Conductor and Conductance and Cell constant Questions in English

(C) The equivalent conductance at infinite dilution for a weak electrolyte like $HF$ cannot be determined by direct extrapolation of its own conductance measurements because the degree of dissociation does not approach $1$ at finite concentrations.
According to Kohlrausch's Law of independent migration of ions,the molar conductivity at infinite dilution can be determined using strong electrolytes.
Specifically,$\Lambda^{\circ}_{m}(HF) = \Lambda^{\circ}_{m}(NaF) + \Lambda^{\circ}_{m}(HCl) - \Lambda^{\circ}_{m}(NaCl)$.
Therefore,the correct approach is to use measurements on dilute solutions of $NaF$,$NaCl$,and $HCl$.

(A) In aqueous solution,cations are hydrated. The extent of hydration is inversely proportional to the size of the bare ion.
Since the size of the bare ion increases in the order $Li^{+} < Na^{+} < K^{+} < Rb^{+}$,the extent of hydration decreases in the same order.
Consequently,the size of the hydrated ion decreases as $Li^{+}(\text{hydrated}) > Na^{+}(\text{hydrated}) > K^{+}(\text{hydrated}) > Rb^{+}(\text{hydrated})$.
Ionic conductance is inversely proportional to the size of the hydrated ion.
Therefore,the ionic conductance increases in the order $Li^{+} < Na^{+} < K^{+} < Rb^{+}$.

(D) The conductance of an electrolytic solution depends on the number of ions present and their mobility.
When the temperature of an electrolytic solution is increased,the kinetic energy of the ions increases,which increases their mobility.
Additionally,for weak electrolytes,the degree of dissociation increases with an increase in temperature,leading to a higher number of ions.
Therefore,the conductance of the solution increases.

(A) Resistance $(R)$ is inversely proportional to the conductance $(G)$ of the solution.
Conductance depends on the number of ions present in the solution.
Lower concentration of the electrolyte results in fewer ions,which leads to lower conductance.
Since $R = \frac{1}{G}$,lower conductance corresponds to higher resistance.
Among the given concentrations,$0.05 \ N$ is the lowest concentration.
Therefore,the $0.05 \ N \ NaCl$ solution will have the fewest ions and thus the highest resistance.

(D) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity of an electrolyte at infinite dilution is the sum of the equivalent conductivities of its constituent ions.
For $BaCl_2$,the equivalent conductivity at infinite dilution is given by:
$\lambda^\infty_{eq}(BaCl_2) = \lambda^\infty_{eq}(1/2 Ba^{2+}) + \lambda^\infty_{eq}(Cl^-)$
Given that $\lambda^\infty_{eq}(Ba^{2+}) = 127 \ ohm^{-1} cm^2 eqv^{-1}$,the equivalent conductivity of the ion $1/2 Ba^{2+}$ is $\frac{127}{2} = 63.5 \ ohm^{-1} cm^2 eqv^{-1}$.
Therefore,$\lambda^\infty_{eq}(BaCl_2) = 63.5 + 76 = 139.5 \ ohm^{-1} cm^2 eqv^{-1}$.

(D) The conductivity of a solution depends on the nature of the electrolyte,the concentration of the solution (which changes with dilution),and the temperature. Since all the factors mentioned in the options affect the conductivity,the correct answer is $D$.

(D) The formula for molar conductance $(\lambda_{m})$ is given by: $\lambda_{m} = \frac{\kappa \times 1000}{M}$
Given: Specific conductance $(\kappa)$ = $6.3 \times 10^{-2} \, \Omega^{-1} cm^{-1}$ and Molarity $(M)$ = $0.1 \, M$.
Substituting the values: $\lambda_{m} = \frac{6.3 \times 10^{-2} \times 1000}{0.1}$
$\lambda_{m} = \frac{63}{0.1} = 630 \, \Omega^{-1} cm^2 mol^{-1}$.

(B) The conductivity $(\kappa)$ of an electrolyte is defined as the conductance of a unit volume of the solution.
On dilution,the number of ions per unit volume decreases.
Since the conductivity depends on the number of ions present in a unit volume,it decreases upon dilution for both strong and weak electrolytes.

(B) The molar conductivity $(\Lambda_m)$ is defined as the conducting power of all the ions produced by dissolving $1 \text{ mole}$ of an electrolyte in $V \text{ mL}$ of solution.
It is related to specific conductivity $(\kappa)$ and molarity $(M)$ by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Since specific resistance (resistivity) $X = \frac{1}{\kappa}$,we have $\kappa = \frac{1}{X}$.
Substituting this into the formula,we get $\Lambda_m = \frac{1000}{MX}$.

