Conductor and Conductance and Cell constant Questions in English

(A) The formula for equivalent conductance is $\Lambda_{eq} = \frac{\kappa \times 1000}{N}$.
First,calculate the conductivity $(\kappa)$: $\kappa = \frac{\text{Cell Constant}}{\text{Resistance}} = \frac{1.15 \ cm^{-1}}{250 \ \Omega} = 0.0046 \ \Omega^{-1} \ cm^{-1}$.
Now,substitute the values into the equivalent conductance formula:
$\Lambda_{eq} = \frac{0.0046 \times 1000}{0.1} = \frac{4.6}{0.1} = 46 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

(C) The molar conductivity at infinite dilution for $AgCl$ is given by: $\lambda_m^\infty (AgCl) = \lambda_{Ag^+}^\infty + \lambda_{Cl^-}^\infty = 61.9 + 76.3 = 138.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
The solubility $(S)$ in $mol \ L^{-1}$ is calculated using the formula: $S = \frac{\kappa \times 1000}{\lambda_m^\infty} = \frac{1.382 \times 10^{-6} \times 1000}{138.2} = 10^{-5} \ mol \ L^{-1}$.
To convert solubility to $g \ L^{-1}$,multiply by the molar mass of $AgCl$ $(143.5 \ g \ mol^{-1})$: $S = 10^{-5} \ mol \ L^{-1} \times 143.5 \ g \ mol^{-1} = 1.435 \times 10^{-3} \ g \ L^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$ and specific conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Rearranging for specific conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Given: $\Lambda_m = 1.26 \times 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $M = 0.01 \ mol \ L^{-1}$.
Substituting the values: $\kappa = \frac{1.26 \times 10^{2} \times 0.01}{1000}$.
$\kappa = \frac{1.26 \times 1}{1000} = 1.26 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(C) The equivalent conductance $(\Lambda_{eq})$ of an electrolyte solution is related to its concentration $(C)$ by the relation $\Lambda_{eq} = \frac{\kappa \times 1000}{C}$,where $\kappa$ is the specific conductance.
As the concentration of the electrolyte increases,the number of ions per unit volume increases,but the inter-ionic attractions also increase,which significantly reduces the mobility of ions.
Consequently,the equivalent conductance decreases as the concentration of the electrolyte increases.
Among the given options,the concentration $1 \ M$ is the highest.
Therefore,the $1 \ M$ $KCl$ solution will have the lowest equivalent conductance.

(C) The cell constant $(G^*)$ is given by the formula: $G^* = \frac{\ell}{A} = \frac{2 \ cm}{4 \ cm^2} = 0.5 \ cm^{-1}$.
The conductivity $(\kappa)$ is related to resistance $(R)$ by the formula: $\kappa = \frac{1}{R} \times G^*$.
Rearranging for resistance $(R)$: $R = \frac{G^*}{\kappa} = \frac{0.5 \ cm^{-1}}{8 \times 10^{-7} \ S \ cm^{-1}}$.
$R = \frac{0.5}{8} \times 10^7 \ \Omega = 0.0625 \times 10^7 \ \Omega = 6.25 \times 10^5 \ \Omega$.

(D) The equivalent weight of $H_2SO_4$ is $49 \, g/eq$.
Given concentration $C = 49 \, g/L$,so the normality $N = \frac{49 \, g/L}{49 \, g/eq} = 1 \, N$.
Given specific conductance $\kappa = 18.4 \, \Omega^{-1} \, cm^{-1}$.
The equivalent conductance at concentration $C$ is $\lambda_{eq}^C = \frac{1000 \times \kappa}{N} = \frac{1000 \times 18.4}{1} = 18400 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.
Wait,checking the provided values: If $\kappa = 18.4$ is actually specific resistance (resistivity $\rho$),then $\kappa = 1/\rho = 1/18.4 \approx 0.0543 \, \Omega^{-1} \, cm^{-1}$.
Then $\lambda_{eq}^C = \frac{1000 \times 0.0543}{1} = 54.3 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.
The degree of dissociation $\alpha = \frac{\lambda_{eq}^C}{\lambda_{eq}^\infty} = \frac{54.3}{384} \approx 0.1414$.
Therefore,$\alpha \% = 14.14 \% \approx 14 \% $.

