Conductor and Conductance and Cell constant Questions in English

(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:
$\wedge_{C_6H_5COOH}^{\infty} = 240 + 349 - 229$
$= 589 - 229$
$= 360 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$

(C) Equivalent conductivity,$\wedge_{eq} = \frac{\kappa \times 1000}{C}$,where $\kappa$ is specific conductance.
Since specific conductance $\kappa = \frac{1}{R}$ (where $R$ is specific resistance,often denoted as resistivity $\rho$ in some contexts,but here $R$ represents the specific resistance value).
Substituting the value of $\kappa$ into the formula:
$\wedge_{eq} = \frac{1}{R} \times \frac{1000}{C} = \frac{1000}{R C}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_m^{\circ} Ba(OH)_2 = \lambda_m^{\circ} BaCl_2 + 2 \lambda_m^{\circ} NaOH - 2 \lambda_m^{\circ} NaCl$
Substituting the given values:
$\lambda_m^{\circ} Ba(OH)_2 = (280 \times 10^{-4}) + 2(248 \times 10^{-4}) - 2(126 \times 10^{-4})$
$\lambda_m^{\circ} Ba(OH)_2 = (280 + 496 - 252) \times 10^{-4} \ S \ m^2 \ mol^{-1}$
$\lambda_m^{\circ} Ba(OH)_2 = 524 \times 10^{-4} \ S \ m^2 \ mol^{-1}$

(C) The specific conductivity $(\kappa)$ is given by the formula: $\kappa = G \times \frac{l}{A}$, where $G$ is conductance, $l$ is thickness, and $A$ is the area of the electrode.
Given: Conductance $G = \frac{314}{5} \ S = 62.8 \ S$.
Diameter $= 20 \ mm = 2 \ cm$, so radius $r = 1 \ cm$.
Area $A = \pi r^2 = 3.14 \times (1 \ cm)^2 = 3.14 \ cm^2$.
Thickness $l = 20 \ \mu m = 20 \times 10^{-4} \ cm$.
Substituting these values: $\kappa = 62.8 \times \frac{20 \times 10^{-4}}{3.14} \ S \ cm^{-1}$.
$\kappa = 20 \times 20 \times 10^{-4} \ S \ cm^{-1} = 400 \times 10^{-4} \ S \ cm^{-1} = 0.04 \ S \ cm^{-1}$.
Converting to $mS \ cm^{-1}$: $0.04 \ S \ cm^{-1} \times 1000 \ mS/S = 40 \ mS \ cm^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_{m(NH_4OH)}^{\infty} = \lambda_{m(NH_4Cl)}^{\infty} + \lambda_{m(Ba(OH)_2)}^{\infty} / 2 - \lambda_{m(BaCl_2)}^{\infty} / 2$
$\lambda_{m(NH_4OH)}^{\infty} = 129.8 + 523.28 / 2 - 280.0 / 2$
$\lambda_{m(NH_4OH)}^{\infty} = 129.8 + 261.64 - 140.0$
$\lambda_{m(NH_4OH)}^{\infty} = 251.44 \ S \ cm^2 \ mol^{-1}$

(A) Step $1$: Calculate the molar conductivity $(\lambda_m)$ at the given concentration.
$\lambda_m = \frac{k \times 1000}{C} = \frac{1.65 \times 10^{-4} \times 1000}{0.02} = 8.25 \ S \ cm^2 \ mol^{-1}$.
Step $2$: Calculate the molar conductivity at infinite dilution $(\lambda_m^{\infty})$ using Kohlrausch's law.
$\lambda_m^{\infty}(CH_3COOH) = \lambda_{H^{+}}^{\infty} + \lambda_{CH_3COO^{-}}^{\infty} = 349.1 + 40.9 = 390.0 \ S \ cm^2 \ mol^{-1}$.
Step $3$: Calculate the degree of dissociation $(\alpha)$.
$\alpha = \frac{\lambda_m}{\lambda_m^{\infty}} = \frac{8.25}{390.0} = 0.02115 \approx 0.021$.

