Conductor and Conductance and Cell constant Questions in English

(D) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
$\kappa = \frac{1}{\rho}$

(C) Key Idea: Kohlrausch law of independent migration of ions can be used to calculate $\Delta_m^o$ for weak electrolytes such as acetic acid.
Given:
$\Delta_{m(NaAc)}^o = 91 \ S \ cm^2 \ mol^{-1}$
$\Delta_{m(HCl)}^o = 425.9 \ S \ cm^2 \ mol^{-1}$
$\Delta_{m(NaCl)}^o = 126.4 \ S \ cm^2 \ mol^{-1}$
Using Kohlrausch law,the molar conductivity of acetic acid is calculated as:
$\Delta_{m(CH_3COOH)}^o = \Delta_{m(CH_3COONa)}^o + \Delta_{m(HCl)}^o - \Delta_{m(NaCl)}^o$
$= (91.0 + 425.9 - 126.4) \ S \ cm^2 \ mol^{-1}$
$= 390.5 \ S \ cm^2 \ mol^{-1}$

(B) Given: Cell constant,$\frac{l}{A} = 1.25 \ cm^{-1}$,Resistance,$R = 2.5 \times 10^3 \ \Omega$,Molarity,$M = 0.1 \ M$.
Conductivity,$\kappa = \frac{\text{Cell constant}}{R} = \frac{1.25}{2.5 \times 10^3} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \ S \ cm^{-1}$.
Molar conductivity,$\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{5 \times 10^{-4} \times 1000}{0.1} = \frac{0.5}{0.1} = 5.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) Conductivity $(\kappa)$ is defined as the conductance of a solution contained between two electrodes of unit area of cross-section separated by a unit distance.
On dilution,the total number of ions in the solution remains the same,but the number of ions per unit volume decreases.
Since conductivity is directly proportional to the number of ions per unit volume,it decreases upon dilution.

(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
Since the $SI$ unit of resistivity is $\Omega \ m$,the unit of conductivity is $\Omega^{-1} \ m^{-1}$.
Since $\Omega^{-1}$ is represented as Siemens $(S)$,the $SI$ unit of conductivity is $S \ m^{-1}$.

(B) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\wedge_m^c}{\wedge_m^0}$
Given: $\wedge_m^c = 7.92 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\wedge_m^0 = 232.7 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Substituting the values: $\alpha = \frac{7.92}{232.7} = 0.034035... \approx 0.0341$

(C) The conductivity of an electrolytic solution depends on the nature of the electrolyte,the concentration of the electrolyte,the nature of the solvent and its viscosity,and the temperature.
As the temperature increases,the kinetic energy of the ions increases and the viscosity of the solvent decreases,which leads to an increase in the conductivity of the solution.
Therefore,the statement that conductivity does not depend upon temperature is incorrect.

(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = G^* \times G$,where $G^*$ is the cell constant and $G$ is the conductance.
Conductance $(G)$ is the reciprocal of resistance $(R)$: $G = \frac{1}{R} = \frac{1}{100 \ \Omega} = 0.01 \ S$.
Given conductivity $\kappa = 1.29 \ S/m$.
Substituting the values into the formula: $1.29 \ S/m = G^* \times 0.01 \ S$.
Therefore,$G^* = \frac{1.29}{0.01} \ m^{-1} = 129 \ m^{-1}$.

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(C)$ for strong electrolytes is given by the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$.
In this equation,$A$ is a constant,and the slope of the graph of $\Lambda_m$ versus $\sqrt{C}$ is $-A$,which is negative.
Among the given options,$CH_3COONa$ (Sodium acetate) is a strong electrolyte,whereas $NH_4OH$,$CH_3COOH$,and $H_2O$ are weak electrolytes.
Therefore,the graph for Sodium acetate shows a negative slope.