(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: $\text{Unit of } K = \frac{S \times m}{m^2 \times (mol \, m^{-3})} = \frac{S \times m}{mol \times m^{-1}} = S \, m^2 \, mol^{-1}$.

(B) The conductivity $(\kappa)$ of an electrolytic solution is defined as the conductance of a unit volume of the solution.
It depends on the concentration of ions present in the solution.
Therefore,the conductivity of a solution is directly proportional to the number of ions present per unit volume.

(D) On dilution,the volume of the solution increases,which reduces the inter-ionic attractions.
Additionally,for weak electrolytes,the degree of dissociation increases with dilution.
However,the primary reason for the increase in equivalent conductance for both strong and weak electrolytes is the increase in the degree of dissociation (for weak electrolytes) and the reduction in inter-ionic attraction (for strong electrolytes).
Among the given options,the increase in the degree of ionisation is the most significant factor for the observed increase in equivalent conductance.

(B) The formula for equivalent conductivity is $\Lambda_{eq} = \frac{\kappa \times 1000}{C}$.
Here,$\kappa$ is the specific conductivity $(ohm^{-1} \ cm^{-1})$ and $C$ is the concentration in $gm \ equivalent \ L^{-1}$ (or $gm \ equivalent \ cm^{-3} \times 1000$).
Substituting the units: $\Lambda_{eq} = \frac{ohm^{-1} \ cm^{-1}}{gm \ equivalent \ cm^{-3}} = ohm^{-1} \ cm^2 \ (gm \ equivalent)^{-1}$.
Therefore,the correct unit is $ohm^{-1} \ cm^2 \ (gm \ equivalent)^{-1}$.

(B) In electrolytic conduction,as the temperature increases,the kinetic energy of the ions increases,and the viscosity of the solvent decreases.
These factors lead to an increase in the mobility of ions,which results in a decrease in resistance.
Therefore,the correct option is $(b)$.

(D) Electrolytic conductance is defined as the ease with which current flows through an electrolytic solution.
It is the reciprocal of resistance $(G = 1/R)$.
Since conductance depends on the number of ions present in the solution,it is a direct measure of the degree of dissociation of the electrolyte.
Therefore,the correct option is $(D)$.

(C) . The conductivity $(\kappa)$ of an electrolyte is defined as the conductance of a unit volume of the solution.
On dilution,the number of ions per unit volume decreases significantly.
Therefore,the conductivity of a strong electrolyte decreases on dilution.
Note: While molar conductivity $(\Lambda_m)$ increases on dilution due to the increase in volume containing one mole of electrolyte,the specific conductivity $(\kappa)$ decreases.

(A) Molar conductivity $(\Lambda_m)$ is defined as the conductance of all the ions produced by one mole of an electrolyte in a given volume of solution.
It is related to the concentration $(C)$ by the expression $\Lambda_m = \frac{\kappa}{C}$,where $\kappa$ is the specific conductivity.
As the concentration of the solution decreases,the volume containing one mole of the electrolyte increases,leading to an increase in molar conductivity.
Therefore,molar conductivity is inversely proportional to the concentration.
Among the given options,the lowest concentration is $0.001 \ M$,which will result in the maximum molar conductivity.

(A) Molar conductivity $(\Lambda_m)$ is defined as the conductivity $(\kappa)$ divided by the molar concentration $(C)$: $\Lambda_m = \frac{\kappa}{C}$.
The unit of conductivity $(\kappa)$ is $\Omega^{-1} cm^{-1}$ (or $S \ cm^{-1}$).
The unit of molar concentration $(C)$ is $mol \ cm^{-3}$.
Therefore,the unit of molar conductivity is $\frac{\Omega^{-1} cm^{-1}}{mol \ cm^{-3}} = \Omega^{-1} cm^{2} mol^{-1}$.