(B) According to Kohlrausch's law of independent migration of ions, the equivalent conductance at infinite dilution for acetic acid is given by:
$\wedge^{0}_{CH_{3}COOH} = \wedge^{0}_{CH_{3}COONa} + \wedge^{0}_{HCl} - \wedge^{0}_{NaCl}$
Substituting the given values:
$\wedge^{0}_{CH_{3}COOH} = 91 + 426.16 - 126.45$
$\wedge^{0}_{CH_{3}COOH} = 390.71 \ \Omega^{-1} \ cm^{2} \ eq^{-1}$

(B) At infinite dilution, the equivalent conductance depends on the ionic mobility of the ions in the solution.
In aqueous solution, the $Li^+$ ion has the smallest size and the highest charge density, causing it to be highly hydrated.
Due to this high degree of hydration, the effective size of the hydrated $Li^+$ ion is the largest, which results in the lowest ionic mobility.
Conversely, the $K^+$ ion has the largest size and the lowest charge density, resulting in the least hydration and the smallest effective size.
Therefore, the ionic mobility follows the order $K^+ > Na^+ > Li^+$.
Since equivalent conductance at infinite dilution is directly proportional to ionic mobility, the order is $KCl > NaCl > LiCl$.

(A) The relationship between equivalent conductance $(\lambda_{eq})$, specific conductance $(K)$, and concentration $(C)$ is given by:
$\lambda_{eq} = \frac{K \times 1000}{C}$
Rearranging the formula to find the ratio of equivalent conductance to specific conductance:
$\frac{\lambda_{eq}}{K} = \frac{1000}{C}$
Given that the concentration $C = 0.01 \ N$:
$\frac{\lambda_{eq}}{K} = \frac{1000}{0.01} = 10^5 \ cm^3 \ eq^{-1}$
Therefore, the correct option is $A$.

(B) The equivalent conductance $(\Lambda_{eq})$ at infinite dilution depends on the ionic mobility of the ions in the solvent.
In water, $H^{+}$ and $HO^{-}$ ions exhibit exceptionally high ionic mobilities due to the Grotthuss mechanism (proton hopping).
Among the given ions, $H^{+}$ has the highest mobility, followed by $HO^{-}$.
$K^{+}$ is a simple hydrated cation with moderate mobility, while $CH_{3}COO^{-}$ is a large polyatomic anion with relatively lower mobility.
Therefore, the correct order of equivalent conductances at infinite dilution is $H^{+} > HO^{-} > K^{+} > CH_{3}COO^{-}$.

(C) Given, specific conductance $k = 1.25 \times 10^{-3} \ S \ cm^{-1}$.
Resistance $R = 800 \ \Omega$.
The relationship between specific conductance $(k)$, resistance $(R)$, and cell constant $(G^*)$ is given by the formula: $k = (1/R) \times G^*$.
Rearranging the formula to solve for the cell constant: $G^* = k \times R$.
Substituting the given values: $G^* = (1.25 \times 10^{-3} \ S \ cm^{-1}) \times (800 \ \Omega)$.
$G^* = 1.00 \ cm^{-1}$.

(A) The molar conductivity at infinite dilution for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is the sum of the molar conductivities of its constituent ions: $\lambda_{m}^{\infty} = \lambda_{m}^{\infty} (oxalate^{2-}) + \lambda_{m}^{\infty} (Na^{+}) + \lambda_{m}^{\infty} (K^{+})$.
Substituting the given values: $\lambda_{m}^{\infty} = 148.2 + 73.5 + 50.1 = 271.8 \ S \ cm^{2} \ mol^{-1}$.
The $n$-factor for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is $2$ because the oxalate ion has a charge of $-2$.
Therefore, the equivalent conductivity at infinite dilution is $\lambda_{eq}^{\infty} = \frac{\lambda_{m}^{\infty}}{n-factor} = \frac{271.8}{2} = 135.9 \ S \ cm^{2} \ eq^{-1}$.

