Emptying tank through diffusion

Water ($$40 ^{\circ}$$, $$1.0\, {\rm atm}$$) slowly and steadily evaporates into nitrogen at the same temperature and pressure. from the bottom of a cylindrical tank as shown in the figure below. A stream of dry nitrogen flows slowly past the open tank. The mole fraction of water in the gas at the top opening of the tank is 0.02. How long will it take for the tank to be emptied just via diffusion, using quasi-steady state approximation?

My attempt:

I start off with a mole balance which gives me: $$N_{A,z} = {\rm const}$$, $$\forall z$$ at a given time $$t$$ ... (1)

Next, I applied Fick's law, and cancelled out the $$N_B$$ term: $$N_{A,z} = x_AN_{A,z} - cD_{AB}\frac{dx}{dz}$$

I imposed the following BCs on the differential equation: $$z=z_2, x_A = x_{A\infty}$$ and $$z = z_1(t), x_A = x_{A*}$$

I am throwing in the time dependence because height of liquid is dropping with time. I said that it was valid integrating $$N_{A,z}$$ wrt $$z$$ as a constant because we are integrating over $$z>z_1(t)$$.

So we have $$N_{A,z} = cD_{AB} \frac{\ln (x_{b\infty}/x_b*))}{(z_2-z_1(t))}$$ ... agrees with the statement (1) Then I impose the mass quasi-steady state mass balance (no flow, no reaction, only diffusion), with my system as the liquid:

$$\frac{dN_A}{dt} = N_{A,z}S$$ $$\Rightarrow \frac{dN_A}{dt} = cD_{AB} \frac{\ln(x_{b\infty}/x_b*)}{z_2-z_1(t)}$$... (2)

Now I impose $$z_1(t) = M_A/(\rho S) = N_A\cdot MW_A /(\rho S)$$

Plugging the above relation in (2) and solving the diff eq in terms of $$N_A$$ and $$t$$, and imposing the BC $$N_A = N_{A0} = \rho S h_0/MW_A$$ at $$t = 0$$, we get $$z_2N_A - \frac{N_A ^2 MW_A}{2\rho S} = cD_{AB} \ln(x_{b\infty}/x_b *)St + z_2N_{A0} - \frac{N_{A0} ^2 MW_A}{2\rho S}$$

at $$t_{empty}, N_A = 0$$

$$\Rightarrow t_{empty} = z_2N_{A_0} - \frac{N_{A0}^2 MW_A}{2\rho S}$$

Then I reimpose $$N_{A0} = h_0\rho S/MW$$.

However, it seems like I am using quasi-steady state approximation incorrectly. Where am I going wrong?