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 $15$.