Conductor and Conductance and Cell constant Questions in English

(A) 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 $(A)$,i.e.,$G^* = l/A$.
Since it is difficult to measure $l$ and $A$ accurately for many cells,the cell constant is determined by measuring the resistance $(R)$ of the cell when filled with a standard solution of known conductivity $(\kappa)$,such as $KCl$ solution.
The relationship is given by: $\kappa = G^* / R$,which implies $G^* = \kappa \times R$.

(N/A) For determining the conductivity of an electrolytic solution,$AC$ (Alternating Current) is used instead of $DC$ (Direct Current).
This is because $DC$ causes electrolysis,which leads to a change in the concentration of the solution near the electrodes.
This change in concentration results in polarization effects,making it difficult to measure the true resistance of the solution.
$AC$ prevents these changes in concentration and polarization,allowing for an accurate measurement of the solution's conductivity.

(A) In a conductivity cell,the electrodes are typically made of platinum $(Pt)$ coated with finely divided platinum black.
This coating of $Pt$ particles significantly increases the effective surface area of the electrodes.
By increasing the surface area,the polarization effect at the electrode-electrolyte interface is minimized,which ensures more accurate and stable measurements of the electrolytic conductivity.

(B) To convert $S \ cm^{-1}$ to $S \ m^{-1}$,we use the conversion factor $1 \ cm = 10^{-2} \ m$.
Therefore,$1 \ cm^{-1} = (10^{-2} \ m)^{-1} = 10^2 \ m^{-1} = 100 \ m^{-1}$.
Given value: $0.0129 \ S \ cm^{-1}$.
Calculation: $0.0129 \ S \ cm^{-1} = 0.0129 \times 100 \ S \ m^{-1} = 1.29 \ S \ m^{-1}$.

(N/A) The conductivity $(\kappa)$ is calculated using the formula: $\kappa = G \times G^*$.
Here,$G$ is the conductance,which is the reciprocal of resistance $(G = \frac{1}{R})$,and $G^*$ is the cell constant,defined as the ratio of the distance between electrodes $(l)$ to the area of cross-section $(A)$,i.e.,$G^* = \frac{l}{A}$.

(C) We know that $1\, m = 100\, cm$.
Therefore,$1\, m^2 = (100\, cm)^2 = 10000\, cm^2 = 10^4\, cm^2$.
Given value is $0.01\, S\, m^2\, mol^{-1}$.
Substituting the conversion factor: $0.01 \times 10^4\, S\, cm^2\, mol^{-1}$.
$= 10^{-2} \times 10^4\, S\, cm^2\, mol^{-1} = 10^2\, S\, cm^2\, mol^{-1} = 100\, S\, cm^2\, mol^{-1}$.

(B) The cross-sectional area $(A)$ of a cylindrical conductor is the area of the circle formed by its cross-section.
For a circle with radius $(r)$,the area is given by the formula $A = \pi r^2$.

(D) The relationship between conductance $(G)$ and conductivity $(k)$ is given by the formula: $G = k \times (A/l)$,where $A$ is the area of cross-section and $l$ is the length of the conductor.
For $G = k$,the cell constant $(l/A)$ must be equal to $1 \ cm^{-1}$.
This occurs when the area of cross-section $A = 1 \ cm^2$ and the length $l = 1 \ cm$.
Since these conditions define a unit cube of solution,the volume is $1 \ cm^3$.
Therefore,all the given options are correct.

(N/A) The term $\Lambda _m$ represents the molar conductivity of an electrolytic solution.
It is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution of volume $V \ mL$.
In the given formula,$\kappa$ is the conductivity (specific conductance) in $S \ cm^{-1}$,$C$ is the molar concentration in $mol \ L^{-1}$,and the factor $1000$ is used to convert the concentration from $mol \ L^{-1}$ to $mol \ cm^{-3}$ (since $1 \ L = 1000 \ cm^3$).