(A) The cell constant is given by $G^* = l/a = 0.5 \ cm^{-1}$.
Resistance $R = 50 \ \Omega$.
The conductivity $\kappa$ is given by $\kappa = \frac{1}{R} \times \frac{l}{a} = \frac{1}{50} \times 0.5 = 0.01 \ \Omega^{-1} \ cm^{-1}$.
The equivalent conductance $\Lambda_{eq}$ is calculated using the formula $\Lambda_{eq} = \frac{\kappa \times 1000}{N}$.
Substituting the values: $\Lambda_{eq} = \frac{0.01 \times 1000}{1.0} = 10 \ \Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law:
$\Lambda_{eq}^o (C_6H_5COOH) = \Lambda_{eq}^o (C_6H_5COO^-) + \Lambda_{eq}^o (H^+) = 42 + 288.42 = 330.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.
The degree of dissociation $(\alpha)$ is given by the ratio of equivalent conductance at a given concentration $(\Lambda_{eq}^c)$ to the equivalent conductance at infinite dilution $(\Lambda_{eq}^o)$:
$\alpha = \frac{\Lambda_{eq}^c}{\Lambda_{eq}^o} = \frac{12.8}{330.42} \approx 0.03874$.
To express this as a percentage:
$\alpha \% = 0.03874 \times 100 = 3.874 \% \approx 3.9 \%$.

(D) Conductance is defined as the reciprocal of resistance.
$Conductance = \frac{1}{Resistance} = \frac{1}{ohm} = ohm^{-1}$ or $mho$ (also known as $Siemens$,$S$).
Therefore,the unit $ohm^{-1}$ represents conductance.

(B) The relationship between specific conductance $(K)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula:
$K = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant:
$G^* = K \times R$
Given:
$K = 0.012 \ \Omega^{-1} \ cm^{-1}$
$R = 55 \ \Omega$
Calculation:
$G^* = 0.012 \ \Omega^{-1} \ cm^{-1} \times 55 \ \Omega = 0.66 \ cm^{-1}$
Therefore,the cell constant is $0.66 \ cm^{-1}$.

(C) The relationship between conductivity $(K)$,conductance $(C)$,and cell constant $(G^*)$ is given by the formula: $K = C \times G^*$.
Therefore,the cell constant is calculated as: $G^* = \frac{K}{C}$.
Given: $K = 0.2 \ \Omega^{-1} cm^{-1}$ and $C = 0.04 \ \Omega^{-1}$.
$G^* = \frac{0.2}{0.04} = 5 \ cm^{-1}$.

(A) The relationship between conductance $(C)$,specific conductance $(\kappa)$,and cell constant $(G^*)$ is given by the formula: $\kappa = C \times G^*$.
Given that $C = 1$ and $\kappa = 1$,we have: $1 = 1 \times G^*$.
Therefore,$G^* = 1$.

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^o(CH_3COOH) = \Lambda_m^o(CH_3COONa) + \Lambda_m^o(HCl) - \Lambda_m^o(NaCl)$
Substituting the given values:
$\Lambda_m^o(CH_3COOH) = 91 + 426.16 - 126.45$
$\Lambda_m^o(CH_3COOH) = 390.71 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(C) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(l)$ to the cross-sectional area of the electrodes $(a)$.
$G^* = \frac{l}{a}$
Given:
$l = 3 \text{ cm}$
$a = 4 \text{ cm}^2$
Therefore,$G^* = \frac{3 \text{ cm}}{4 \text{ cm}^2} = 0.75 \text{ cm}^{-1}$ or $3/4 \text{ cm}^{-1}$.
Thus,the correct option is $(C)$.

(D) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(l)$ to the area of cross-section of the electrodes $(A)$.
$G^* = \frac{l}{A}$
Since the unit of length $(l)$ is $cm$ and the unit of area $(A)$ is $cm^2$,the unit of cell constant is $\frac{cm}{cm^2} = cm^{-1}$.

(C) The formula for cell constant $(G^*)$ is given by:
$G^* = \kappa \times R$
Where $\kappa$ is the specific conductivity and $R$ is the resistance.
Given:
$\kappa = 0.002765\,S\,cm^{-1}$
$R = 400\,\Omega$
Calculation:
$G^* = 0.002765\,S\,cm^{-1} \times 400\,\Omega = 1.106\,cm^{-1}$
Therefore,the correct option is $(c)$.

(C) According to Kohlrausch's law of independent migration of ions:
$\wedge _{NaCl}^0 = \lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0 = 126 \ S \ cm^2 \ mol^{-1}$ $(1)$
$\wedge _{KBr}^0 = \lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0 = 152 \ S \ cm^2 \ mol^{-1}$ $(2)$
$\wedge _{KCl}^0 = \lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0 = 150 \ S \ cm^2 \ mol^{-1}$ $(3)$
To find $\wedge _{NaBr}^0 = \lambda _{Na^{+}}^0 + \lambda _{Br^{-}}^0$,we perform the operation: $(1)$ + $(2)$ - $(3)$
$\wedge _{NaBr}^0 = (\lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0) + (\lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0) - (\lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0)$
$\wedge _{NaBr}^0 = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte at infinite dilution can be calculated using the molar conductivities of strong electrolytes.
$\Lambda _{HOAc}^{\infty } = \Lambda _{NaOAc}^{\infty } + \Lambda _{HCl}^{\infty } - \Lambda _{NaCl}^{\infty }$
Substituting the given values:
$\Lambda _{HOAc}^{\infty } = 91.0 + 426.2 - 126.5$
$\Lambda _{HOAc}^{\infty } = 390.7 \ S \ cm^2 \ mol^{-1}$