(D) Conductivity depends on the number of ions and their ionic mobility.
$CH_{3}COOH$ is a weak electrolyte and dissociates poorly.
$NaCl$, $KNO_{3}$, and $HCl$ are strong electrolytes.
Among these, $HCl$ provides $H^{+}$ ions, which possess the highest ionic mobility in aqueous solution compared to $Na^{+}$, $K^{+}$, $Cl^{-}$, and $NO_{3}^{-}$ ions.
Therefore, $0.1 \ M \ HCl$ will have the highest conductivity.

(D) For metallic conductors, resistance $R$ increases with temperature $T$ $(R \propto T)$. Since conductivity is the reciprocal of resistivity, conductivity decreases as temperature increases.
For semiconductors, the number of charge carriers (electron-hole pairs) increases significantly with temperature, which leads to an increase in conductivity.

(A) For a weak acid, $K_a = C\alpha^2$ and $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.
$K_a = C \left(\frac{\Lambda_m}{\Lambda_m^0}\right)^2$
$pK_a = -\log K_a = -\log C - 2\log \Lambda_m + 2\log \Lambda_m^0$
$pK_a(HQ) - pK_a(HZ) = \log \left(\frac{C_{HZ}}{C_{HQ}}\right) + 2\log \left(\frac{\Lambda_{m(HZ)}}{\Lambda_{m(HQ)}}\right)$
Given $\frac{\Lambda_{m(HQ)}}{\Lambda_{m(HZ)}} = \frac{1}{30}$, so $\frac{\Lambda_{m(HZ)}}{\Lambda_{m(HQ)}} = 30$.
Since $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$ and $\lambda_{H^+}^0$ is the same for both, $\Lambda_{m(HQ)}^0 = \Lambda_{m(HZ)}^0$.
$\Delta pK_a = \log \left(\frac{0.02}{0.18}\right) + 2\log(30)$
$\Delta pK_a = \log \left(\frac{1}{9}\right) + 2\log(30) = -2\log 3 + 2(\log 3 + \log 10) = -2\log 3 + 2\log 3 + 2 = 2$.

(B) The given equation is $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$.
Using the data for $c = 0.04 \ mol \ L^{-1}$ and $\Lambda_{m} = 96.1 \ S \ cm^2 \ mol^{-1}$:
$96.1 = \Lambda_{m}^{\circ} - A \times \sqrt{0.04} = \Lambda_{m}^{\circ} - 0.2A \quad \dots(I)$
Using the data for $c = 0.09 \ mol \ L^{-1}$ and $\Lambda_{m} = 95.7 \ S \ cm^2 \ mol^{-1}$:
$95.7 = \Lambda_{m}^{\circ} - A \times \sqrt{0.09} = \Lambda_{m}^{\circ} - 0.3A \quad \dots(II)$
Subtracting equation $(II)$ from equation $(I)$:
$(96.1 - 95.7) = (-0.2A) - (-0.3A)$
$0.4 = 0.1A$
$A = \frac{0.4}{0.1} = 4$
Thus, the value of constant $A$ is $4$.

(B) Step $1$: Calculate the conductivity $(\kappa)$ of the solution.
$\kappa = G \times (l/A) = 1.9 \times 10^{-3} \text{ S} \times 1.3 \text{ cm}^{-1} = 2.47 \times 10^{-3} \text{ S cm}^{-1}$.
Step $2$: Calculate the molarity $(M)$ of the solution using the molar conductivity formula.
$\Lambda_m = (\kappa \times 1000) / M$
$123.5 = (2.47 \times 10^{-3} \times 1000) / M$
$M = 2.47 / 123.5 = 0.02 \text{ mol L}^{-1}$.
Step $3$: Calculate the mass of the solute in $1 \text{ L}$ of solution.
Since density is $1.0 \text{ g mL}^{-1}$, $1 \text{ L}$ of solution weighs $1000 \text{ g}$.
Mass of $MX = \text{moles} \times \text{molar mass} = 0.02 \text{ mol} \times 75 \text{ g mol}^{-1} = 1.5 \text{ g}$.
Step $4$: Calculate the percentage by weight $(w/w)$.
$\% (w/w) = (\text{mass of solute} / \text{mass of solution}) \times 100 = (1.5 / 1000) \times 100 = 0.15$.
$0.15 = 15 \times 10^{-2}$.
Therefore, the value of $x$ is $15$.