(B) $1$. First,calculate the cell constant $(G^*)$ using the data for the $0.2 \, M$ solution:
$G^* = \kappa \times R = 1.3 \, S \, m^{-1} \times 50 \, \Omega = 65 \, m^{-1}$.
$2$. Now,calculate the conductivity $(\kappa)$ for the $0.4 \, M$ solution using the same cell constant:
$\kappa = \frac{G^*}{R} = \frac{65 \, m^{-1}}{260 \, \Omega} = 0.25 \, S \, m^{-1}$.
$3$. Calculate the molar conductivity $(\Lambda_m)$ using the formula $\Lambda_m = \frac{\kappa}{C}$:
Given $C = 0.4 \, M = 0.4 \times 10^3 \, mol \, m^{-3} = 400 \, mol \, m^{-3}$.
$\Lambda_m = \frac{0.25 \, S \, m^{-1}}{400 \, mol \, m^{-3}} = 6.25 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

(C) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution is the sum of the equivalent conductances of the individual ions.
$\Lambda^0_{eq}(BaCl_2) = \lambda^0_{eq}(Ba^{2+}) + \lambda^0_{eq}(Cl^{-})$
Given:
$\lambda^0_{eq}(Ba^{2+}) = 127 \ \Omega^{-1} \ cm^2 \ eq^{-1}$
$\lambda^0_{eq}(Cl^{-}) = 76 \ \Omega^{-1} \ cm^2 \ eq^{-1}$
Therefore,
$\Lambda^0_{eq}(BaCl_2) = 127 + 76 = 203 \ \Omega^{-1} \ cm^2 \ eq^{-1}$
Note: If the values provided are molar conductances,the calculation would differ,but for equivalent conductance,we sum the values directly.

(B) The degree of dissociation $\alpha$ is given by the formula: $\alpha = \frac{\lambda^{c}}{\lambda^{\infty}}$
$(i)$ Calculate the limiting molar conductivity of acetic acid:
$\lambda^{\infty} (CH_{3}COOH) = \lambda^{\infty} (CH_{3}COO^{-}) + \lambda^{\infty} (H^{+})$
$\lambda^{\infty} (CH_{3}COOH) = 40.9 + 349.8 = 390.7 \ S \ cm^{2} \ eq.^{-1}$
$(ii)$ Calculate the degree of dissociation $\alpha$:
$\alpha = \frac{5.20}{390.7} \approx 0.0133$
$(iii)$ Convert to percentage:
$\alpha \% = 0.0133 \times 100 = 1.33 \% \approx 1.3 \%$

(C) According to Kohlrausch's law of independent migration of ions:
$\mu^{\infty}_{NH_4OH} = \mu^{\infty}_{NH_4Cl} + \mu^{\infty}_{NaOH} - \mu^{\infty}_{NaCl}$
Substituting the given values:
$\mu^{\infty}_{NH_4OH} = 129.8 + 248.1 - 126.4$
$\mu^{\infty}_{NH_4OH} = 251.5 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda^o_{KCl} = \Lambda^o_{K^+} + \Lambda^o_{Cl^-} = 152 \ S \ cm^2 \ mol^{-1}$
$\Lambda^o_{NaCl} = \Lambda^o_{Na^+} + \Lambda^o_{Cl^-} = 128 \ S \ cm^2 \ mol^{-1}$
$\Lambda^o_{KNO_3} = \Lambda^o_{K^+} + \Lambda^o_{NO_3^-} = 111 \ S \ cm^2 \ mol^{-1}$
To find $\Lambda^o_{NaNO_3}$,we use the expression:
$\Lambda^o_{NaNO_3} = \Lambda^o_{NaCl} + \Lambda^o_{KNO_3} - \Lambda^o_{KCl}$
$\Lambda^o_{NaNO_3} = 128 + 111 - 152$
$\Lambda^o_{NaNO_3} = 87 \ S \ cm^2 \ mol^{-1}$

