Yes, it is hard to deliver that kind of power with a charge pump. If you want to boost the voltage from 9 V (RMS) to 170 V (DC) with a CW charge pump you would need approximately 14 stages using this equation (if we forget about the voltage drop due to the load current and forward voltage of the diodes).
Eout = Ein * sqrt(2) * n
Each stage consists of two diodes and two capacitors. In your case you would also have to take into consideration the total voltage drop of all diodes, that would reduce the output voltage. You also have to know that the output voltage drops with a higher load current.
Vdiode_drop = 2 * n * Vf
Vdiode_drop is the voltage drop of the diodes
Vf is the foward voltage of the diodes
n is the number of stages
Vload_drop = I / ( f * C ) * (2/3 * n³ + n² / 2 - n / 6
Vload_drop is the voltage drop
I is the current drawn in amperes
f is the frequency in hertz
n is the number of stages
C is the size of the capacitors used in farads
In conclusion, CW multiplier is not practical for applications requiring high output currents and high number of multiplication stages. You would get better results using a standard boost converter or a transformer.