(A) According to the Debye-$H$ückel-Onsager equation for a strong electrolyte: $\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$.
Given, slope $A = 90.0 \ S \ cm^2 \ mol^{3/2} \ L^{1/2}$, concentration $C = 0.01 \ M$, and $\Lambda_m = 145.5 \ S \ cm^2 \ mol^{-1}$.
Substituting the values: $145.5 = \Lambda_m^\circ - 90.0 \times \sqrt{0.01}$.
$145.5 = \Lambda_m^\circ - 90.0 \times 0.1$.
$145.5 = \Lambda_m^\circ - 9.0$.
$\Lambda_m^\circ = 145.5 + 9.0 = 154.5 \ S \ cm^2 \ mol^{-1}$.
By Kohlrausch's law of independent migration of ions: $\Lambda_m^\circ(XY) = \lambda^\circ_{X^+} + \lambda^\circ_{Y^-}$.
$154.5 = 74.0 + \lambda^\circ_{Y^-}$.
$\lambda^\circ_{Y^-} = 154.5 - 74.0 = 80.5 \ S \ cm^2 \ mol^{-1}$.
Rounding to the nearest provided option, the value is $80.0 \ S \ cm^2 \ mol^{-1}$.

(A) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$
Given: $\kappa = 2.50 \times 10^{-4} \text{ S cm}^{-1}$ and $M = 0.0020 \text{ mol L}^{-1}$.
Substituting the values: $\Lambda_m = \frac{2.50 \times 10^{-4} \times 1000}{0.0020}$
$\Lambda_m = \frac{0.25}{0.0020} = \frac{2500}{20} = 125 \text{ S cm}^2 \text{mol}^{-1}$.

(C) $1$. Molar conductivity $(\Lambda_m)$ is defined as $\Lambda_m = \frac{\kappa}{C}$. Upon dilution, the concentration $(C)$ decreases, which leads to an increase in the volume containing $1 \text{ mole}$ of electrolyte, thereby increasing $\Lambda_m$.
$2$. Conductivity $(\kappa)$ is defined as the conductance of $1 \text{ cm}^3$ of solution. Upon dilution, the number of ions per unit volume decreases, which results in a decrease in $\kappa$.
$3$. Therefore, $\Lambda_m$ increases and $\kappa$ decreases.

(D) Step $1$: The statement describes the behavior of ions in an electrolytic solution at infinite dilution.
Step $2$: According to Kohlrausch's law of independent migration of ions, at infinite dilution, each ion makes a definite contribution to the molar conductivity of an electrolyte, which is independent of the presence of the other ion.
Step $3$: Therefore, the correct law is Kohlrausch's law of independent migration of ions.

(C) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given: Conductivity $\kappa = 0.0248 \text{ }\Omega^{-1} \text{cm}^{-1}$ and Concentration $C = 0.2 \text{ M}$.
Substituting the values: $\Lambda_m = \frac{0.0248 \times 1000}{0.2}$.
$\Lambda_m = \frac{24.8}{0.2} = 124 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$, where $M$ is the molarity.
Given: $\Lambda_m = 200 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and $M = 0.04 \text{ M}$.
Rearranging the formula to solve for $\kappa$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Substituting the values: $\kappa = \frac{200 \times 0.04}{1000}$.
$\kappa = \frac{8}{1000} = 0.008 \text{ }\Omega^{-1} \text{cm}^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(C)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given: $\Lambda_m = 120 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1}$, $\kappa = 0.0024 \text{ ohm}^{-1} \text{cm}^{-1}$.
Rearranging the formula for concentration $(C)$: $C = \frac{\kappa \times 1000}{\Lambda_m}$.
Substituting the values: $C = \frac{0.0024 \times 1000}{120}$.
$C = \frac{2.4}{120} = 0.02 \text{ M}$.

