(B) The given relationship is $K \propto \frac{A \times C}{l}$,where $K$ is the constant of proportionality. Thus,$K = \frac{K_{cond} \times l}{A \tim

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(B) The given relationship is $K \propto \frac{A \times C}{l}$,where $K$ is the constant of proportionality. Thus,$K = \frac{K_{cond} \times l}{A \times C}$,where $K_{cond}$ is conductivity $(S)$,$l$ is length $(m)$,$A$ is area $(m^2)$,and $C$ is concentration $(mol \, m^{-3})$. Substituting the units: $\text{Unit of } K = \frac{S \times m}{m^2 \times (mol \, m^{-3})} = \frac{S \times m}{mol \times m^{-1}} = S \, m^2 \, mol^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Difficult

Conductivity (unit $S$) is directly proportional to the area of the vessel $(A)$ and the concentration of the solution $(C)$ in it,and is inversely proportional to the length of the vessel $(l)$. The unit of the constant of proportionality is:

AIEEE 2002 All AIEEE

A
$S \, m \, mol^{-1}$
B
$S \, m^2 \, mol^{-1}$
C
$S^{-2} \, m^2 \, mol$
D
$S^2 \, m^2 \, mol^{-2}$

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Class 12 Questions

Class 12

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Chapter

Electrochemistry

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Conductor and Conductance and Cell constant

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AIEEE 2002 Paper All AIEEE years →

Calculate $\Lambda _m^o$ for $CaCl_2$ and $MgSO_4$ from the data given in the table below:

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$

$H^{+} : 349.6$ $OH^{-} : 199.1$

$Na^{+} : 50.1$ $Cl^{-} : 76.3$

$K^{+} : 73.5$ $Br^{-} : 78.1$

$Ca^{2+} : 119.0$ $CH_3COO^{-} : 40.9$

$Mg^{2+} : 106.0$ $SO_4^{2-} : 160.0$

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

Medium

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$0.0129 \ S \ cm^{-1}$ is equivalent to how many $S \ m^{-1}$ (in $S \ m^{-1}$)?

Medium

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What is the ionic mobility of $Ag^{+}$? Given $\lambda^{Ag^{+}} = 5 \times 10^{-4} \ \Omega^{-1} \ cm^{2} \ eq^{-1}$.

Easy

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Find out the ionisation constant of a weak acid $(HA)$ in terms of $\Lambda_m^o$ and $\Lambda_m^c$. (Given $\alpha$ cannot be ignored with respect to $1$)

Medium

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Which of the following is not a correct relationship between $\lambda_c$ and $\lambda_\infty$ for $NaCl$? Here,$\lambda_c$ and $\lambda_\infty$ are the equivalent conductance at concentration $C$ and at infinite dilution,respectively.

Medium

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Conductivity (unit $S$) is directly proportional to the area of the vessel $(A)$ and the concentration of the solution $(C)$ in it,and is inversely proportional to the length of the vessel $(l)$. The unit of the constant of proportionality is:

Similar Questions

$0.0129 \ S \ cm^{-1}$ is equivalent to how many $S \ m^{-1}$ (in $S \ m^{-1}$)?

What is the ionic mobility of $Ag^{+}$? Given $\lambda^{Ag^{+}} = 5 \times 10^{-4} \ \Omega^{-1} \ cm^{2} \ eq^{-1}$.

Find out the ionisation constant of a weak acid $(HA)$ in terms of $\Lambda_m^o$ and $\Lambda_m^c$. (Given $\alpha$ cannot be ignored with respect to $1$)

Which of the following is not a correct relationship between $\lambda_c$ and $\lambda_\infty$ for $NaCl$? Here,$\lambda_c$ and $\lambda_\infty$ are the equivalent conductance at concentration $C$ and at infinite dilution,respectively.

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