(A) The conductivity $(\kappa)$ is given by: $\kappa = \frac{\text{cell constant}}{\text{resistance}} = \frac{0.66 \ cm^{-1}}{210 \ \Omega} \approx 0.00314 \ S \ cm^{-1}$.
Equivalent conductivity $(\Lambda_{eq})$ is calculated as: $\Lambda_{eq} = \frac{\kappa \times 1000}{N}$.
$\Lambda_{eq} = \frac{0.00314 \times 1000}{0.01} = 314 \ mho \ cm^2 \ eq^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $NaNO_3$ can be calculated as:
$\Lambda^\circ_{m(NaNO_3)} = \Lambda^\circ_{m(AgNO_3)} + \Lambda^\circ_{m(NaCl)} - \Lambda^\circ_{m(AgCl)}$
Substituting the given values:
$\Lambda^\circ_{m(NaNO_3)} = 116.5 + 110.3 - 121.6$
$\Lambda^\circ_{m(NaNO_3)} = 226.8 - 121.6 = 105.2 \ S \ cm^2 \ mol^{-1}$

(C) The degree of dissociation $\alpha$ is given by the ratio of equivalent conductance at concentration $C$ to equivalent conductance at infinite dilution: $\alpha = \frac{\wedge_{eq}^c}{\wedge_{eq}^0}$.
We know that $\wedge_{eq}^c = \frac{\kappa \times 1000}{N}$,where $\kappa$ is the conductivity and $N$ is the normality.
Given: $\alpha = 0.90$,$\wedge_{eq}^0 = 425 \ \Omega^{-1} \ cm^2 \ eq^{-1}$,and $\kappa = 3.825 \ \Omega^{-1} \ cm^{-1}$.
Substituting the values: $0.9 = \frac{3.825 \times 1000}{N \times 425}$.
$0.9 = \frac{3825}{N \times 425}$.
$0.9 = \frac{9}{N}$.
$N = \frac{9}{0.9} = 10 \ N$.

(A) Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions,multiplied by their respective stoichiometric coefficients.
For the electrolyte $A_2B$,it dissociates as follows:
$A_2B \rightarrow 2A^+ + B^{2-}$
Therefore,the limiting molar conductivity $\Lambda_m^\infty$ is given by:
$\Lambda_m^\infty(A_2B) = 2\lambda _{A^+}^\infty + \lambda _{B^{2-}}^\infty$

(A) The equivalent conductivity $(\lambda_{eq})$ is inversely proportional to the concentration of the solution.
As the concentration increases,the number of ions per unit volume increases,but the mobility of ions decreases due to interionic attractions.
Therefore,the equivalent conductivity decreases as the concentration increases.
Among the given options,$1 \ M$ has the highest concentration,and thus it will have the lowest equivalent conductivity.

(A) Specific conductivity $\kappa = \frac{1}{R} \times \text{cell constant} = \frac{1}{250} \times 1.15 \, \Omega^{-1} \, \text{cm}^{-1}$.
Equivalent conductivity $\Lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{1.15}{250} \times 1000$.
$\Lambda_{eq} = 4.6 \, \Omega^{-1} \, \text{cm}^2 \, \text{equiv}^{-1}$.

(A) Conductivity (or specific conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
On dilution,the number of ions per unit volume decreases,which leads to a decrease in the conductivity of the solution.
Therefore,the statement 'The conductivity of an electrolytic solution increases with dilution' is incorrect.