(A) The relationship between conductivity $(\kappa)$, resistance $(R)$, and cell constant $(G^*)$ is given by the formula: $\kappa = \frac{G^*}{R}$.
Rearranging for the cell constant: $G^* = \kappa \times R$.
Given: $\kappa = 0.0123 \text{ ohm}^{-1} \text{ cm}^{-1}$ and $R = 120 \text{ ohm}$.
Substituting the values: $G^* = 0.0123 \text{ ohm}^{-1} \text{ cm}^{-1} \times 120 \text{ ohm} = 1.476 \text{ cm}^{-1}$.

(D) $1$. According to the Kohlrausch law, for strong electrolytes, the variation of molar conductivity $(\Lambda_m)$ with concentration $(c)$ is given by the Debye-$H$ückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$.
$2$. This equation represents a straight line with a negative slope and an intercept equal to $\Lambda_m^0$ on the y-axis.
$3$. Weak electrolytes like $CH_3COOH$, $NH_4OH$, and $HCOOH$ show a non-linear increase in $\Lambda_m$ with dilution, especially at low concentrations.
$4$. $CH_3COONa$ is a strong electrolyte, which follows the linear relationship.
$5$. Therefore, '$X$' is $CH_3COONa$.

(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{\text{Cell constant}}{\text{Resistance}}$
Given: Cell constant $(G^*)$ = $1.2 \text{ cm}^{-1}$, Resistance $(R)$ = $95.0 \text{ }\Omega$.
Substituting the values: $\kappa = \frac{1.2 \text{ cm}^{-1}}{95.0 \text{ }\Omega} = 0.01263 \text{ }\Omega^{-1} \text{ cm}^{-1}$.
Expressing in scientific notation: $\kappa = 1.26 \times 10^{-2} \text{ }\Omega^{-1} \text{ cm}^{-1}$.

(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Given: $\Lambda_m = 240 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$, $\kappa = 0.012 \text{ }\Omega^{-1}\text{cm}^{-1}$.
Rearranging the formula for $M$: $M = \frac{\kappa \times 1000}{\Lambda_m}$.
Substituting the values: $M = \frac{0.012 \times 1000}{240}$.
$M = \frac{12}{240} = \frac{1}{20} = 0.05 \text{ M}$.

(A) According to Kohlrausch's law, the molar conductivity at infinite dilution $(\Lambda^0_m)$ can be expressed as the sum of the ionic conductivities.
Given:
$\Lambda^0_m(H_2SO_4) = 2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x$
$\Lambda^0_m(K_2SO_4) = 2\lambda^0(K^+) + \lambda^0(SO_4^{2-}) = y$
$\Lambda^0_m(CH_3COOK) = \lambda^0(CH_3COO^-) + \lambda^0(K^+) = z$
We need to find $\Lambda^0_m(CH_3COOH) = \lambda^0(H^+) + \lambda^0(CH_3COO^-)$.
From the given equations:
$\frac{\Lambda^0_m(H_2SO_4) - \Lambda^0_m(K_2SO_4)}{2} = \frac{(2\lambda^0(H^+) + \lambda^0(SO_4^{2-})) - (2\lambda^0(K^+) + \lambda^0(SO_4^{2-}))}{2} = \lambda^0(H^+) - \lambda^0(K^+) = \frac{x - y}{2}$.
Adding this to $\Lambda^0_m(CH_3COOK)$:
$\frac{x - y}{2} + z = (\lambda^0(H^+) - \lambda^0(K^+)) + (\lambda^0(CH_3COO^-) + \lambda^0(K^+)) = \lambda^0(H^+) + \lambda^0(CH_3COO^-) = \Lambda^0_m(CH_3COOH)$.
Thus, the correct option is $A$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given:
Conductivity $(\kappa)$ = $4.37 \times 10^{-4} \text{ }\Omega^{-1} \text{cm}^{-1}$
Concentration $(C)$ = $0.02 \text{ M}$
Substituting the values:
$\Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02}$
$\Lambda_m = \frac{0.437}{0.02}$
$\Lambda_m = 21.85 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
Thus, the correct option is $B$.