(B) Molar conductivity $(\Lambda_m)$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution of volume $V$.
As the concentration of the solution increases,the number of ions per unit volume increases,but the volume of the solution containing one mole of electrolyte decreases.
Due to the decrease in volume and increased inter-ionic attractions at higher concentrations,the mobility of ions decreases.
Consequently,the molar conductivity $(\Lambda_m)$ decreases with an increase in the concentration of the solution.

(D) $(i)$ Conductivity $(\kappa)$ is defined as the conductance of $1 \ cm^3$ of solution. As concentration increases,the number of ions per unit volume increases,so conductivity increases. Thus,statement $(i)$ is False.
$(ii)$ As concentration decreases,the number of ions per unit volume decreases,so conductivity decreases. Thus,statement $(ii)$ is False.
$(iii)$ As temperature increases,the kinetic energy of ions increases and the viscosity of the solvent decreases,which increases the mobility of ions,thereby increasing conductivity. Conversely,if temperature is less,conductivity is less. Thus,statement $(iii)$ is False.
Therefore,all statements are False.

(B) $(i)$ For electrolytic solutions,conductivity increases with an increase in temperature because the kinetic energy of ions increases and viscosity of the solvent decreases,leading to higher ionic mobility. Thus,statement $(i)$ is $T$.
$(ii)$ $Cu$ is a metallic conductor. For metallic conductors,conductivity is inversely proportional to temperature due to increased scattering of electrons by vibrating metal ions. Thus,statement $(ii)$ is $F$.
$(iii)$ As temperature increases,the resistance of metallic conductors like $Cu$ increases,which means conductivity decreases. Thus,statement $(iii)$ is $T$.
Therefore,the correct sequence is $(i) T, (ii) F, (iii) T$.

(A) The molar conductivity at infinite dilution $\Lambda _m^o$ for $NaCl$ can be calculated using Kohlrausch's law of independent migration of ions.
$\Lambda _m^o (NaCl) = \lambda ^o (Na^+) + \lambda ^o (Cl^-)$.
Given the standard values at $298 \ K$:
$\lambda ^o (Na^+) = 50.1 \ S \ cm^2 \ mol^{-1}$
$\lambda ^o (Cl^-) = 76.3 \ S \ cm^2 \ mol^{-1}$
Therefore,$\Lambda _m^o (NaCl) = 50.1 + 76.3 = 126.4 \ S \ cm^2 \ mol^{-1}$.

(N/A) Alternating current $(AC)$ is used to prevent the occurrence of electrolysis in the electrolytic cell. If direct current $(DC)$ were used,it would lead to the migration of ions towards the electrodes and cause chemical changes,thereby altering the concentration of the solution. By using $AC$,the direction of the current changes rapidly,which prevents the net migration of ions and keeps the concentration of the solution constant,allowing for an accurate measurement of resistance.

(B) Electrolyte $B$ is a strong electrolyte because,upon dilution,the number of ions remains constant,and only the interionic attraction decreases,leading to a small increase in $\Lambda_m$.
Electrolyte $A$ is a weak electrolyte because it does not undergo complete ionization. Upon dilution,the degree of dissociation increases significantly,leading to a large increase in the number of ions,which causes a drastic increase ($25$ times) in its $\Lambda_m$ value.
Therefore,$B$ is the strong electrolyte.

(B) Specific conductivity (kappa) of a solution decreases with dilution.
On the addition of water to the electrolytic solution,the number of ions per unit volume decreases.
Since specific conductivity is defined as the conductance of $1 \ cm^3$ of the solution,a decrease in the number of ions per unit volume leads to a decrease in specific conductivity.

(D) The conductance of a saturated solution depends on both the ionic mobility and the solubility of the salt.
For $NaCl$,which is highly soluble,the solubility does not change significantly with temperature,but ionic mobility increases with $T$,so $C_{NaCl}$ increases.
For $BaSO_4$,although ionic mobility increases with $T$,the solubility of $BaSO_4$ is very low and its dissolution is endothermic. However,the increase in ionic mobility generally dominates the conductance behavior in dilute solutions.
Wait,re-evaluating: The solubility of $BaSO_4$ increases with temperature. Therefore,both the number of ions and the ionic mobility increase with $T$.
Actually,the statement that $C_{BaSO_4}$ decreases is false because both factors (solubility and mobility) increase with temperature for $BaSO_4$.

