The conductivity of a 0.05 , M solution of a | vedclass

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

Question

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 	imes \kappa}{M}$$\lambda_{m} = \frac{1000

Vedclass · Accepted Answer

$8 	imes 10^{-5}$

Vedclass · Answer

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 	imes \kappa}{M}$$\lambda_{m} = \frac{1000 	imes 10^{-3}}{0.05} = 20 \, S \, cm^{2} \, mol^{-1}$Next,calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{20}{500} = 0.04$Finally,calculate the dissociation constant $(K_{a})$ using the formula: $K_{a} = C \alpha^{2}$$K_{a} = 0.05 	imes (0.04)^{2} = 0.05 	imes 0.0016 = 8 	imes 10^{-5}$

The conductivity of a $0.05 \, M$ solution of a weak monobasic acid is $10^{-3} \, S \, cm^{-1}$. If $\lambda_{m}^{\infty}$ for the weak acid is $500 \, S \, cm^{2} \, mol^{-1}$,calculate the $K_{a}$ of the weak monobasic acid.

Similar Questions

Discuss the construction and uses of a conductivity cell.

Calculate the conductivity of $0.02 \ M$ electrolyte solution if its molar conductivity is $407.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

For an aqueous solution of $CaCl_2$ electrolyte,the graph between molar conductivity $(\lambda_m)$ and $(\text{concentration})^{1/2}$ is:

The conductivity of a strong electrolyte:

The molar conductivity for electrolytes $A$ and $B$ are plotted against $C^{1/2}$ as shown below. Electrolytes $A$ and $B$ respectively are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module