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(c)$ for strong electrolytes is given by the Kohlrausch equation:
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
Where:
$1$. $\Lambda_m$ is the molar conductivity at a given concentration.
$2$. $\Lambda_m^0$ is the molar conductivity at infinite dilution.
$3$. $A$ is a constant that depends on the nature of the solvent and temperature.
$4$. $c$ is the concentration of the electrolyte.
Thus, option $B$ is correct.

(A) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$
Given:
$\kappa = 2.50 \times 10^{-4} \text{ S cm}^{-1}$
$M = 0.0020 \text{ mol L}^{-1}$
Substituting the values:
$\Lambda_m = \frac{2.50 \times 10^{-4} \times 1000}{0.0020}$
$\Lambda_m = \frac{0.25}{0.0020}$
$\Lambda_m = 125 \text{ S cm}^2 \text{mol}^{-1}$

(C) $1$. Molar conductivity $(\Lambda_m)$ is defined as $\Lambda_m = \frac{\kappa}{C}$. As the solution is diluted, the concentration $(C)$ decreases, which leads to an increase in the number of ions per unit volume available for conduction, causing $\Lambda_m$ to increase.
$2$. Conductivity $(\kappa)$ is defined as the conductance of a solution contained between two electrodes of unit area and unit distance apart. Upon dilution, the number of ions per unit volume decreases, which leads to a decrease in conductivity $(\kappa)$.
$3$. Therefore, upon dilution, $\Lambda_m$ increases and $\kappa$ decreases.

(D) Step $1$: The statement describes the behavior of ions in an electrolytic solution at infinite dilution.
Step $2$: According to Kohlrausch's law of independent migration of ions, at infinite dilution, where dissociation is complete, each ion makes a definite contribution to the molar conductivity of the electrolyte, which is independent of the presence of other ions.
Step $3$: Therefore, the correct law is Kohlrausch's law.

(C) The formula for molar conductivity $(\Lambda_m)$ is: $\Lambda_m = \frac{1000 \times k}{M}$
Given: Conductivity $(k)$ = $0.0248 \text{ } \Omega^{-1} \text{cm}^{-1}$, Molarity $(M)$ = $0.2 \text{ M}$.
Substituting the values: $\Lambda_m = \frac{1000 \times 0.0248}{0.2}$
$\Lambda_m = \frac{24.8}{0.2} = 124 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Rearranging for conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Given: $\Lambda_m = 200 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$, $M = 0.04 \text{ M}$.
Substituting the values: $\kappa = \frac{200 \times 0.04}{1000}$.
$\kappa = \frac{8}{1000} = 0.008 \text{ } \Omega^{-1} \text{ cm}^{-1}$.

(C) Step $1$: Calculate the degree of dissociation $(\alpha)$ using the formula $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{15}{375} = 0.04$.
Step $2$: Calculate the concentration of $[H^+]$ using the relation $[H^+] = C \times \alpha$.
Given $C = 0.01 \text{ M}$.
$[H^+] = 0.01 \times 0.04 = 4 \times 10^{-4} \text{ M}$.

(B) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(k)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{1000 \times k}{M}$.
Rearranging for molarity: $M = \frac{1000 \times k}{\Lambda_m}$.
Given: $k = 0.0024 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 120 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
Substituting the values: $M = \frac{1000 \times 0.0024}{120}$.
$M = \frac{2.4}{120} = 0.02 \text{ M}$.

(D) $1$. According to the Kohlrausch law and the Debye-$H$ückel-Onsager equation, strong electrolytes show a linear relationship between molar conductivity $(\Lambda_m)$ and $\sqrt{c}$ given by $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$, where $\Lambda_m^\circ$ is the intercept on the y-axis.
$2$. Weak electrolytes like $CH_3COOH$, $NH_4OH$, and $HCOOH$ show a non-linear curve that increases sharply as concentration approaches zero.
$3$. $CH_3COONa$ is a strong electrolyte, which dissociates completely in solution, thus following the linear relationship.
$4$. Therefore, '$X$' is $CH_3COONa$.

