Solutions of hydrochloric acid and potassium chloride, each of 0.00100 mol/L, are so dilute that…

Step 1. The limiting values. $\Lambda_m^\circ(\text{HCl}) = 349.6 + 76.3 = 425.9\ \text{S cm}^2/\text{mol}$ and $\Lambda_m^\circ(\text{KCl}) = 73.5 + 76.3 = 149.8\ \text{S cm}^2/\text{mol}$.

Step 2. The conductivities. From $\Lambda_m = \dfrac{10\,\kappa}{c}$, we have $\kappa = \dfrac{\Lambda_m c}{10}$. For the acid, $\kappa = \dfrac{425.9 \times 0.00100}{10} = 0.0426\ \text{S/m}$. For the salt, $\kappa = \dfrac{149.8 \times 0.00100}{10} = 0.0150\ \text{S/m}$.

Step 3. The ratio. $\dfrac{0.0426}{0.0150} = 2.84$. Since the concentrations are equal, this is also $\dfrac{425.9}{149.8}$.