(A) The given graph shows a sharp increase in molar conductivity with a decrease in concentration (as $\sqrt{c}$ approaches zero). This behavior is characteristic of a weak electrolyte,where the degree of dissociation increases significantly upon dilution.
Among the given options,$CH_{3}COOH$ is a weak electrolyte,while $KNO_{3}$,$HCl$,and $NaCl$ are strong electrolytes which show a linear variation according to the Kohlrausch equation: $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$.
Therefore,the correct electrolyte is $CH_{3}COOH$.

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 \times \kappa}{M}$
$\lambda_{m} = \frac{1000 \times 10^{-3}}{0.05} = 20 \, S \, cm^{2} \, mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{20}{500} = 0.04$
Finally,calculate the dissociation constant $(K_{a})$ using the formula: $K_{a} = C \alpha^{2}$
$K_{a} = 0.05 \times (0.04)^{2} = 0.05 \times 0.0016 = 8 \times 10^{-5}$

(C) Given: Concentration $(C) = 5.0 \, mmol \, dm^{-3} = 5.0 \times 10^{-3} \, mol \, L^{-1} = 5.0 \, mol \, m^{-3}$.
Conductance $(G) = 0.55 \, mS = 0.55 \times 10^{-3} \, S$.
Cell constant $(G^*) = 1.3 \, cm^{-1} = 130 \, m^{-1}$.
Conductivity $(\kappa) = G \times G^* = 0.55 \, mS \times 130 \, m^{-1} = 71.5 \, mS \, m^{-1} = 0.0715 \, S \, m^{-1}$.
Molar conductivity $(\lambda_m) = \frac{\kappa}{C} = \frac{0.0715 \, S \, m^{-1}}{5.0 \, mol \, m^{-3}} = 0.0143 \, S \, m^2 \, mol^{-1} = 14.3 \, mS \, m^2 \, mol^{-1}$.
Rounding to the nearest integer,the value is $14$.

(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$
For the same conductivity cell,$G^{*}$ is constant,therefore $\kappa \cdot R = G^{*} = \text{constant}$.
For the $KCl$ solution: $\kappa_{KCl} \cdot R_{KCl} = 0.14 \times 4.19 = 0.5866 \, S$.
For the $HCl$ solution: $\kappa_{HCl} \cdot R_{HCl} = 0.5866 \, S$.
$\kappa_{HCl} = \frac{0.5866}{1.03} \approx 0.5695 \, S m^{-1}$.
Converting to the required units: $0.5695 \, S m^{-1} = 56.95 \times 10^{-2} \, S m^{-1}$.
Rounding to the nearest integer,we get $57 \times 10^{-2} \, S m^{-1}$.

(D) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\infty}(SO_{4}^{2-})$
We can express this in terms of the given electrolytes:
$\Lambda_{m}^{\infty}(BaSO_{4}) = \Lambda_{m}^{\infty}(BaCl_{2}) + \Lambda_{m}^{\infty}(H_{2}SO_{4}) - 2\Lambda_{m}^{\infty}(HCl)$
Substituting the given values:
$\Lambda_{m}^{\infty}(BaSO_{4}) = 280 + 860 - 2(426)$
$= 1140 - 852$
$= 288 \, S \, cm^{2} \, mol^{-1}$

(D) Statement $I$: The limiting molar conductivity $(\Lambda_{m}^{\infty})$ is the sum of the ionic conductivities of the constituent ions. For $CH_{3}COOH$,$\Lambda_{m}^{\infty} = \lambda^{\infty}(H^{+}) + \lambda^{\infty}(CH_{3}COO^{-}) \approx 349.8 + 40.9 = 390.7 \, S \, cm^{2} \, mol^{-1}$. For $KCl$,$\Lambda_{m}^{\infty} = \lambda^{\infty}(K^{+}) + \lambda^{\infty}(Cl^{-}) \approx 73.5 + 76.3 = 149.8 \, S \, cm^{2} \, mol^{-1}$. Since $390.7 > 149.8$,Statement $I$ is false.
Statement $II$: Molar conductivity $(\Lambda_{m})$ is defined as $\Lambda_{m} = \frac{\kappa}{c}$. As concentration $(c)$ decreases,the dilution increases. For both strong and weak electrolytes,the decrease in concentration leads to an increase in molar conductivity because the interionic attractions decrease (for strong electrolytes) or the degree of dissociation increases (for weak electrolytes). Thus,Statement $II$ is false.