(D) The graph of $\sqrt{C}$ versus $\Lambda_{m}$ is a straight line for strong electrolytes,as they dissociate completely in solution and follow the Kohlrausch equation: $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{C}$.
$HCl$,$NaCN$,and $NaCl$ are strong electrolytes,so they yield a straight line.
$HCN$ is a weak electrolyte,which does not dissociate completely at all concentrations. Therefore,its graph of $\sqrt{C}$ versus $\Lambda_{m}$ is not a straight line,but a curve that approaches $\Lambda_{m}^{\circ}$ asymptotically as $C \rightarrow 0$.

(C) The correct answer is $C$.
Metallic conductivity (or electronic conductivity) is due to the movement of electrons through the metal lattice.
As the temperature of a metal increases,the positive ions (kernels) in the lattice begin to vibrate more vigorously.
This increased vibration causes more collisions between the moving electrons and the vibrating ions,which increases the resistance of the metal.
Therefore,the electrical conductivity of a metal decreases,not increases,with an increase in temperature.

(A) Electronic conductance (metallic conductance) depends on the nature and structure of the metal,the number of valence electrons per atom,and temperature.
It does not depend on the concentration of the electrolyte,as electronic conductance is a property of metals,not electrolytic solutions.
Therefore,the correct option is $A$.

(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
We are given:
$\Lambda_{m}^0(NaCl) = 126.4 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{m}^0(HCl) = 425.9 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{m}^0(NaAc) = 91.0 \ S \ cm^2 \ mol^{-1}$
We need to find $\Lambda_{m}^0(HAc)$,which is given by:
$\Lambda_{m}^0(HAc) = \Lambda_{m}^0(H^+) + \Lambda_{m}^0(Ac^-)$
Using the given values:
$\Lambda_{m}^0(HAc) = \Lambda_{m}^0(HCl) + \Lambda_{m}^0(NaAc) - \Lambda_{m}^0(NaCl)$
$\Lambda_{m}^0(HAc) = 425.9 + 91.0 - 126.4$
$\Lambda_{m}^0(HAc) = 516.9 - 126.4 = 390.5 \ S \ cm^2 \ mol^{-1}$
Therefore,the correct option is $D$.

(B) The molar conductivity at infinite dilution for methanoic acid $(HCOOH)$ is calculated using Kohlrausch's law: $\Lambda_m^0(HCOOH) = \lambda^0(H^+) + \lambda^0(HCOO^-)$.
Substituting the given values: $\Lambda_m^0(HCOOH) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a given concentration $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^0)$: $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.
$\alpha = \frac{46.1}{404.2} \approx 0.114$.

(D) Electronic conductance in metals is primarily determined by the number of valence electrons per atom,the nature and structure of the metal,and temperature (as temperature increases,the vibration of metal ions increases,causing more scattering of electrons and thus decreasing conductance). Pressure has a negligible effect on the electronic conductance of metals. Therefore,the correct option is $D$.

(B) 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.
For acetic acid $(HAc)$,the expression is:
$\Lambda_{m(HAc)}^0 = \lambda_{H^+} + \lambda_{Ac^-}$
Using strong electrolytes $HCl$,$NaAc$,and $NaCl$:
$\Lambda_{m(HCl)}^0 = \lambda_{H^+} + \lambda_{Cl^-}$
$\Lambda_{m(NaAc)}^0 = \lambda_{Na^+} + \lambda_{Ac^-}$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+} + \lambda_{Cl^-}$
Therefore,$\Lambda_{m(HCl)}^0 + \Lambda_{m(NaAc)}^0 - \Lambda_{m(NaCl)}^0 = (\lambda_{H^+} + \lambda_{Cl^-}) + (\lambda_{Na^+} + \lambda_{Ac^-}) - (\lambda_{Na^+} + \lambda_{Cl^-}) = \lambda_{H^+} + \lambda_{Ac^-} = \Lambda_{m(HAc)}^0$.
Thus,the correct option is $B$.

(C) The resistance $R$ of a conductor is directly proportional to its length $l$ and inversely proportional to its area of cross-section $A$.
Mathematically,this is expressed as $R \propto \frac{l}{A}$.
Therefore,the correct relation is $R \propto \frac{l}{A}$.

(B) Electronic conductance is the property of metallic conductors. It depends on the number of valence electrons per atom,the structure of the metal,and the temperature.
Electronic conductance $\propto \frac{\text{number of valence electrons per atom}}{\text{Temperature}}$.