(C) The resistance $(R)$ of an electrolyte is inversely proportional to its specific conductivity $(\kappa)$, given by the relation $R = \frac{l}{\kappa A}$.
Since the geometry $(l/A)$ is assumed to be constant for all, the electrolyte with the lowest specific conductivity will offer the maximum resistance.
Comparing the given values:
$P = 5.0 \times 10^{-5}$
$Q = 7.0 \times 10^{-8}$
$R = 1.0 \times 10^{-10}$
$S = 9.2 \times 10^{-3}$
The lowest value is $1.0 \times 10^{-10}$ for electrolyte $R$.
Therefore, $R$ offers the maximum resistance.

(A) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^\infty(Ba(OH)_2) = \lambda^\infty(Ba^{2+}) + 2\lambda^\infty(OH^-) = 523.28 \ \dots (i)$
$\Lambda_m^\infty(BaCl_2) = \lambda^\infty(Ba^{2+}) + 2\lambda^\infty(Cl^-) = 280.0 \ \dots (ii)$
$\Lambda_m^\infty(NH_4Cl) = \lambda^\infty(NH_4^+) + \lambda^\infty(Cl^-) = 129.8 \ \dots (iii)$
We need to find $\Lambda_m^\infty(NH_4OH) = \lambda^\infty(NH_4^+) + \lambda^\infty(OH^-)$.
Using the equations: $\Lambda_m^\infty(NH_4OH) = \Lambda_m^\infty(NH_4Cl) + \frac{1}{2}\Lambda_m^\infty(Ba(OH)_2) - \frac{1}{2}\Lambda_m^\infty(BaCl_2)$
$\Lambda_m^\infty(NH_4OH) = 129.8 + \frac{523.28}{2} - \frac{280.0}{2}$
$\Lambda_m^\infty(NH_4OH) = 129.8 + 261.64 - 140.0 = 251.44 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte can be calculated using the limiting molar conductivities of strong electrolytes.
For $NH_4OH$,the expression is: $\lambda^{\infty}_m(NH_4OH) = \lambda^{\infty}_m(NH_4Cl) + \lambda^{\infty}_m(NaOH) - \lambda^{\infty}_m(NaCl)$.
Substituting the given values: $\lambda^{\infty}_m(NH_4OH) = 129.8 + 248.1 - 126.4$.
$\lambda^{\infty}_m(NH_4OH) = 377.9 - 126.4 = 251.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(B) The molar conductivity at infinite dilution for benzoic acid is calculated using Kohlrausch's law:
$\Lambda^0_m (C_6H_5COOH) = \Lambda^0_m(C_6H_5COO^-) + \Lambda^0_m(H^+) = 42 + 288.42 = 330.42 \, Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$.
The degree of dissociation $(\alpha)$ is given by the ratio of equivalent conductance at a given concentration to the equivalent conductance at infinite dilution:
$\alpha = \frac{\Lambda_m}{\Lambda^0_m} = \frac{12.8}{330.42} \approx 0.0387$.
To express this as a percentage:
$\alpha \% = 0.0387 \times 100 = 3.87 \% \approx 3.9 \% $.

(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for a weak electrolyte like acetic acid $(HOAc)$ can be calculated using the values of strong electrolytes.
The relation is: $\Lambda^{0}_{HOAc} = \Lambda^{0}_{NaOAc} + \Lambda^{0}_{HCl} - \Lambda^{0}_{NaCl}$.
Therefore,the value of $\Lambda^{0}_{NaCl}$ is required.

(A) The formula for equivalent conductance $(\Lambda_{eq})$ is given by: $\Lambda_{eq} = \frac{\kappa \times 1000}{N}$
Given:
Specific conductance $(\kappa)$ = $0.0014 \, \Omega^{-1} \, cm^{-1}$
Normality $(N)$ = Molarity $(M)$ for $KCl$ (since $n$-factor = $1$) = $0.01 \, N$
Substituting the values:
$\Lambda_{eq} = \frac{0.0014 \times 1000}{0.01} = \frac{1.4}{0.01} = 140 \, \Omega^{-1} \, cm^{2} \, eq^{-1}$

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

(A) The ionic mobility $(u)$ is related to the ionic conductance $(\lambda)$ by the formula: $u = \frac{\lambda}{F}$,where $F$ is the Faraday constant $(96500 \ C \ eq^{-1})$.
Substituting the given values:
$u = \frac{5 \times 10^{-4} \ \Omega^{-1} \ cm^{2} \ eq^{-1}}{96500 \ C \ eq^{-1}}$
$u \approx 5.18 \times 10^{-9} \ cm^{2} \ s^{-1} \ V^{-1}$.

