Please help me with this chemistry problem!!!! exam tomorrow :(

i dont know how to do this question off my practice exam:
What is the molality of an aqueous solution of urea, CO(NH2)2, if the vapor pressure
above the solution is 22.83 mmHg at 25°C? Assume that urea is non-volatile. The
vapor pressure of pure water is 23.77 mmHg at 25°C

The correct answer is: 2.31 m

THANKS SOOOOO MUCHH! i really do appreciate it :)

Psoln = P° times (mole fraction of solvent)

22.83 = (23.77) (x)

x = 0.960 (mole fraction of water)

mole fraction of urea = 0.04

Let us change 0.960 mole to grams of water:

0.960 times 18.015 = 17.2944 g

calculate molality:

0.04 mol / 0.0172944 kg = 2.31 m