(A) The conductivity $\kappa$ is given by $\kappa = \frac{1}{R} \times \left(\frac{\ell}{A}\right)$.
Given $R = 1500 \, \Omega$ and cell constant $\frac{\ell}{A} = 1.14 \, cm^{-1}$.
$\kappa = \frac{1}{1500} \times 1.14 = 7.6 \times 10^{-4} \, S \, cm^{-1}$.
Molar conductivity $\wedge_{m}$ is calculated as $\wedge_{m} = \frac{1000 \times \kappa}{C}$.
$\wedge_{m} = \frac{1000 \times 7.6 \times 10^{-4}}{0.001} = \frac{0.76}{0.001} = 760 \, S \, cm^{2} \, mol^{-1}$.

(A) The cell constant is defined as the ratio of the distance between electrodes $(\ell)$ to the area of cross-section $(A)$,so its unit is $m^{-1}$.
Molar conductivity $(\Lambda_{m})$ is defined as the conductance of all ions produced from one mole of electrolyte,with units $S\, m^{2}\, mol^{-1}$ (or $S\, cm^{2}\, mol^{-1}$).
Conductivity $(\kappa)$ is the reciprocal of resistivity,with units $\Omega^{-1}\, m^{-1}$ or $S\, m^{-1}$.
Degree of dissociation $(\alpha)$ is the ratio of the number of moles dissociated to the total number of moles,which is a dimensionless quantity.
Therefore,the correct matching is $a-iii, b-i, c-iv, d-ii$.

(C) Given: Conductivity $\kappa = 1.07 \times 10^{6} \, S \, m^{-1}$ and Resistance $R = 0.243 \, \Omega$.
The relationship between conductivity $(\kappa)$,conductance $(G)$,and cell constant $(G^{*})$ is $\kappa = G \times G^{*}$.
Since $G = \frac{1}{R}$,we have $\kappa = \frac{1}{R} \times G^{*}$.
Rearranging for the cell constant: $G^{*} = \kappa \times R$.
Substituting the values: $G^{*} = (1.07 \times 10^{6} \, S \, m^{-1}) \times (0.243 \, \Omega) = 0.26001 \times 10^{6} \, m^{-1}$.
Expressing in the form $x \times 10^{4} \, m^{-1}$: $G^{*} = 26.001 \times 10^{4} \, m^{-1}$.
Therefore,the value of $x$ is $26$.

(C) First,calculate the limiting molar conductivity of acetic acid:
$\Lambda_{m}^{\circ}(CH_{3}COOH) = \Lambda_{H^{+}}^{\circ} + \Lambda_{CH_{3}COO^{-}}^{\circ} = 350 + 50 = 400 \ S \ cm^{2} \ mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{400} = 0.05$
Finally,calculate the dissociation constant $(K_{a})$ using the formula $K_{a} = C \alpha^{2} / (1 - \alpha)$. Since $\alpha$ is very small,we use $K_{a} \approx C \alpha^{2}$:
$K_{a} = 0.007 \times (0.05)^{2}$
$K_{a} = 7 \times 10^{-3} \times 25 \times 10^{-4}$
$K_{a} = 175 \times 10^{-7} = 1.75 \times 10^{-5} \ mol \ L^{-1}$
Thus,the value is $1.75$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $AgCl$ can be calculated as:
$\Lambda_{m}^{\infty}(AgCl) = \Lambda_{m}^{\infty}(Ag^{+}) + \Lambda_{m}^{\infty}(Cl^{-})$
Using the given values:
$\Lambda_{m}^{\infty}(AgCl) = \Lambda_{m}^{\infty}(AgNO_{3}) + \Lambda_{m}^{\infty}(KCl) - \Lambda_{m}^{\infty}(KNO_{3})$
Substituting the values:
$= 133.4 + 149.9 - 144.9$
$= 283.3 - 144.9$
$= 138.4 \, S \, cm^{2} \, mol^{-1}$
Rounding to the nearest integer,the value is $138 \, S \, cm^{2} \, mol^{-1}$.

