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 wate r at this temperature is 0.038atm. please 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
1 answer:
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. Answer: 0.035 atm
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