(C) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte can be calculated as follows:
$\lambda^{0}_{NaBr} = \lambda^{0}_{Na^+} + \lambda^{0}_{Br^-}$
We can express this in terms of the given values:
$\lambda^{0}_{NaBr} = \lambda^{0}_{NaCl} + \lambda^{0}_{KBr} - \lambda^{0}_{KCl}$
Substituting the given values:
$\lambda^{0}_{NaBr} = 126 + 152 - 150$
$\lambda^{0}_{NaBr} = 128 \ S \ cm^{2} \ mol^{-1}$

(A) For nitric acid $(HNO_3)$,the normality $(N)$ is equal to the molarity $(M)$ because the $n$-factor is $1$. Thus,$M = 0.1 \, M$.
The formula for molar conductivity $(\Lambda_m)$ is: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Given $\kappa = 6.2 \times 10^{-2} \, \Omega^{-1} \, cm^{-1}$ and $M = 0.1 \, M$.
$\Lambda_m = \frac{6.2 \times 10^{-2} \times 1000}{0.1} = \frac{62}{0.1} = 620 \, \Omega^{-1} \, cm^{2} \, mol^{-1}$.

(A) The cell constant $G^* = \frac{L}{A} = \frac{2.1 \, cm}{4.2 \, cm^2} = 0.5 \, cm^{-1}$.
The specific conductivity $\kappa$ is given by $\kappa = \frac{1}{R} \times G^* = \frac{1}{50 \, \Omega} \times 0.5 \, cm^{-1} = 0.01 \, S \, cm^{-1}$.
The equivalent conductivity $\lambda_{eq}$ is calculated using the formula $\lambda_{eq} = \frac{\kappa \times 1000}{N}$.
Substituting the values: $\lambda_{eq} = \frac{0.01 \times 1000}{1} = 10 \, S \, cm^2 \, eq^{-1}$.

(A) According to Kohlrausch's law,the equivalent conductivity of an electrolyte at infinite dilution is the sum of the equivalent conductivities of its constituent ions.
For $BaCl_2$,the formula is:
$(\wedge_{eq}^{\infty})_{BaCl_2} = (\lambda_{eq}^{\infty})_{Ba^{2+}} + (\lambda_{eq}^{\infty})_{Cl^-}$
Given that the values provided are already equivalent conductivities of the ions:
$(\wedge_{eq}^{\infty})_{BaCl_2} = 127 + 76 = 203 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

(D) Specific conductivity $(\kappa)$ is defined as the conductance of a solution of $1 \ cm^3$ volume.
As the concentration of an electrolyte solution increases,the number of ions per unit volume increases.
Since specific conductivity depends on the number of ions present in a unit volume,it increases with an increase in concentration.
Among the given options,the solution with the highest concentration $(1 \ N)$ will have the maximum number of ions per unit volume.
Therefore,$1 \ N$ $NaCl$ solution has the maximum specific conductivity.

(A) Specific conductance $(\kappa) = \text{Conductivity} \times \text{Cell constant} = \frac{1}{R} \times \frac{l}{A} = \frac{1}{2.5 \times 10^{3}} \times 1.15 = 4.6 \times 10^{-4} \, \Omega^{-1} \, cm^{-1}$.
Equivalent conductance $(\Lambda_{eq}) = \frac{\kappa \times 1000}{N} = \frac{4.6 \times 10^{-4} \times 1000}{0.1} = 4.6 \, \Omega^{-1} \, cm^{2} \, eq^{-1}$.

