(A) Given: $K_a = 1.75 \times 10^{-5}$,$C = 0.01 \, M$,$\Lambda _m^o = 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1} = 370.6 \, S \, cm^2 \, mol^{-1}$.

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$1.55 \times 10^{-4} \, S \, cm^{-1}$

(A) Given: $K_a = 1.75 \times 10^{-5}$,$C = 0.01 \, M$,$\Lambda _m^o = 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1} = 370.6 \, S \, cm^2 \, mol^{-1}$.Degree of dissociation $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.75 \times 10^{-5}}{0.01}} = \sqrt{1.75 \times 10^{-3}} \approx 0.0418$.Molar conductivity $\Lambda _m = \alpha \times \Lambda _m^o = 0.0418 \times 370.6 \approx 15.5 \, S \, cm^2 \, mol^{-1}$.Specific conductance $\kappa = \frac{\Lambda _m \times C}{1000} = \frac{15.5 \times 0.01}{1000} = 1.55 \times 10^{-4} \, S \, cm^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

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The dissociation constant of acetic acid is $1.75 \times 10^{-5}$ and $\Lambda _{CH_3COOH}^o = 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1}$. The specific conductance of $0.01 \, M$ acetic acid solution will be:

A
$1.55 \times 10^{-4} \, S \, cm^{-1}$
B
$1.55 \times 10^{-5} \, S \, cm^{-1}$
C
$1.55 \times 10^{-6} \, S \, cm^{-1}$
D
$1.55 \times 10^{-8} \, S \, cm^{-1}$

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Conductor and Conductance and Cell constant

$A$ $KCl$ solution of conductivity $0.14 \, S m^{-1}$ shows a resistance of $4.19 \, \Omega$ in a conductivity cell. If the same cell is filled with an $HCl$ solution,the resistance drops to $1.03 \, \Omega$. The conductivity of the $HCl$ solution is $....... \, \times 10^{-2} \, S m^{-1}$. (Round off to the Nearest Integer).

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At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

DifficultTS EAMCET 2009

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

EasyMHT CET 2021

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The specific conductance of a saturated solution of silver bromide is $\kappa \ S \ cm^{-1}$. The limiting ionic conductances for $Ag^{+}$ and $Br^{-}$ are $x$ and $y$ $S \ cm^2 \ mol^{-1}$ respectively. Then the solubility of silver bromide (in $g/L$) will be $(Ag = 108, Br = 80)$.

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What must be the molarity of $BaCl_2$ solution to have molar conductivity $240 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$ and conductivity $0.012 \text{ }\Omega^{-1}\text{cm}^{-1}$ at $25 \text{ }^\circ\text{C}$ (in M)?

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The dissociation constant of acetic acid is $1.75 \times 10^{-5}$ and $\Lambda _{CH_3COOH}^o = 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1}$. The specific conductance of $0.01 \, M$ acetic acid solution will be:

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At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

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

The specific conductance of a saturated solution of silver bromide is $\kappa \ S \ cm^{-1}$. The limiting ionic conductances for $Ag^{+}$ and $Br^{-}$ are $x$ and $y$ $S \ cm^2 \ mol^{-1}$ respectively. Then the solubility of silver bromide (in $g/L$) will be $(Ag = 108, Br = 80)$.

What must be the molarity of $BaCl_2$ solution to have molar conductivity $240 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$ and conductivity $0.012 \text{ }\Omega^{-1}\text{cm}^{-1}$ at $25 \text{ }^\circ\text{C}$ (in M)?

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