(C) The equation $\lambda_{m} = \lambda_{m}^{\circ} - A \sqrt{C}$ is the Debye-Huckel-Onsager equation.
In this equation,'$A$' is a constant that depends on the nature of the solvent and the temperature,but it also depends on the type of electrolyte (i.e.,the charges on the ions,such as $1:1$,$1:2$,$2:1$,or $2:2$ electrolytes).
$NaCl$ and $KBr$ are both $1:1$ electrolytes.
Since they are of the same type,they will have the same value of '$A$'.

(D) Given: Concentration $C = 0.1 \ M$,Molar conductivity at infinite dilution $\Lambda_{m}^{\circ} = 390 \ S \ cm^2 \ mol^{-1}$,Resistance $R = 2 \times 10^3 \ \Omega$,Cell constant $G^* = 0.78 \ cm^{-1}$.
First,calculate the conductivity $K$:
$K = \frac{G^*}{R} = \frac{0.78}{2 \times 10^3} = 3.9 \times 10^{-4} \ S \ cm^{-1}$.
Next,calculate the molar conductivity $\Lambda_{m}$:
$\Lambda_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{-4} \times 1000}{0.1} = 3.9 \ S \ cm^2 \ mol^{-1}$.
Calculate the degree of dissociation $\alpha$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{3.9}{390} = 0.01 = 10^{-2}$.
Calculate the concentration of $H^{+}$ ions:
$[H^{+}] = C \times \alpha = 0.1 \times 10^{-2} = 10^{-3} \ M$.
Finally,calculate the $pH$:
$pH = -\log[H^{+}] = -\log(10^{-3}) = 3$.

(C) We know that,molar conductivity,$\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
This implies that $\Lambda_{m}$ is inversely proportional to the molarity $(M)$.
The solution with the lowest concentration will have the maximum molar conductivity.
Comparing the given concentrations,$0.001 \ M$ is the lowest concentration.
Therefore,the solution with $0.001 \ M$ concentration has the maximum molar conductivity.

(D) Given,specific conductance,$\kappa = 6.3 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$.
$HNO_3$ concentration,$c = 0.1 \ M$.
Molar conductance formula is $\Lambda_m = \frac{\kappa \times 1000}{c}$.
Substituting the values: $\Lambda_m = \frac{6.3 \times 10^{-2} \times 1000}{0.1} = \frac{63}{0.1} = 630 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) Given: Resistance $(R) = 1500 \ \Omega$
Conductivity $(\kappa) = 0.1466 \times 10^{-3} \ S \ cm^{-1}$
The formula for cell constant $(G^*)$ is:
$G^* = \kappa \times R$
Substituting the values:
$G^* = (0.1466 \times 10^{-3} \ S \ cm^{-1}) \times (1500 \ \Omega)$
$G^* = 0.2199 \ \approx 0.219 \ cm^{-1}$

(B) The Debye-Huckel-Onsager equation is given by $\Lambda_{m} = \Lambda_{m}^{\circ} - A \sqrt{C}$.
In this equation,the constant $A$ depends on the stoichiometry of the electrolyte,which determines the charge type (e.g.,$1:1, 1:2, 2:2$ electrolytes).
Electrolytes with the same charge type (e.g.,both are $1:1$ electrolytes) will have the same value for the constant $A$.
$NH_{4}Cl$ dissociates into $NH_{4}^{+}$ and $Cl^{-}$ ($1:1$ electrolyte).
$NaBr$ dissociates into $Na^{+}$ and $Br^{-}$ ($1:1$ electrolyte).
Since both are $1:1$ electrolytes,they possess the same value for the constant $A$.

(A) The molar conductance $(\Lambda_{m})$ is related to specific conductance $(K)$ by the formula: $\Lambda_{m} = \frac{K \times 1000}{C}$.
Rearranging the formula to find the ratio of molar conductance to specific conductance: $\frac{\Lambda_{m}}{K} = \frac{1000}{C}$.
Given the concentration $C = 0.01 \ M$,we substitute the value into the equation:
$\frac{\Lambda_{m}}{K} = \frac{1000}{0.01} = 10^{5} \ cm^{3} \ mol^{-1}$.