(D) The cell constant $(G^*)$ is given by the formula: $G^* = \kappa \times R$
Given conductivity $(\kappa)$ = $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$
Given resistance $(R)$ = $120 \text{ } \Omega$
$G^* = 0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1} \times 120 \text{ } \Omega$
$G^* = 1.476 \text{ cm}^{-1}$

(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{G^*}{R}$
Given: Cell constant $(G^*)$ = $1.2 \text{ cm}^{-1}$, Resistance $(R)$ = $95.0 \text{ } \Omega$.
Substituting the values: $\kappa = \frac{1.2 \text{ cm}^{-1}}{95.0 \text{ } \Omega} = 0.01263 \text{ } \Omega^{-1} \text{cm}^{-1}$.
Therefore, $\kappa = 1.263 \times 10^{-2} \text{ } \Omega^{-1} \text{cm}^{-1}$.

(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by the formula:
$\Lambda_m = \frac{1000 \times \kappa}{M}$
Rearranging the formula to solve for molarity $(M)$:
$M = \frac{1000 \times \kappa}{\Lambda_m}$
Substitute the given values: $\kappa = 0.012 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 240 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
$M = \frac{1000 \times 0.012}{240}$
$M = \frac{12}{240} = 0.05 \text{ M}$

(A) According to Kohlrausch's law:
$\Lambda^0_m(CH_3COOH) = \Lambda^0_m(CH_3COOK) + \Lambda^0_m(H^+) - \Lambda^0_m(K^+)$
We know that $\Lambda^0_m(H^+) = \frac{1}{2} \Lambda^0_m(H_2SO_4) = \frac{x}{2}$ and $\Lambda^0_m(K^+) = \frac{1}{2} \Lambda^0_m(K_2SO_4) = \frac{y}{2}$.
Substituting these values:
$\Lambda^0_m(CH_3COOH) = z + \frac{x}{2} - \frac{y}{2} = \frac{(x - y)}{2} + z$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given conductivity $(\kappa)$ = $4.37 \times 10^{-4} \text{ } \Omega^{-1} \text{cm}^{-1}$
Given concentration $(C)$ = $0.02 \text{ M}$
Substituting the values: $\Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02}$
$\Lambda_m = \frac{0.437}{0.02} = 21.85 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^0(NH_4OH) = \lambda^0(NH_4^+) + \lambda^0(OH^-)$
To obtain this, we use strong electrolytes:
$\Lambda_m^0(NH_4Cl) = \lambda^0(NH_4^+) + \lambda^0(Cl^-)$
$\Lambda_m^0(NaOH) = \lambda^0(Na^+) + \lambda^0(OH^-)$
$\Lambda_m^0(NaCl) = \lambda^0(Na^+) + \lambda^0(Cl^-)$
Adding the first two and subtracting the third:
$\Lambda_m^0(NH_4Cl) + \Lambda_m^0(NaOH) - \Lambda_m^0(NaCl) = (\lambda^0(NH_4^+) + \lambda^0(Cl^-)) + (\lambda^0(Na^+) + \lambda^0(OH^-)) - (\lambda^0(Na^+) + \lambda^0(Cl^-))$
$= \lambda^0(NH_4^+) + \lambda^0(OH^-) = \Lambda_m^0(NH_4OH)$.

(B) Given:
Conductivity $(\kappa) = 0.021\text{ }\Omega^{-1}\text{ cm}^{-1}$
Resistance $(R) = 60\text{ }\Omega$
Cell constant $(G^*) = \kappa \times R$
$G^* = 0.021\text{ }\Omega^{-1}\text{ cm}^{-1} \times 60\text{ }\Omega = 1.26\text{ cm}^{-1}$.