(C) The electrical conductivity of metals depends on the mobility of free electrons. Among the given options,$Silver$ $(Ag)$ has the highest electrical conductivity at room temperature,making it the best conductor of electricity.

(C) Conductivity depends on the number of ions produced per formula unit in the solution.
$A$: $K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$ $(4 \, \text{ions})$
$B$: $K_2[Ni(CN)_4] \rightarrow 2K^+ + [Ni(CN)_4]^{2-}$ $(3 \, \text{ions})$
$C$: $FeSO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \rightarrow Fe^{2+} + 2Al^{3+} + 4SO_4^{2-}$ $(7 \, \text{ions})$
$D$: $Na_2[Ag(S_2O_3)_2] \rightarrow 2Na^+ + [Ag(S_2O_3)_2]^{2-}$ $(3 \, \text{ions})$
Since $FeSO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ produces the highest number of ions $(7)$,it exhibits the maximum conductivity.

(B) All the given solutions are $1 \, N$ (normal) solutions of strong electrolytes.
Since they are strong electrolytes,they dissociate completely in the solution.
However,the conductivity depends on the mobility of the ions.
Among the given options,$KCl$ provides $K^+$ and $Cl^-$ ions,which have higher ionic mobility compared to the ions produced by $CH_3COONa$ $(CH_3COO^-)$ and $NH_4Cl$ $(NH_4^+)$.
Therefore,$N \, KCl$ exhibits the maximum conductivity.

(D) The molar ionic conductivity of an ion in an aqueous solution depends on its size and charge.
Smaller ions with higher charge density generally have higher conductivity due to their higher mobility in water.
Among the given options,the hydrogen ion $(H^{+})$ has the smallest size and exhibits a unique mechanism of transport known as the Grotthuss mechanism (proton hopping) in water,which results in a significantly higher molar ionic conductivity compared to other ions like $K^{+}$,$Cl^{-}$,or $Ca^{2+}$.

(D) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity at infinite dilution is the sum of the equivalent conductivities of the individual ions.
$\lambda_{eq}^{\infty} (Al_2(SO_4)_3) = \lambda_{eq}^{\infty} (Al^{3+}) + \lambda_{eq}^{\infty} (SO_4^{2-})$
Given $\lambda_{eq}^{\infty} (Al^{3+}) = 189 \ \Omega^{-1} cm^2 eq^{-1}$ and $\lambda_{eq}^{\infty} (SO_4^{2-}) = 160 \ \Omega^{-1} cm^2 eq^{-1}$.
Therefore,$\lambda_{eq}^{\infty} (Al_2(SO_4)_3) = 189 + 160 = 349 \ \Omega^{-1} cm^2 eq^{-1}$.
Wait,the standard formula for equivalent conductivity at infinite dilution is $\lambda_{eq}^{\infty} = \lambda_{+}^{\infty} + \lambda_{-}^{\infty}$.
Thus,$189 + 160 = 349 \ \Omega^{-1} cm^2 eq^{-1}$.
Note: The provided solution in the prompt used a molar conductivity calculation approach,but the question specifically asks for equivalent conductivity.

(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for an electrolyte is the sum of the equivalent conductances of its constituent ions.
For $CH_3COONa$: $\Lambda^\infty_{CH_3COONa} = \lambda^\infty_{CH_3COO^-} + \lambda^\infty_{Na^+} = 91 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ $(i)$
For $HCl$: $\Lambda^\infty_{HCl} = \lambda^\infty_{H^+} + \lambda^\infty_{Cl^-} = 426 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ (ii)
For $CH_3COOH$: $\Lambda^\infty_{CH_3COOH} = \lambda^\infty_{CH_3COO^-} + \lambda^\infty_{H^+} = 391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ (iii)
We need to find $\Lambda^\infty_{NaCl} = \lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}$.
By performing the operation $(i)$ + (ii) - (iii):
$\Lambda^\infty_{NaCl} = (\lambda^\infty_{CH_3COO^-} + \lambda^\infty_{Na^+}) + (\lambda^\infty_{H^+} + \lambda^\infty_{Cl^-}) - (\lambda^\infty_{CH_3COO^-} + \lambda^\infty_{H^+})$
$\Lambda^\infty_{NaCl} = \lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}$
$\Lambda^\infty_{NaCl} = 91 + 426 - 391 = 126 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