(B) In metallic conductors,the resistance increases with an increase in temperature due to the increased vibration of metal ions,which hinders the flow of electrons.
In electrolytic conductors,the resistance decreases with an increase in temperature because the viscosity of the solvent decreases and the kinetic energy of the ions increases,leading to higher ionic mobility.

(A) Given: Resistance $(R)$ = $200 \ \Omega$,Cell constant $(G^*)$ = $1 \ cm^{-1}$,Normality $(N)$ = $0.01 \ N$.
First,calculate the conductivity $(\kappa)$:
$\kappa = \frac{G^*}{R} = \frac{1 \ cm^{-1}}{200 \ \Omega} = 0.005 \ \Omega^{-1} cm^{-1}$.
Next,calculate the equivalent conductance $(\lambda_{eq})$:
$\lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{0.005 \times 1000}{0.01} = \frac{5}{0.01} = 500 \ \Omega^{-1} cm^2 eq^{-1}$.
This can be expressed as $5 \times 10^2 \ \Omega^{-1} cm^2 eq^{-1}$.

(A) The relationship between specific conductivity $(K)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula: $K = \frac{1}{R} \times G^*$.
Rearranging the formula to solve for the cell constant: $G^* = K \times R$.
Given values: $K = 0.0212\,ohm^{-1}\,cm^{-1}$ and $R = 55\,ohm$.
Substituting the values: $G^* = 0.0212 \times 55 = 1.166\,cm^{-1}$.

(B) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given,electrolytic conductivity $\kappa = 5.76 \times 10^{-3} \ S \ cm^{-1}$ and Molarity $M = 0.5 \ mol/dm^3 = 0.5 \ mol/L$.
Substituting the values:
$\Lambda_{m} = \frac{5.76 \times 10^{-3} \ S \ cm^{-1} \times 1000 \ cm^3/L}{0.5 \ mol/L} = 11.52 \ S \ cm^2 \ mol^{-1}$.

(A) The degree of ionization $(\alpha)$ is given by the ratio of molar conductance at a specific concentration $(\lambda^c_m)$ to the molar conductance at infinite dilution $(\lambda^\infty_m)$:
$\alpha = \frac{\lambda^c_m}{\lambda^\infty_m} = \frac{9.54}{238} \approx 0.04008$
To express this as a percentage:
$\% \alpha = \alpha \times 100 = 0.04008 \times 100 = 4.008 \%$

(D) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte can be calculated using the limiting molar conductivities of strong electrolytes.
For $NH_4OH$,we can express it as:
$\Lambda ^o_{m(NH_4OH)} = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-}$
Using strong electrolytes $NH_4Cl$,$NaOH$,and $NaCl$:
$\Lambda ^o_{m(NH_4Cl)} = \lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}$
$\Lambda ^o_{m(NaOH)} = \lambda ^o_{Na^+} + \lambda ^o_{OH^-}$
$\Lambda ^o_{m(NaCl)} = \lambda ^o_{Na^+} + \lambda ^o_{Cl^-}$
By performing the operation $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$,we get:
$(\lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}) + (\lambda ^o_{Na^+} + \lambda ^o_{OH^-}) - (\lambda ^o_{Na^+} + \lambda ^o_{Cl^-}) = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-} = \Lambda ^o_{m(NH_4OH)}$
Therefore,the correct expression is $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$.

