Which of the following has maximum molar cond | vedclass

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

Question

(A) For $(1)$,molar conductivity $\lambda_M = \frac{\kappa 	imes 1000}{M} = \frac{2 	imes 10^{-2} 	imes 1000}{0.08} = 250 \, \Omega^{-1} \, cm^2 \,

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) For $(1)$,molar conductivity $\lambda_M = \frac{\kappa 	imes 1000}{M} = \frac{2 	imes 10^{-2} 	imes 1000}{0.08} = 250 \, \Omega^{-1} \, cm^2 \, mol^{-1}$.For $(2)$,molar conductivity $\lambda_M = \frac{\kappa 	imes 1000}{M}$,where $\kappa = \frac{1}{ho} = \frac{1}{50} \, \Omega^{-1} \, cm^{-1}$.Thus,$\lambda_M = \frac{1}{50} 	imes \frac{1000}{0.1} = 200 \, \Omega^{-1} \, cm^2 \, mol^{-1}$.Comparing the two,the solution in $(1)$ has a higher molar conductivity than the solution in $(2)$.

Which of the following has maximum molar conductivity?

Similar Questions

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

The conductivity of $0.04 \ M \ BaCl_2$ solution is $0.0112 \ \Omega^{-1} \ cm^{-1}$ at $25^{\circ} C$. What is its molar conductivity?

If the resistance of a $0.01 \, M$ electrolyte in a cell is $40 \, \Omega$ and the cell constant is $0.4 \, \text{cm}^{-1}$,what is the molar conductivity in $\Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$?

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

Resistance of a decimolar solution between two electrodes $0.02 \ m$ apart and $0.0004 \ m^2$ in area was found to be $50 \ \Omega.$ Specific conductance $(\kappa)$ is .......... $S \ m^{-1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module