Plotting $$1/\Lambda_m$$ against $$c\Lambda_m$$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is

[$$\Lambda_m$$ = molar conductivity, $$\Lambda_m^\circ$$ = limiting molar conductivity, c = molar concentration, $$K_a$$ = dissociation constant of HX]

For a weak monobasic acid $$HX$$, the degree of dissociation $$\alpha$$ is related to the molar conductivity by Ostwald’s dilution law:

$$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$
$$K_a = \frac{c\,\alpha^{2}}{1-\alpha}$$

Substitute $$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ}$$ in the expression of $$K_a$$:

$$K_a = \frac{c\left(\dfrac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1-\dfrac{\Lambda_m}{\Lambda_m^\circ}} = \frac{c\,\Lambda_m^{2}}{\Lambda_m^{\circ 2}-\Lambda_m\,\Lambda_m^\circ}$$

Multiply numerator and denominator by $$\Lambda_m^\circ$$ to simplify:

$$K_a = \frac{c\,\Lambda_m^{2}}{\Lambda_m^\circ\left(\Lambda_m^\circ-\Lambda_m\right)}$$

Re-arrange to isolate the linear relationship we need:

$$K_a\Lambda_m^\circ\left(\Lambda_m^\circ-\Lambda_m\right)=c\,\Lambda_m^{2}$$

Divide both sides by $$\Lambda_m\,\Lambda_m^{\circ 2}$$:

$$K_a\Bigl(\frac{\Lambda_m^\circ}{\Lambda_m}-1\Bigr)=\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

The term in parentheses simplifies since $$\dfrac{\Lambda_m^\circ}{\Lambda_m}-1=\dfrac{\Lambda_m^\circ-\Lambda_m}{\Lambda_m}$$, giving

$$K_a\Bigl(\frac{1}{\Lambda_m}-\frac{1}{\Lambda_m^\circ}\Bigr)=\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

Finally, separate the two fractions on the left:

$$K_a\left(\frac{1}{\Lambda_m}\right)=K_a\left(\frac{1}{\Lambda_m^\circ}\right)+\frac{c\,\Lambda_m}{\Lambda_m^{\circ 2}}$$

Divide every term by $$K_a$$ to obtain the required linear form:

$$\frac{1}{\Lambda_m}= \frac{1}{\Lambda_m^\circ} + \frac{c\,\Lambda_m}{K_a\,\Lambda_m^{\circ 2}}$$

Comparing with the straight-line equation $$y = P + Sx$$, let

$$y = \frac{1}{\Lambda_m}, \qquad x = c\,\Lambda_m$$

Intercept: $$P = \frac{1}{\Lambda_m^\circ}$$
Slope: $$S = \frac{1}{K_a\,\Lambda_m^{\circ 2}}$$

The required ratio is

$$\frac{P}{S}= \frac{\dfrac{1}{\Lambda_m^\circ}}{\dfrac{1}{K_a\,\Lambda_m^{\circ 2}}}=K_a\,\Lambda_m^\circ$$

Hence, $$\dfrac{P}{S}=K_a\Lambda_m^\circ$$.

Option A which is: $$K_a \Lambda_m^\circ$$