The specific conductance (k) of 0.02 M aqueo | vedclass

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

Question

(A) Step $1$: Calculate the molar conductivity $(\lambda_m)$ at the given concentration.$\lambda_m = \frac{k 	imes 1000}{C} = \frac{1.65 	imes 10^{-

Vedclass · Accepted Answer

$0.021$

Vedclass · Answer

(A) Step $1$: Calculate the molar conductivity $(\lambda_m)$ at the given concentration.$\lambda_m = \frac{k 	imes 1000}{C} = \frac{1.65 	imes 10^{-4} 	imes 1000}{0.02} = 8.25 \ S \ cm^2 \ mol^{-1}$.Step $2$: Calculate the molar conductivity at infinite dilution $(\lambda_m^{\infty})$ using Kohlrausch's law.$\lambda_m^{\infty}(CH_3COOH) = \lambda_{H^{+}}^{\infty} + \lambda_{CH_3COO^{-}}^{\infty} = 349.1 + 40.9 = 390.0 \ S \ cm^2 \ mol^{-1}$.Step $3$: Calculate the degree of dissociation $(\alpha)$.$\alpha = \frac{\lambda_m}{\lambda_m^{\infty}} = \frac{8.25}{390.0} = 0.02115 \approx 0.021$.

Similar Questions

Which is the conductivity of $0.02 \ M \ HCl$ solution if molar conductivity of the solution at $25^{\circ} C$ is $412.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$?

Conductivity of a solution is $1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. Calculate molar conductivity for $0.01 \ M$ solution.

$0.2\% (w/v)$ solution of $NaOH$ is measured to have resistivity $870.0 \ m\Omega \ m$. The molar conductivity of the solution will be $........ \times 10^2 \ mS \ dm^2 \ mol^{-1}$. $(Nearest \ integer)$

$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.

The conductivity of a solution of concentration $0.1 \ mol \ L^{-1}$ of a weak monobasic acid $(HA)$ (in $S \ cm^{-1}$) is (Given $\Lambda_{HA}^{\circ}=400 \ S \ cm^2 \ mol^{-1}$ and degree of dissociation $(\alpha)$ of $HA=0.02$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module