(C) Specific conductance (or conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
As the solution is diluted,the number of ions present per unit volume $(1 \ cm^3)$ decreases.
Since the number of charge carriers per unit volume decreases,the specific conductance decreases with dilution.

(D) Specific conductance (also known as conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of an electrolytic solution.
Upon dilution,the number of ions per unit volume $(1 \ cm^3)$ decreases,which leads to a decrease in specific conductance.
Therefore,specific conductance is directly proportional to the concentration of the electrolyte.
Among the given options,$0.002 \ N$ is the lowest concentration,so it will have the least specific conductance.

(B) Given,$\Lambda_{m}^{\circ} = 140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$,$K = 1.85 \times 10^{-5} \ S \ m^{-1}$.
For a sparingly soluble salt,the molar conductivity $\Lambda_{m}^{\circ}$ is related to solubility $S$ (in $mol \ m^{-3}$) by the formula $\Lambda_{m}^{\circ} = \frac{K}{S}$.
$S = \frac{K}{\Lambda_{m}^{\circ}} = \frac{1.85 \times 10^{-5} \ S \ m^{-1}}{140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}} = 1.32 \times 10^{-3} \ mol \ m^{-3}$.
Since $1 \ m^{3} = 1000 \ L$,the solubility in $mol \ L^{-1}$ is $S = \frac{1.32 \times 10^{-3}}{1000} = 1.32 \times 10^{-6} \ mol \ L^{-1}$.
For the salt $AB$,$K_{sp} = S^{2} = (1.32 \times 10^{-6})^{2} = 1.74 \times 10^{-12}$.

(D) The formula for molar conductance $(\Lambda_{m})$ is given by: $\Lambda_{m} = \frac{\kappa \times 1000}{M}$
Here,$\kappa$ (specific conductance) = $0.0115 \ S \ cm^{-1}$
$M$ (molarity) = $0.05 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0115 \times 1000}{0.05}$
$\Lambda_{m} = \frac{11.5}{0.05}$
$\Lambda_{m} = 230 \ S \ cm^2 \ mol^{-1}$
Therefore,the correct option is $D$.

(A) Step $1$: Calculate the cell constant $(G^*)$.
Conductivity $(\kappa) = G^* \times \text{Conductance} = G^* / R$.
For $0.1 \ M$ $KCl$: $\kappa = 1.29 \ S \ m^{-1} = 1.29 \times 10^{-2} \ S \ cm^{-1}$.
$G^* = \kappa \times R = (1.29 \times 10^{-2} \ S \ cm^{-1}) \times (100 \ \Omega) = 1.29 \ cm^{-1}$.
Step $2$: Calculate the conductivity of $0.02 \ M$ $KCl$ solution.
$\kappa = G^* / R = 1.29 \ cm^{-1} / 520 \ \Omega \approx 0.00248 \ S \ cm^{-1}$.
Step $3$: Calculate molar conductivity $(\Lambda_m)$.
$\Lambda_m = (\kappa \times 1000) / M = (0.00248 \times 1000) / 0.02 = 2.48 / 0.02 = 124 \ S \ cm^2 \ mol^{-1}$.

(A) The electrical properties and their corresponding $SI$/common units are as follows:
$1.$ Molar conductivity $(\wedge_m)$ is measured in $S\,cm^2\,mol^{-1}$. Thus,$A-I$.
$2.$ Conductance $(G)$ is the reciprocal of resistance,measured in Siemens $(S)$. Thus,$B-II$.
$3.$ Conductivity $(\kappa)$ is measured in $S\,cm^{-1}$. Thus,$C-III$.
$4.$ Cell constant $(G^*)$ is defined as $l/A$,measured in $cm^{-1}$. Thus,$D-IV$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(A) Molar mass of $BaCl_2 = 137 + 2 \times 35.5 = 208 \ g \ mol^{-1}$.
Number of moles of $BaCl_2 = \frac{2.08 \ g}{208 \ g \ mol^{-1}} = 0.01 \ mol$.
Molarity $(C) = \frac{\text{moles}}{\text{Volume in } L} = \frac{0.01 \ mol}{0.2 \ L} = 0.05 \ M$.
Molar conductivity $(\Lambda_m) = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-3} \ \Omega^{-1} \ cm^{-1} \times 1000}{0.05 \ mol \ L^{-1}} = \frac{6}{0.05} = 120 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Given $\Lambda_m = x \times 10^2$,so $120 = x \times 100$,which gives $x = 1.2$.