(C) The cell constant $(G^*)$ is defined as the product of specific conductivity $(\kappa)$ and resistance $(R)$.
Given:
Specific conductivity $(\kappa) = 0.002765 \, S \, cm^{-1}$
Resistance $(R) = 400 \, \Omega$
Formula:
$G^* = \kappa \times R$
Calculation:
$G^* = 0.002765 \, S \, cm^{-1} \times 400 \, \Omega = 1.106 \, cm^{-1}$

(A) Molarity $(M)$ of $BaCl_2 = \frac{\text{mass} \times 1000}{\text{Molar mass} \times \text{Volume in } mL} = \frac{1.0 \times 1000}{208 \times 200} = 0.02404 \ M \approx 0.024 \ M$.
Normality $(N)$ of $BaCl_2 = M \times \text{valency factor} = 0.024 \times 2 = 0.048 \ N$.
Molar conductivity $(\Lambda_m)$ = $\frac{\kappa \times 1000}{M} = \frac{0.0058 \times 1000}{0.024} = 241.67 \ S \ cm^2 \ mol^{-1}$.
Equivalent conductivity $(\Lambda_{eq})$ = $\frac{\kappa \times 1000}{N} = \frac{0.0058 \times 1000}{0.048} = 120.83 \ S \ cm^2 \ eq^{-1}$.

(B) Equivalent conductance is directly proportional to the ionic character of the compound.
According to Fajan's rule,polarizing power is directly proportional to the covalent character.
The order of polarizing power of cations is $Li^{+} > Na^{+} > K^{+}$.
Therefore,the order of covalent character is $LiCl > NaCl > KCl$.
Since ionic character is the inverse of covalent character,the order of ionic character is $KCl > NaCl > LiCl$.
Thus,the order of equivalent conductance at infinite dilution is $KCl > NaCl > LiCl$.

(A) According to Kohlrausch's law,the limiting molar conductivity of a weak electrolyte like acetic acid $(HOAc)$ can be calculated from the limiting molar conductivities of strong electrolytes:
$\Lambda^{\infty}_{HOAc} = \Lambda^{\infty}_{NaOAc} + \Lambda^{\infty}_{HCl} - \Lambda^{\infty}_{NaCl}$
Substituting the given values:
$\Lambda^{\infty}_{HOAc} = 91.0 + 426.2 - 126.5$
$\Lambda^{\infty}_{HOAc} = 517.2 - 126.5 = 390.7 \, S \, cm^{2} \, mol^{-1}$

(D) Specific conductivity $(\kappa)$ is directly proportional to the concentration of the electrolyte solution.
As the concentration decreases,the number of ions per unit volume decreases,leading to a decrease in specific conductivity.
Therefore,the $0.001 \ M$ $KCl$ solution has the lowest value of specific conductivity among the given options.

(C) Given: Concentration $(C)$ = $0.01 \, \text{M}$,Resistance $(R)$ = $40 \, \Omega$,Cell constant $(G^*)$ = $0.4 \, \text{cm}^{-1}$.
Conductivity $(\kappa)$ = $\frac{G^*}{R} = \frac{0.4 \, \text{cm}^{-1}}{40 \, \Omega} = 0.01 \, \Omega^{-1} \, \text{cm}^{-1}$.
Molar conductivity $(\Lambda_m)$ = $\frac{\kappa \times 1000}{C} = \frac{0.01 \times 1000}{0.01} = 1000 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$.
Therefore,$\Lambda_m = 10^3 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$.

(A) Step $1$: Calculate the cell constant $(G^* = \frac{\ell}{A})$.
Given for $0.1 \ M$ solution: $R = 100 \ \Omega$,$\kappa = 1.29 \ S \ m^{-1}$.
$G^* = \kappa \times R = 1.29 \ S \ m^{-1} \times 100 \ \Omega = 129 \ m^{-1}$.
Step $2$: Calculate conductivity $(\kappa)$ for $0.02 \ M$ solution.
Given: $R = 520 \ \Omega$,$G^* = 129 \ m^{-1}$.
$\kappa = \frac{G^*}{R} = \frac{129}{520} \approx 0.248 \ S \ m^{-1}$.
Step $3$: Calculate molar conductivity $(\Lambda_m)$.
$\Lambda_m = \frac{\kappa}{C} = \frac{0.248 \ S \ m^{-1}}{0.02 \ mol \ L^{-1}} = \frac{0.248 \ S \ m^{-1}}{0.02 \times 10^3 \ mol \ m^{-3}} = \frac{0.248}{20} \ S \ m^2 \ mol^{-1} = 0.0124 \ S \ m^2 \ mol^{-1}$.
Converting to scientific notation: $124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$.

