What is the vapor pressure of an aqueous solution made up of 60.0g of urea (ch4n2o) in 180g of water? the vapor pressure of water at this temperature is 0.038atm.
answer truthfully and explain how you got this answer because i dont know and my head hurts.
also the choices are
0.020atm
0.0025atm
0.035atm
0.00055atm

To get the answer you use the Law of Raoult.

Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.

ΔP = Pa * Xa

Here Pa = 0.038 atm

And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b

Na = mass of urea / molar mass of urea =  60 g / (molar mass of CH4N2O)

molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol

Na = 60 g / 60 g/mol = 1 mol

Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol

Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091

ΔP = Pb * Xa = 0.038 atm * 0.09091 =  0.0035 atm

Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.

  0.035 atm