(C) According to Kohlrausch's law of independent migration of ions:
$\wedge_{m}^{\circ}(NH_4OH) = \wedge_{m}^{\circ}(NH_4^+) + \wedge_{m}^{\circ}(OH^{-})$
We can express this using the given values:
$\wedge_{m}^{\circ}(NH_4OH) = \wedge_{m}^{\circ}(NH_4Cl) + \frac{1}{2} \wedge_{m}^{\circ}(Ba(OH)_2) - \frac{1}{2} \wedge_{m}^{\circ}(BaCl_2)$
Substituting the given values:
$\wedge_{m}^{\circ}(NH_4OH) = 213.0 + (\frac{1}{2} \times 457.0) - (\frac{1}{2} \times 240.6)$
$\wedge_{m}^{\circ}(NH_4OH) = 213.0 + 228.5 - 120.3 = 321.2 \ S \ cm^2 \ mol^{-1}$

(C) Conductivity $(K)$ is defined as the conductance of ions present in a unit volume of the solution.
Upon dilution,the total number of ions remains the same,but the volume of the solution increases.
Therefore,the number of ions per unit volume decreases,which leads to a decrease in conductivity.
Thus,Assertion $(A)$ is correct,but Reason $(R)$ is incorrect because the number of ions per unit volume decreases,not increases.

(B) Conductivity $(\kappa)$ is defined as the conductance of a solution of $1 \ cm^3$ volume.
It depends on the number of ions present per unit volume of the solution.
As the concentration of the electrolyte increases,the number of ions per unit volume increases,leading to an increase in conductivity.
Given molarities are $0.01 \ M$ $(Y)$,$0.1 \ M$ $(X)$,and $1.0 \ M$ $(Z)$.
Since $1.0 \ M > 0.1 \ M > 0.01 \ M$,the order of conductivity is $Z > X > Y$.

(D) The cell constant $G^*$ is given by $G^* = \kappa \times R$.
For the first solution: $G^* = 1.29 \ S \ m^{-1} \times 100 \ \Omega = 129 \ m^{-1}$.
For the second solution: $\kappa_2 = \frac{G^*}{R_2} = \frac{129 \ m^{-1}}{258 \ \Omega} = 0.5 \ S \ m^{-1}$.
Thus,the conductivity of the $0.02 \ mol \ L^{-1}$ $NaCl$ solution is $0.5 \ S \ m^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_{m}^0(KBr) = \lambda_{K^+} + \lambda_{Br^-} = 120.5 \ S \ cm^2 \ mol^{-1}$
$\lambda_{m}^0(HBr) = \lambda_{H^+} + \lambda_{Br^-} = 420.6 \ S \ cm^2 \ mol^{-1}$
$\lambda_{m}^0(KNH_2) = \lambda_{K^+} + \lambda_{NH_2^-} = 90.48 \ S \ cm^2 \ mol^{-1}$
We need to find $\lambda_{m}^0(NH_3) = \lambda_{H^+} + \lambda_{NH_2^-}$.
Using the given values:
$\lambda_{m}^0(NH_3) = \lambda_{m}^0(HBr) + \lambda_{m}^0(KNH_2) - \lambda_{m}^0(KBr)$
$\lambda_{m}^0(NH_3) = 420.6 + 90.48 - 120.5 = 390.58 \ S \ cm^2 \ mol^{-1}$.
Rounding to the nearest provided option,the value is $390.5 \ S \ cm^2 \ mol^{-1}$.

