Crystal Voltage Measurement
Yes, that's correct. You can't easily distinguish between the voltage drop created by the internal resistance and the drop across the (equivalent) inductor/capacitor, you only see the sum of the two drops.

You could of course calculate the voltage drop across the inductor using dI/dt or the phase of the current, but in that case you already have the current anyway so there's no point in measuring the voltage to get the power dissipated in the crystal.

Edit: As brhans noted, the voltage isn't meaningless in general however. For example, crystals have a dielectric breakdown voltage that must not be exceeded. It's just not useful for calculating power dissipation.