(B) The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a specific concentration $(\Lambda_c)$ to the molar conductivity at infinite dilution $(\Lambda_\infty)$.
$\alpha = \frac{\Lambda_c}{\Lambda_\infty}$
Given $\Lambda_c = 80 \, S \, cm^2 \, mol^{-1}$ and $\Lambda_\infty = 400 \, S \, cm^2 \, mol^{-1}$.
$\alpha = \frac{80}{400} = \frac{1}{5} = 0.2$.

(C) The equivalent conductivity of a strong electrolyte increases with dilution. $ \Lambda_{eq} = \kappa \times V $. As dilution increases,the volume $ V $ containing one equivalent of the electrolyte increases,which leads to an increase in equivalent conductivity.

(A) Given: Cell constant $(G^*) = \frac{l}{a} = 0.5 \ cm^{-1}$,Resistance $(R) = 50 \ \Omega$,Normality $(N) = 1 \ N$.
Conductivity $(\kappa) = \frac{1}{R} \times \frac{l}{a} = \frac{1}{50} \times 0.5 = 0.01 \ \Omega^{-1} \ cm^{-1}$.
Equivalent conductivity $(\Lambda_{eq}) = \frac{1000 \times \kappa}{N} = \frac{1000 \times 0.01}{1} = 10 \ \Omega^{-1} \ cm^2 \ (g \ eq)^{-1}$.

(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $HOAc$ (acetic acid) is given by:
$\wedge^o_{HOAc} = \wedge^o_{H^+} + \wedge^o_{OAc^-}$
We are given:
$\wedge^o_{NaOAc} = \wedge^o_{Na^+} + \wedge^o_{OAc^-} = 91.0 \, S \, cm^2/mol$
$\wedge^o_{HCl} = \wedge^o_{H^+} + \wedge^o_{Cl^-} = 426.2 \, S \, cm^2/mol$
To obtain $\wedge^o_{HOAc}$,we perform the operation:
$\wedge^o_{HOAc} = \wedge^o_{NaOAc} + \wedge^o_{HCl} - \wedge^o_{NaCl}$
$\wedge^o_{HOAc} = (\wedge^o_{Na^+} + \wedge^o_{OAc^-}) + (\wedge^o_{H^+} + \wedge^o_{Cl^-}) - (\wedge^o_{Na^+} + \wedge^o_{Cl^-})$
$\wedge^o_{HOAc} = \wedge^o_{H^+} + \wedge^o_{OAc^-}$
Thus,we need to subtract the molar conductivity of $NaCl$ to cancel out the unwanted ions ($Na^+$ and $Cl^-$). The question asks for the value to be added to the sum of $\wedge^o_{NaOAc}$ and $\wedge^o_{HCl}$ to get the result,which is $-\wedge^o_{NaCl}$. However,among the options provided,$NaCl$ is the substance whose conductivity must be subtracted. If the question implies which substance is involved in the calculation,$NaCl$ is the correct choice.

(B) According to Kohlrausch's law of independent migration of ions:
$ᴧ^{0} CH_{3}COOH = ᴧ^{0} CH_{3}COO^{-} + ᴧ^{0} H^{+}$
To obtain this,we use the values of strong electrolytes:
$ᴧ^{0} CH_{3}COOH = ᴧ^{0} CH_{3}COONa + ᴧ^{0} HCl - ᴧ^{0} NaCl$
Therefore,the value of $ᴧ^{0} NaCl$ is required.

(C) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$
Here,$\kappa = 1.061 \times 10^{-4} \ S \ cm^{-1}$ and $M = 0.01 \ M$.
Substituting the values: $\Lambda_m = \frac{1.061 \times 10^{-4} \times 1000}{0.01}$
$\Lambda_m = \frac{1.061 \times 10^{-1}}{10^{-2}}$
$\Lambda_m = 1.061 \times 10^1 = 10.61 \ S \ cm^2 \ mol^{-1}$.