It has been awhile since electronics class so I had to look through my old book. I am not sure if this is right, I was never that good at electronics anyway, but maybe I'll learn something.

The sum of the dc voltages across the capacitor and the resistor equals the source voltage (Vs) at all times.

As the capacitor is charged, the voltage across the capacitor is series- opposing the source voltage. Therefore, the voltage across the resistor at any given moment equals the source voltage minus the capacitor voltage. That is, Vr = Vs - Vc. Thus, as the capacitor charges to the Vs value, the resistor voltage decreases toward zero volts. When the capacitor is fully charged, Vc equals Vs and zero current flows, and no IR drop occurs across the resistor. When the switch is initially closed, the capacitor has zero charge and zero voltage. Therefore, the source voltage is dropped by the resistor (R). Since this is a series circuit, and the charging current is the same throughout a series circuit, Ohm's law reveals what the charging current is at the moment.

I = Vr/R

So I am guessing that the voltage across the capacitor will be eqaul to the source, 9v.