(C) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$k = \frac{1}{R} \times G^*$
Given:
$R = 1750 \, \Omega$
$k = 0.152 \times 10^{-3} \, S \, cm^{-1}$
Substituting the values:
$0.152 \times 10^{-3} = \frac{1}{1750} \times G^*$
$G^* = 0.152 \times 10^{-3} \times 1750$
$G^* = 266 \times 10^{-3} \, cm^{-1}$

(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(Ag^+) + \lambda_{m}^{\infty}(I^-)$
We can express this using the given values:
$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(AgNO_3) + \lambda_{m}^{\infty}(NaI) - \lambda_{m}^{\infty}(NaNO_3)$
Substituting the given values:
$\lambda_{m}^{\infty}(AgI) = 13.3 + 12.7 - 12.0$
$\lambda_{m}^{\infty}(AgI) = 26.0 - 12.0$
$\lambda_{m}^{\infty}(AgI) = 14.0 \, mS \, m^2 \, mol^{-1}$

(A) The formula for molar conductivity is $\Lambda_{m} = \kappa \times \frac{1000}{M}$,where $M$ is the molarity of the solution.
Since the conductivities $\kappa$ are the same,$\Lambda_{m} \propto \frac{1}{M}$.
Calculate the molarity for both solutions:
$M_1 = \frac{n_1}{V_1} = \frac{10 \ mol}{20 \ mL} = 0.5 \ mol/mL$.
$M_2 = \frac{n_2}{V_2} = \frac{20 \ mol}{80 \ mL} = 0.25 \ mol/mL$.
Now,find the ratio:
$\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{M_1}{M_2} = \frac{0.5}{0.25} = 2$.
Therefore,$\Lambda_{m2} = 2 \Lambda_{m1}$.

(A) Cell constant $G^* = \frac{\ell}{A} = 129 \; m^{-1} = 1.29 \; cm^{-1}$.
Conductivity $\kappa = \frac{G^*}{R}$.
For solution $1$ $(74.5 \; ppm)$: $\kappa_1 = \frac{1.29}{100} \; S \; cm^{-1}$.
Concentration $C_1 \propto 74.5 \; ppm$.
For solution $2$ $(149 \; ppm)$: $\kappa_2 = \frac{1.29}{50} \; S \; cm^{-1}$.
Concentration $C_2 \propto 149 \; ppm$.
Since $ppm$ is proportional to molarity $(M)$,$\frac{C_1}{C_2} = \frac{74.5}{149} = \frac{1}{2}$.
Molar conductivity $\wedge_m = \frac{1000 \kappa}{C}$.
$\frac{\wedge_1}{\wedge_2} = \frac{\kappa_1}{\kappa_2} \times \frac{C_2}{C_1} = \frac{1.29/100}{1.29/50} \times \frac{149}{74.5} = \frac{50}{100} \times 2 = 1$.
Given $\frac{\wedge_1}{\wedge_2} = x \times 10^{-3} = 1$.
Therefore,$x = 1000$.

(B) The principle of conductometric titration is based on the fact that during the titration,one of the ions is replaced by another,and these two ions differ in their ionic conductivity.
In the conductometric titration of $0.05 \ M \ H_2SO_4$ with $0.1 \ M \ NH_4OH$,initially,the highly mobile $H^+$ ions are neutralized by $OH^-$ ions to form $H_2O$ and are replaced by less mobile $NH_4^+$ ions. This leads to a decrease in conductance up to the equivalence point.
After the equivalence point,the addition of excess weak electrolyte $NH_4OH$ does not significantly increase the conductance because it is weakly dissociated.
Therefore,the plot showing a decrease in conductance followed by a nearly constant conductance is the correct representation.