(D) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of a weak electrolyte can be calculated using the molar conductivities of strong electrolytes.
$\Lambda ^o_m(CH_3COOH) = \Lambda ^o_m(CH_3COONa) + \Lambda ^o_m(HCl) - \Lambda ^o_m(NaCl)$
Substituting the given values:
$\Lambda ^o_m(CH_3COOH) = 91.0 + 425.9 - 126.4$
$\Lambda ^o_m(CH_3COOH) = 516.9 - 126.4 = 390.5 \ S \ cm^2 \ mol^{-1}$

(A) For a strong electrolyte,the number of ions is already fixed because it is completely dissociated at all concentrations.
Equivalent conductance $(\lambda_{eq})$ is defined as $\lambda_{eq} = \kappa \times V$,where $\kappa$ is the specific conductance and $V$ is the volume containing $1 \ g$-equivalent of the electrolyte.
Upon dilution,the inter-ionic attractions decrease,which leads to an increase in the ionic mobility of the ions.
Since the number of ions remains constant for a strong electrolyte,the increase in equivalent conductance is primarily due to the increase in ionic mobility.

(A) The equivalent conductance at infinite dilution $(\Lambda_{eq}^{\infty})$ of an electrolyte is the sum of the equivalent conductances of its constituent ions.
By definition,the equivalent conductance of any electrolyte at infinite dilution is the sum of the equivalent conductances of its cation and anion.
For $Al_2(SO_4)_3$,the equivalent conductance at infinite dilution is given by:
$\Lambda_{eq}^{\infty} = \Lambda_{Al^{3+}}^o + \Lambda_{SO_4^{2-}}^o$
This is because the equivalent conductance of an ion is defined as the molar conductance divided by its charge (valence factor),which inherently accounts for the stoichiometry of the salt.

(A) Kohlrausch's law states that "the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductances of the component ions."
$\lambda_{\infty} = \lambda_{a} + \lambda_{c}$
Where,$\lambda_{a} = \text{equivalent conductance of the anion}$,$\lambda_{c} = \text{equivalent conductance of the cation}$.
Each ion has the same constant ionic conductance at a fixed temperature,regardless of which electrolyte it forms a part of.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte like acetic acid $(CH_3COOH)$ at infinite dilution can be calculated using the molar conductivities of strong electrolytes.
The expression is:
$\Lambda_{CH_3COOH}^o = \Lambda_{CH_3COONa}^o + \Lambda_{HCl}^o - \Lambda_{NaCl}^o$
Given that we have $\Lambda_{HCl}^o$,we need $\Lambda_{CH_3COONa}^o$ and $\Lambda_{NaCl}^o$ to complete the calculation. However,if the question implies we are starting from a set of values,the missing component to relate these to acetic acid is $\Lambda_{CH_3COONa}^o$ and $\Lambda_{NaCl}^o$. Based on the provided options,$\Lambda_{NaCl}^o$ is the standard additional value required to cancel out the ions not present in the target electrolyte.

(B) Step $1$: Calculate the cell constant $(G^* = l/a)$.
Given for $0.1 \, M$ solution: $R = 100 \, \Omega$ and $\kappa = 1.29 \, S \, m^{-1}$.
$G^* = \kappa \times R = 1.29 \, S \, m^{-1} \times 100 \, \Omega = 129 \, m^{-1}$.
Step $2$: Calculate conductivity $(\kappa)$ for $0.2 \, M$ solution.
Given $R = 520 \, \Omega$ for $0.2 \, M$ solution.
$\kappa = \frac{G^*}{R} = \frac{129 \, m^{-1}}{520 \, \Omega} \approx 0.248 \, S \, m^{-1}$.
Step $3$: Calculate molar conductivity $(\Lambda_m)$.
$\Lambda_m = \frac{\kappa}{C} = \frac{0.248 \, S \, m^{-1}}{0.2 \, mol \, L^{-1}} = \frac{0.248 \, S \, m^{-1}}{0.2 \times 10^3 \, mol \, m^{-3}} = 1.24 \times 10^{-3} \, S \, m^2 \, mol^{-1}$.
Converting to the required units $(\times 10^{-4} \, S \, m^2 \, mol^{-1})$:
$1.24 \times 10^{-3} = 12.4 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions, the molar conductivity of a weak electrolyte at infinite dilution can be determined using the molar conductivities of strong electrolytes.
For acetic acid $(CH_3COOH)$, the expression is:
$\Lambda_{CH_3COOH}^o = \Lambda_{CH_3COONa}^o + \Lambda_{HCl}^o - \Lambda_{NaCl}^o$
Given the values for $\Lambda_{CH_3COONa}^o$ and $\Lambda_{HCl}^o$, we require the value of $\Lambda_{NaCl}^o$ to complete the calculation.
Thus, the correct option is $(B)$.