(B) According to Kohlrausch's law:
$\lambda_{m}^{\circ}(BaSO_4) = \lambda_{m}^{\circ}(Ba^{2+}) + \lambda_{m}^{\circ}(SO_4^{2-})$
$\lambda_{m}^{\circ}(BaSO_4) = 110 + 136.6 = 246.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
The conductivity of $BaSO_4$ is:
$\kappa_{BaSO_4} = \kappa_{\text{solution}} - \kappa_{\text{water}}$
$\kappa_{BaSO_4} = (3.648 \times 10^{-6}) - (1.25 \times 10^{-6}) = 2.398 \times 10^{-6} \ S \ cm^{-1}$
Solubility in $mol \ L^{-1}$ is given by:
$\text{Solubility} = \frac{\kappa \times 1000}{\lambda^{\circ}_{m}} = \frac{2.398 \times 10^{-6} \times 1000}{246.6} \approx 9.724 \times 10^{-6} \ mol \ L^{-1}$
Molar mass of $BaSO_4 = 137 + 32 + (4 \times 16) = 233 \ g \ mol^{-1}$
Solubility in $g \ L^{-1} = 9.724 \times 10^{-6} \ mol \ L^{-1} \times 233 \ g \ mol^{-1} \approx 2.266 \times 10^{-3} \ g \ L^{-1}$

(A) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by: $\kappa = \frac{1}{R} \times G^*$
Therefore,$G^* = \kappa \times R$
Given values are: $\kappa = 0.013 \ S \ cm^{-1}$ and $R = 300 \ \Omega$
Substituting these values: $G^* = 0.013 \ S \ cm^{-1} \times 300 \ \Omega = 3.9 \ cm^{-1}$.

(B) The formula for molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Given,conductivity $\kappa = 0.024 \ S \ cm^{-1}$ and molar concentration $M = 0.5 \ M$.
Substituting the values: $\Lambda_m = \frac{0.024 \times 1000}{0.5} = \frac{24}{0.5} = 48 \ S \ cm^2 \ mol^{-1}$.

(A) According to Kohlrausch law,the molar conductivity of an electrolyte can be expressed as the sum of the molar conductivities of its constituent ions.
$\Lambda_{m}^{\circ}(NaNO_3) = \lambda^{\circ}(Na^+) + \lambda^{\circ}(NO_3^-)$
We can express this in terms of the given electrolytes:
$\Lambda_{m}^{\circ}(NaNO_3) = \Lambda_{m}^{\circ}(NaCl) + \Lambda_{m}^{\circ}(KNO_3) - \Lambda_{m}^{\circ}(KCl)$
Substituting the given values:
$\Lambda_{m}^{\circ}(NaNO_3) = 120 + 90 - 100$
$\Lambda_{m}^{\circ}(NaNO_3) = 210 - 100 = 110 \ S \ cm^2 \ mol^{-1}$.

(A) $1$. Calculate the molar conductivity $(\wedge_m)$ using the formula: $\wedge_m = \frac{\kappa \times 1000}{C} = \frac{5.07 \times 10^{-5} \times 1000}{0.001} = 50.7 \ S \ cm^2 \ mol^{-1}$.
$2$. Calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\wedge_m}{\wedge_m^0} = \frac{50.7}{390} = 0.13$.
$3$. Calculate the dissociation constant $(K_a)$ using the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.
$4$. Substituting the values: $K_a = \frac{0.001 \times (0.13)^2}{1 - 0.13} = \frac{0.001 \times 0.0169}{0.87} \approx 1.94 \times 10^{-5}$.

(D) Given: Concentration $C = 0.02 \ M = 0.02 \ mol \ L^{-1} = 0.02 \times 1000 \ mol \ m^{-3} = 20 \ mol \ m^{-3}$.
Cell constant $G^* = 129 \ m^{-1}$.
Molar conductivity $\Lambda_m = 124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$.
Formula: $\Lambda_m = \frac{\kappa}{C}$,where $\kappa$ is conductivity.
$\kappa = \Lambda_m \times C = (124 \times 10^{-4} \ S \ m^2 \ mol^{-1}) \times (20 \ mol \ m^{-3}) = 2480 \times 10^{-4} \ S \ m^{-1} = 0.248 \ S \ m^{-1}$.
Resistance $R = \frac{G^*}{\kappa} = \frac{129 \ m^{-1}}{0.248 \ S \ m^{-1}} \approx 520.16 \ \Omega$.
Rounding to the nearest integer,$R = 520 \ \Omega$.

