1. For a strong electrolyte, the molar conductivity follows:

$\Lambda_m = \Lambda_m^0 - K\sqrt{c}$

So, in a plot of $\Lambda_m$ vs. $\sqrt{c}$, the slope is $-K$.

2. Therefore, the unit of slope is:

$\text{unit of slope} = \frac{\text{unit of }\Lambda_m}{\text{unit of }\sqrt{c}}$

3. Unit of molar conductivity $\Lambda_m$:

$\Lambda_m : \mathrm{S\,cm^2\,mol^{-1}}$

4. Unit of concentration $c$:

$c : \mathrm{mol\,L^{-1}}$

Hence,

$\sqrt{c} : (\mathrm{mol\,L^{-1}})^{1/2} = \mathrm{mol^{1/2}L^{-1/2}}$

5. Now divide:

$\text{unit of slope} = \frac{\mathrm{S\,cm^2\,mol^{-1}}}{\mathrm{mol^{1/2}L^{-1/2}}}$

$= \mathrm{S\,cm^2\,mol^{-1} \cdot mol^{-1/2} \cdot L^{1/2}}$

$= \mathrm{S\,cm^2\,mol^{-3/2}L^{1/2}}$

6. Comparing with the options:

• A: Incorrect
• B: Incorrect
• C: Incorrect
• D: Correct

Therefore, the correct answer is:

$\boxed{\mathrm{S\,cm^2\,mol^{-3/2}L^{1/2}}}$

So, option $\boxed{D}$ is correct.