(A) For the $0.2 \, M$ solution:
$R_1 = 50 \, \Omega$,$\kappa_1 = 1.4 \, S \, m^{-1}$.
Cell constant $G^* = R_1 \times \kappa_1 = 50 \times 1.4 = 70 \, m^{-1}$.
For the $0.5 \, M$ solution:
$R_2 = 280 \, \Omega$.
Specific conductance $\kappa_2 = \frac{G^*}{R_2} = \frac{70}{280} = 0.25 \, S \, m^{-1}$.
Molar conductivity $\Lambda_m = \frac{\kappa_2}{C} = \frac{0.25 \, S \, m^{-1}}{0.5 \, mol \, L^{-1}}$.
Convert concentration to $mol \, m^{-3}$: $0.5 \, mol \, L^{-1} = 0.5 \times 1000 \, mol \, m^{-3} = 500 \, mol \, m^{-3}$.
$\Lambda_m = \frac{0.25}{500} = 0.0005 = 5 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

(C) According to the Debye-$H$ückel-Onsager equation,the variation of molar (or equivalent) conductivity with concentration for strong electrolytes is given by the relation:
$\lambda_C = \lambda_{\infty} - B \sqrt{C}$
Here,$\lambda_C$ is the equivalent conductance at concentration $C$,$\lambda_{\infty}$ is the equivalent conductance at infinite dilution,and $B$ is a constant that depends on the nature of the solvent and temperature.

(B) Kohlrausch's law of independent migration of ions states that at infinite dilution,where dissociation is complete,each ion makes a definite contribution to the molar conductance of an electrolyte,regardless of the nature of the other ion with which it is associated.

(A) According to Kohlrausch's law of independent migration of ions:
$\wedge_{BA}^0 = \lambda_B^+ + \lambda_A^- = 140 \ S \ cm^2 \ eq^{-1}$
$\wedge_{CA}^0 = \lambda_C^+ + \lambda_A^- = 120 \ S \ cm^2 \ eq^{-1}$
$\wedge_{BX}^0 = \lambda_B^+ + \lambda_X^- = 198 \ S \ cm^2 \ eq^{-1}$
We need to find $\wedge_{CX}^0 = \lambda_C^+ + \lambda_X^-$.
Using the relation: $\wedge_{CX}^0 = \wedge_{CA}^0 + \wedge_{BX}^0 - \wedge_{BA}^0$
$\wedge_{CX}^0 = 120 + 198 - 140 = 178 \ S \ cm^2 \ eq^{-1}$

(B) The molar conductivity at infinite dilution $(\Lambda _m^\infty)$ is related to the equivalent conductivity at infinite dilution $(\lambda _{eq}^\infty)$ by the formula: $\Lambda _m^\infty = n \times \lambda _{eq}^\infty$,where $n$ is the valency factor (total positive or negative charge per formula unit).
For $CaF_2$,the dissociation is $CaF_2 \rightarrow Ca^{2+} + 2F^-$.
The valency factor $n$ for $CaF_2$ is $2$.
Thus,$\Lambda _{m_{CaF_2}}^\infty = 2 \times \lambda _{eq_{CaF_2}}^\infty$.
According to Kohlrausch's law,$\lambda _{eq_{CaF_2}}^\infty = \lambda _{eq_{Ca^{2+}}}^\infty + \lambda _{eq_{F^{-}}}^\infty$.
Substituting this,we get $\Lambda _{m_{CaF_2}}^\infty = 2(\lambda _{eq_{Ca^{2+}}}^\infty + \lambda _{eq_{F^{-}}}^\infty)$.