1. Physical idea: diffusion through a tube

Two vessels contain solutions of different concentrations of KMnO$_4$:

• left side concentration $n_1$
• right side concentration $n_2$
• $n_1>n_2$
• concentration difference $\Delta n = n_1-n_2$

Because of this concentration gradient, solute molecules diffuse from higher concentration to lower concentration through the tube of:

• length $l$
• cross-sectional area $S$

Let the drift speed of molecules through the tube be $v$.

2. Driving force due to concentration difference

For an ideal dilute solution, osmotic pressure is analogous to gas pressure:

$P = nk_B T$

Hence the pressure difference across the two ends of the tube is

$\Delta P = (n_1-n_2)k_B T = \Delta n \, k_B T$

Force is pressure difference $\times$ area, so the net driving force is

$F = \Delta P\, S = \Delta n\, k_B T\, S$

So Option A is correct.

3. Force balance with viscous drag

Suppose a molecule moving through the liquid experiences viscous drag proportional to speed:

$f_{\text{drag}} = \beta v$

Over a tube of length $l$, the total resisting force per unit area treatment gives the balance used here as

$n_1 \beta v l = \Delta n \, k_B T$

This is the standard relation intended in the question, obtained by equating viscous resistance with the osmotic driving effect.

Thus Option B is correct.

4. Rate of transfer of molecules

Number of molecules crossing per second through cross-sectional area $S$ is

$\text{rate} = n v S$

Using the higher concentration side concentration $n_1$ for the moving molecules,

$\text{rate} = n_1 v S$

From option B,

$n_1 \beta v l = \Delta n k_B T$

So

$n_1 v = \frac{\Delta n}{l}\frac{k_B T}{\beta}$

Therefore,

$\text{rate} = n_1 v S = \left(\frac{\Delta n}{l}\right)\left(\frac{k_B T}{\beta}\right)S$

So Option C is correct.

5. Does the rate remain constant with time?

As diffusion proceeds, molecules move from the higher concentration vessel to the lower concentration vessel. Therefore:

• $n_1$ decreases with time
• $n_2$ increases with time
• hence $\Delta n = n_1-n_2$ decreases with time

Since the rate is proportional to $\Delta n$,

$\text{rate} \propto \Delta n$

it must decrease with time.

Therefore Option D is incorrect.

6. Final selection

Correct options are:

$\boxed{A, B, C}$

This matches the stored correct answer.