(A) The limiting molar conductivities of the given electrolytes are calculated using Kohlrausch's law:
$\lambda^{\circ}_{KCl} = \lambda^{\circ}_{K^{+}} + \lambda^{\circ}_{Cl^{-}} = 73.5 + 76.3 = 149.8 \ S \ cm^{2} \ mol^{-1}$
$\lambda^{\circ}_{CaCl_{2}} = \lambda^{\circ}_{Ca^{2+}} + 2\lambda^{\circ}_{Cl^{-}} = 119.0 + 2 \times 76.3 = 271.6 \ S \ cm^{2} \ mol^{-1}$
$\lambda^{\circ}_{K_{2}SO_{4}} = 2\lambda^{\circ}_{K^{+}} + \lambda^{\circ}_{SO_{4}^{2-}} = 2 \times 73.5 + 160.0 = 307.0 \ S \ cm^{2} \ mol^{-1}$
Comparing the values,the order is: $\lambda^{\circ}_{KCl} < \lambda^{\circ}_{CaCl_{2}} < \lambda^{\circ}_{K_{2}SO_{4}}$.

(A) Given,specific conductance $(\kappa) = 1.65 \times 10^{-4} \ S \ cm^{-1}$ and molarity $(M) = 0.02 \ M$.
Molar conductance $(\lambda_{m})$ is calculated as:
$\lambda_{m} = \frac{1000 \times \kappa}{M} = \frac{1000 \times 1.65 \times 10^{-4}}{0.02} = 8.25 \ S \ cm^{2} \ mol^{-1}$.
Molar conductance at infinite dilution $(\lambda_{m}^{\infty})$ is:
$\lambda_{m}^{\infty} = \lambda_{m(H^{+})}^{\infty} + \lambda_{m(CH_{3}COO^{-})}^{\infty} = 349.1 + 40.9 = 390 \ S \ cm^{2} \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is given by:
$\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{8.25}{390} \approx 0.0211$.

(A) Molar conductivity at infinite dilution $(\Lambda_{m}^{\circ})$ depends on the mobility of ions in the solution.
$HCl$ is a strong acid and dissociates completely into $H^{ }$ and $Cl^{-}$ ions. $H^{ }$ ions have the highest ionic mobility.
$NaCl$ is a strong electrolyte,dissociating into $Na^{ }$ and $Cl^{-}$ ions.
$CH_{3}COONa$ is a strong electrolyte,dissociating into $CH_{3}COO^{-}$ and $Na^{ }$ ions.
$CH_{3}COOH$ is a weak acid and dissociates only partially,resulting in fewer ions in the solution compared to strong electrolytes.
Since $HCl$ provides the highly mobile $H^{ }$ ion,it has the highest molar conductivity.
$NaCl$ has higher conductivity than $CH_{3}COONa$ because the $Cl^{-}$ ion has higher mobility than the $CH_{3}COO^{-}$ ion.
$CH_{3}COOH$ has the lowest molar conductivity due to its weak dissociation.
Thus,the correct order is $HCl > NaCl > CH_{3}COONa > CH_{3}COOH$.

(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $KOH$ can be expressed as:
$\lambda^{\infty}_{KOH} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{OH^-}$
Given values:
$\lambda^{\infty}_{NaCl} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$
$\lambda^{\infty}_{KCl} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$
$\lambda^{\infty}_{NaOH} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{OH^-} = 250 \ S \ cm^2 \ mol^{-1}$
To obtain $\lambda^{\infty}_{KOH}$,we perform the operation:
$\lambda^{\infty}_{KOH} = \lambda^{\infty}_{KCl} + \lambda^{\infty}_{NaOH} - \lambda^{\infty}_{NaCl}$
$\lambda^{\infty}_{KOH} = 150 + 250 - 126$
$\lambda^{\infty}_{KOH} = 400 - 126 = 274 \ S \ cm^2 \ mol^{-1}$
Therefore,the correct option is $D$.