(D) The molar conductivity $\Lambda_{m}$ is related to the degree of dissociation $\alpha$ and limiting molar conductivity $\Lambda_{m}^{\circ}$ by the formula: $\Lambda_{m} = \alpha \times \Lambda_{m}^{\circ}$.
Substituting the given values: $\Lambda_{m} = 0.02 \times 400 = 8.0 \ S \ cm^2 \ mol^{-1}$.
The conductivity $\kappa$ is related to molar conductivity $\Lambda_{m}$ and concentration $C$ (in $mol \ L^{-1}$) by: $\kappa = \frac{\Lambda_{m} \times C}{1000}$.
Substituting the values: $\kappa = \frac{8.0 \times 0.1}{1000} = 0.8 \times 10^{-3} = 8 \times 10^{-4} \ S \ cm^{-1}$.

(B) Acetic acid $(CH_3COOH)$ is a weak electrolyte.
For weak electrolytes,the degree of dissociation increases with dilution (decrease in concentration).
As a result,the molar conductivity $(\lambda_{m})$ increases sharply as the concentration $(C)$ approaches zero.
The plot of $\lambda_{m}$ versus $\sqrt{C}$ for a weak electrolyte shows a steep increase as $\sqrt{C}$ decreases,and it does not intersect the y-axis at finite concentration.
Graph $B$ correctly represents this behavior,where $\lambda_{m}$ increases as $\sqrt{C}$ decreases.

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^{\circ}(NH_4OH) = \Lambda_m^{\circ}(NH_4Cl) + \Lambda_m^{\circ}(KOH) - \Lambda_m^{\circ}(KCl)$
$= 152.8 + 272.6 - 149.8 = 275.6 \ S \ cm^2 \ mol^{-1}$
Degree of dissociation $(\alpha)$ is given by:
$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{25.1}{275.6} \approx 0.091$
Percentage dissociation $= \alpha \times 100 = 0.091 \times 100 = 9.1 \%$

(A) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{\infty}(NH_4Cl) = \Lambda_{\infty}(NH_4^+) + \Lambda_{\infty}(Cl^-) = 130 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(i)$
$\Lambda_{\infty}(NaOH) = \Lambda_{\infty}(Na^+) + \Lambda_{\infty}(OH^-) = 217 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(ii)$
$\Lambda_{\infty}(NaCl) = \Lambda_{\infty}(Na^+) + \Lambda_{\infty}(Cl^-) = 109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(iii)$
To find $\Lambda_{\infty}(NH_4OH)$,we perform the operation: $(i) + (ii) - (iii)$
$\Lambda_{\infty}(NH_4OH) = \Lambda_{\infty}(NH_4^+) + \Lambda_{\infty}(OH^-) = \Lambda_{\infty}(NH_4Cl) + \Lambda_{\infty}(NaOH) - \Lambda_{\infty}(NaCl)$
$\Lambda_{\infty}(NH_4OH) = 130 + 217 - 109 = 238 \ ohm^{-1} \ cm^2 \ equiv^{-1}$

(C) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for benzoic acid $(C_6H_5COOH)$ can be calculated as follows:
$\wedge_{C_6H_5COOH}^{\infty} = \wedge_{C_6H_5COONa}^{\infty} + \wedge_{HCl}^{\infty} - \wedge_{NaCl}^{\infty}$
Given values:
$\wedge_{C_6H_5COONa}^{\infty} = 240 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
$\wedge_{HCl}^{\infty} = 349 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
$\wedge_{NaCl}^{\infty} = 229 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
Substituting the values:
$\wedge_{C_6H_5COOH}^{\infty} = 240 + 349 - 229$
$= 589 - 229$
$= 360 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$