Well it's plausible but the numbers seem suspicious (I think). Your meter's manual refers to measuring capacitance/resistance in "parallel mode" by default, but that this can be changed to "series mode". This suggests that it is trying to compute the equivalent parallel resistance of the network. The real part of the impedance of a parallel RC network is
\$\frac{R}{1+\omega^2R^2C^2}\$
... and this is frequency dependant. The equivalent parallel resistance in this case is constant with frequency (by definition it is R) and this is what I would expect the meter to display. However in a real capacitor, the equivalent shunt resistance is not formed by a real resistor.
It would be interesting to switch to "series mode" if possible and see what the numbers are then.
Similarly, how to calculate the reactance for a capacitor when a
square wave is passing through it. What is the formula?
There isn't a capacitive reactance associated with a square wave. The very concept of reactance depends on the context of sinusoidal excitation.
When we solve AC circuits in the phasor domain, it is taken for granted that the circuit is in sinusoidal steady state, i.e., all sources are sinusoidal of the same frequency and all transients have decayed away.
That fact is this: one cannot meaningfully sum phasors or reactances for sinusoids of different frequencies.
Now, that doesn't mean that you can't apply the concept of reactance to find the capacitor voltage across for a square wave current through.
Since (ideal) capacitors are linear, we can decompose the square wave into sinusoidal components, find the associated sinusoidal voltage for each component, and then sum to voltage components to find the total voltage.
Recall the fundamental phasor domain relationship for capacitor voltage and current:
$$\vec V_c = \dfrac{1}{j \omega C}\vec I_c $$
where \$\omega\$ is the angular frequency of the associated sinusoid.
Now, let
$$i_C(t) = a_1 \cos(\omega t + \phi_1) + a_2 \cos(2\omega t + \phi_2) + a_3 \cos(3\omega t + \phi_3) + ...$$
For each sinusoidal component, there is an associated phasor. For example, for the first component, the associated phasor is
$$\vec I_{c_1} = a_1 e^{j\phi_1}$$
Thus
$$\vec V_{c_1} = \dfrac{a_1 e^{j\phi_1}}{j \omega C}$$
so that
$$v_{C_1}(t) = \dfrac{a_1}{\omega C}\cos(\omega t +\phi_1 - \frac{\pi}{2}) $$
Repeat for each term in the series and then sum to find the total capacitor voltage.
Note that we have not defined a reactance to the entire current waveform nor can we define such a thing. Instead, we
(1) found the reactance to each sinusoidal component
(2) converted each resulting phasor voltage back into the time domain
(3) summed the individual time domain voltage components
Best Answer
Some authors specify the reactance of basic circuit elements as an absolute value. Although this is confusing it is not so uncommon. The "trick" is to remember that if you define reactances as:
\[ X_L = \omega L \qquad X_C = \frac{1}{\omega C} \]
then the impedance for an inductor and a capacitor are:
\[ Z_L = j X_L = j \omega L
\qquad
Z_C = -j X_C = \frac{- j}{\omega C} = \frac {1}{j \omega C} \]
The problem with this approach is that you must always remember that the reactance as the imaginary part of a generic impedance (i.e. X = Im(z)) is not the same reactance you speak of when talking about "pure" capacitors (there the sign of the reactance is embedded in the value of X).