(A) The correct answer is $A$.
Conductivity of an electrolytic solution depends on the nature of the electrolyte,the size of the ions produced,their solvation,the nature of the solvent,and its viscosity.
Option $A$ is incorrect because conductivity is inversely proportional to the viscosity of the solvent.
Option $C$ is also technically incorrect as molar conductivity decreases with an increase in concentration,but specific conductivity increases; however,in the context of general electrolytic conductivity,$A$ is the most fundamentally incorrect statement regarding the physical properties of the solution.

(B) Electrolyte $B$ is a strong electrolyte,and electrolyte $A$ is a weak electrolyte.
Statement $(A)$ is incorrect because $\Lambda_m^0$ for a weak electrolyte $(A)$ cannot be obtained by extrapolation; it is calculated using Kohlrausch's law.
Statement $(B)$ is correct because strong electrolytes follow the Debye-$H$ückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$,which is a straight line.
Statement $(C)$ is incorrect because at infinite dilution,the degree of dissociation $(\alpha)$ for any electrolyte approaches $1$ (or $100\%$),not zero.
Statement $(D)$ is correct because Kohlrausch's law of independent migration of ions states that $\Lambda_m^0$ can be calculated for any electrolyte using the limiting molar conductivities of its individual ions.
Therefore,statements $(A)$ and $(C)$ are incorrect. The total number of incorrect statements is $2$.

(C) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given that conductivity $\kappa = \frac{1}{\rho}$,where $\rho = 5 \times 10^{-3} \ \Omega \ cm$.
Substituting the values: $\Lambda_{m} = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}$.
$\Lambda_{m} = 200 \times 1250 = 250,000 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Expressing in the required form: $250,000 = 25 \times 10^4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Thus,the nearest integer is $25$.

(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k \times 1000}{C}$
Given $k = 5 \times 10^{-5} \ S \ cm^{-1}$ and $C = 0.0025 \ M$.
$\wedge_m = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{0.05}{0.0025} = 20 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $\alpha$ is given by $\alpha = \frac{\wedge_m}{\wedge_m^\circ} = \frac{20}{400} = 0.05$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha}$.
Since $\alpha$ is very small,$1 - \alpha \approx 1$,so $K_a \approx C \alpha^2 = 0.0025 \times (0.05)^2 = 0.0025 \times 0.0025 = 6.25 \times 10^{-6} = 62.5 \times 10^{-7}$.
Rounding to the nearest integer,we get $63 \times 10^{-7}$.

(B) Conductivity $(\kappa)$ decreases with dilution because the number of ions per unit volume decreases. This is true for both strong and weak electrolytes. Statement $(A)$ is correct.
$(B)$ On dilution,the volume of the solution increases,so the number of ions per unit volume decreases. Statement $(B)$ is incorrect.
$(C)$ Molar conductivity $(\Lambda_m)$ increases with decrease in concentration (dilution) because the total volume containing one mole of electrolyte increases. Statement $(C)$ is correct.
$(D)$ For strong electrolytes,$\Lambda_m$ increases slowly with dilution,whereas for weak electrolytes,it increases sharply due to an increase in the degree of dissociation. Statement $(D)$ is correct.
$(E)$ For weak electrolytes,the change in molar conductivity with dilution is due to an increase in the degree of dissociation $(\alpha)$,not a decrease. Statement $(E)$ is incorrect.
The correct statements are $(A)$,$(C)$,and $(D)$. Therefore,the total number of correct statements is $3$.

(D) The relationship between conductivity $(k)$,conductance $(G)$,and cell constant $(G^*)$ is given by:
$k = G \times G^*$
Since conductance $G = \frac{1}{R}$,where $R$ is the resistance:
$k = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant $(G^*)$:
$G^* = k \times R$
Given:
$k = 0.0210 \, \Omega^{-1} cm^{-1}$
$R = 60 \, \Omega$
Calculation:
$G^* = 0.0210 \, \Omega^{-1} cm^{-1} \times 60 \, \Omega = 1.26 \, cm^{-1}$

(B) The electrolytic conductance of a solution depends on the nature of the electrolyte (degree of dissociation),the concentration of the electrolyte,the temperature,and the nature of the solvent (viscosity and dielectric constant).
The nature of the electrode used does not affect the conductance of the solution itself,as conductance is a property of the electrolyte solution.

