You are right, actually $\frac{\langle v^2 \rangle}{\langle v \rangle^2}=8\pi/3$.
For what the professor wrote you do not need to assume that $\langle v^2 \rangle=\langle v \rangle^2 $, but only that $\frac{\langle v^2 \rangle}{\langle v \rangle^2}=\frac{\langle v_r^2 \rangle}{\langle v_r \rangle^2}=C $, with $C$ arbitrary.
Even if this assumption might look more plausible, it is still unjustified, so this is still not a rigorous demonstration. The right way to do it is to use the speed distribution and calculate it by brute force. I tried a couple of times but ended up with terrible integrals. I am not sure now how straightforward it is to do it. So do not be discouraged if you try and fail.
This is all about diffusion. The speed of individual molecules is not relevant, because they collide with one another and change direction so frequently (at least at standard temperature and pressure) that this speed does not at all characterize the diffusion of one species into another.
Engineers have tabulated rate coefficients that describe the rate of diffusion of various gases through air, for example: https://www.engineeringtoolbox.com/air-diffusion-coefficient-gas-mixture-temperature-d_2010.html
This doesn't give the rate you'd want, but we can get the ballpark studying a similar rate, Argon diffusing through air.
Say you've got a can with Argon on the bottom, air on top, and a 1 cm mixed layer between otherwise pure gases.
J = (D = 0.189 cm^2/s) * (1.7 kg/m^3 Argon at STP)/(1cm) = 31.8 kg/cm^2/s
This is the mass flux of argon through the barrier. Multiply by some area A, Divide by argon density at STP (1.7 kg/m^3), and divide by A again to get argon flow per unit area, areas cancel and we have: = 1.89 cm/s. Note that the Argon mass actually canceled out here too, basically the mix rate just relates to how thick the boundary layer starts out. Initially, when it is infinitesimal, the rate is infinite, since the rate is just D=0.189 cm^2/s divided by the boundary layer thickness L, which I assumed to start at 1 cm.
This means that the pure argon below diffuses up into the pure air through the boundary at like 2 cm/s. Of course one second later the boundary layer is 3 cm thick instead of 1, so the rate slows 3x. Three seconds later it is five centimeters thick. You have to solve a differential equation to really get your answer of how long, and the notion of a firm boundary between pure and boundary layer is just an approximation. But roughly... continuing this pattern you hit 21 cm thick "boundary layer" after 100 seconds, which I'm guessing is close to your tank size. Double or triple that for the boundary layer to further mix up to your .1% requirement, and we're at 5 minutes.
Notably, given this surprisingly slow timescale, it probably does help to shake up the tank. I suspect that Argon is a slightly worse case than N2 and O2, but I don't really know. Comparing other gases on the engineering toolbox link, seems like D roughly goes with mass, lighter mass higher D, but Argon isn't very different from air anyway (40 v 29).
Best Answer
For an ideal gas the sound speed can be written as
$$c = \sqrt{\frac{k P}{\rho}}$$
where $\rho$ is the mass density. The constant $k$ is present because the expansion and compression of the gas is supposed to be adiabatic, and $k$ is the adiabatic index. For a gas mixture with $n_i$ moles of the $i$-th gas we have
$$k=\frac{\sum_i n_i c_{p,i}}{\sum_i n_i c_{v,i}} = \frac{\sum_i n_i \frac{k_i}{k_i-1}}{\sum_i n_i \frac{1}{k_i-1}} $$
where $k_i$ is the adiabatic index for the $i$-th gas. The mass density is
$$\rho=V^{-1} \sum_i M_i n_i$$
where $M_i$ is the $i$-th molar mass. Inserting in the expression for the sound speed we get
$$c = \sqrt{\left(\frac{1}{\sum_i x_i \frac{1}{k_i-1}}+1\right)\frac{ RT}{\sum_i M_i x_i}}$$
while
$$c_i = \sqrt{\frac{k_i RT}{M_i}}$$
In the general case it is not possible to express $c$ as a function of the ratios $k_i/M_i$ only, so it is not possible to express $c$ as a function of the $x_i$ and $c_i$ only.
A particular case is when all the gases in the mixture have the same adiabatic index (for example, they can be all biatomic). Then
$$c = \sqrt{\frac{k RT}{\sum_i M_i x_i}}$$
and
$$c_i = \sqrt{\frac{k RT}{M_i}}$$
then
$$c = \left(\sum_i \frac{x_i}{c_i^2} \right)^{-1/2}$$