Step-by-Step Solution

The problem asks for the value of Z, where the limiting molar conductivity ($\Lambda_m^o$) of a weak monobasic acid is expressed as $Z \times 10^2 Sc{m^2}mo{l^{ - 1}}$. We are given the concentration, conductance, cell dimensions, and pH of the solution.

Step 1: Calculate the cell constant ($G^*$) and conductivity ($\kappa$)

First, we calculate the cell constant of the conductivity cell. The cell constant is defined as the ratio of the distance between the electrodes ($l$) to their area of cross-section ($A$).

Given:

• Distance, $l = 120 cm$
• Area, $A = 1 cm^2$

Cell constant, $G^* = \frac{l}{A} = \frac{120 \, cm}{1 \, cm^2} = 120 \, cm^{-1}$

Next, we calculate the conductivity (specific conductance, $\kappa$) of the solution using its relationship with conductance ($G$) and the cell constant ($G^*$).

Given:

• Conductance, $G = 5 \times 10^{-7} S$

Conductivity, $\kappa = G \times G^*$ $\kappa = (5 \times 10^{-7} \, S) \times (120 \, cm^{-1})$ $\kappa = 600 \times 10^{-7} \, S \, cm^{-1} = 6 \times 10^{-5} \, S \, cm^{-1}$

Step 2: Calculate the molar conductivity ($\Lambda_m$)

Molar conductivity at a given concentration is calculated using the formula: ${\Lambda _m} = \frac{{\kappa \times 1000}}{C}$ where $\kappa$ is in $S \, cm^{-1}$ and concentration $C$ is in $mol \, L^{-1}$ (M).

Given:

• Concentration, $C = 0.0015 M = 1.5 \times 10^{-3} M$

${\Lambda _m} = \frac{{(6 \times 10^{-5} \, S \, cm^{-1}) \times 1000 \, cm^3 \, L^{-1}}}{{1.5 \times 10^{-3} \, mol \, L^{-1}}}$ ${\Lambda _m} = \frac{{6 \times 10^{-2}}}{{1.5 \times 10^{-3}}} \, S \, cm^2 \, mol^{-1}$ ${\Lambda _m} = \frac{6}{1.5} \times 10^{(-2 - (-3))} = 4 \times 10^1 = 40 \, S \, cm^2 \, mol^{-1}$

Step 3: Calculate the degree of dissociation ($\alpha$)

The acid is a weak monobasic acid (HA). Its dissociation in water is: $HA \rightleftharpoons H^+ + A^-$

The pH of the solution is given as 4. We can find the hydrogen ion concentration, $[H^+]$, from the pH. $pH = - \log [H^+]$ $4 = - \log [H^+]$ $[H^+] = 10^{-4} M$

For a weak electrolyte, the concentration of hydrogen ions is related to the initial concentration ($C$) and the degree of dissociation ($\alpha$) by the equation: $[H^+] = C\alpha$

Now we can calculate $\alpha$: $\alpha = \frac{{[H^+]}}{C} = \frac{{10^{-4} \, M}}{{1.5 \times 10^{-3} \, M}}$ $\alpha = \frac{1}{1.5} \times 10^{-1} = \frac{1}{15}$

Step 4: Calculate the limiting molar conductivity ($\Lambda_m^o$)

The degree of dissociation ($\alpha$) is also related to the molar conductivity ($\Lambda_m$) and the limiting molar conductivity ($\Lambda_m^o$) by the Arrhenius equation: $\alpha = \frac{{{\Lambda _m}}}{{{\Lambda _m^o}}}$

We can rearrange this to find ${\Lambda _m^o}$: ${\Lambda _m^o} = \frac{{{\Lambda _m}}}{\alpha }$

Substituting the values we calculated: ${\Lambda _m^o} = \frac{{40 \, S \, cm^2 \, mol^{-1}}}{{1/15}}$ ${\Lambda _m^o} = 40 \times 15 = 600 \, S \, cm^2 \, mol^{-1}$

Step 5: Determine the value of Z

The problem states that the limiting molar conductivity is $Z \times 10^2 \, S \, cm^2 \, mol^{-1}$. We equate this to our calculated value: $Z \times 10^2 = 600$ $Z = \frac{600}{100}$ $Z = 6$

The value of Z is 6.