(B) Materials with high conductivity (typically $> 10^2 \ S \ m^{-1}$) are classified as conductors.
Given values in $S \ m^{-1}$ are: $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$.
$1$. $2.1 \times 10^3$ (Conductor)
$2$. $1.0 \times 10^{-16}$ (Insulator)
$3$. $1.2 \times 10$ (Semiconductor)
$4$. $3.91$ (Semiconductor)
$5$. $1.5 \times 10^{-2}$ (Semiconductor)
$6$. $1 \times 10^{-1}$ (Semiconductor)
$7$. $1.0 \times 10^3$ (Conductor)
There are $2$ materials with high conductivity values ($2.1 \times 10^3$ and $1.0 \times 10^3$).
Therefore,the number of conductors is $2$.

(A) The Kohlrausch equation for strong electrolytes is given by: $\Lambda_{m} = \Lambda_{m}^{\circ} - A \sqrt{C}$.
Here,$\Lambda_{m}$ is the molar conductivity,$C$ is the concentration,and $A$ is a constant.
The unit of $\Lambda_{m}$ is $S \ cm^2 \ mol^{-1}$.
The unit of $\sqrt{C}$ is $(mol \ L^{-1})^{1/2} = mol^{1/2} \ L^{-1/2}$.
Since the slope $A = \frac{\Lambda_{m}}{\sqrt{C}}$,the unit of $A$ is $\frac{S \ cm^2 \ mol^{-1}}{mol^{1/2} \ L^{-1/2}} = S \ cm^2 \ mol^{-3/2} \ L^{1/2}$.

(B) 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 ionic conductivities of its constituent ions.
For an electrolyte $CA$ where $C$ is a divalent cation $(C^{2+})$ and $A$ is a divalent anion $(A^{2-})$,the molar conductivity is given by:
$\Lambda_m^\circ = \lambda_+ + \lambda_-$
Given:
$\lambda_{C^{2+}} = 57 \ S \ cm^2 \ mol^{-1}$
$\lambda_{A^{2-}} = 73 \ S \ cm^2 \ mol^{-1}$
Therefore:
$\Lambda_m^\circ = 57 + 73 = 130 \ S \ cm^2 \ mol^{-1}$

(B) The molar conductivity at infinite dilution,denoted as $\Lambda_m^\circ$,represents the limiting value of molar conductivity when the concentration of the electrolyte approaches zero (i.e.,at infinite dilution).
Since the solution is already at infinite dilution,the degree of dissociation of the weak electrolyte is already at its maximum value.
Adding more water to an already infinitely dilute solution does not change the concentration of ions or the degree of dissociation.
Therefore,the molar conductivity remains constant and does not change.

(C) For strong electrolytes,molar conductivity $(\Lambda_m)$ varies linearly with $C^{1/2}$ according to the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A \sqrt{C}$. This is represented by the straight line $B$.
For weak electrolytes,molar conductivity increases sharply with dilution (decrease in $C^{1/2}$) due to an increase in the degree of dissociation,as shown by the curve $A$.

(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{\alpha^2 C}{1 - \alpha}$.
Since $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$,we substitute this into the expression:
$K_{a} = \frac{(\Lambda_{m} / \Lambda_{m}^{\circ})^2 C}{1 - (\Lambda_{m} / \Lambda_{m}^{\circ})} = \frac{\Lambda_{m}^2 C}{\Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m})}$.
Rearranging this gives: $K_{a} \Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m}) = \Lambda_{m}^2 C$.
$K_{a} (\Lambda_{m}^{\circ})^2 - K_{a} \Lambda_{m} \Lambda_{m}^{\circ} = \Lambda_{m}^2 C$.
Therefore,the correct equation is $\Lambda_{m}^2 C + K_{a} \Lambda_{m} \Lambda_{m}^{\circ} - K_{a} (\Lambda_{m}^{\circ})^2 = 0$.