^{-1}$).

The equivalent conductance at infinite dilution of a weak acid such as $HF$:

$\Lambda_{m}^0$ for $NaCl$,$HCl$ and $NaAc$ are $126.4 \ S \ cm^2 \ mol^{-1}$,$425.9 \ S \ cm^2 \ mol^{-1}$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda^0$ for $HAc$.

Electrical conductance due to all the ions in $1 \ cm^3$ of a given solution is called as

Vedclass Test Series

Exam Paper Generator

Online Exam Module