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
V1 = 30 mL
P1 = 760 torr
P2 = 1520 torr
V2 = ?
applying Boyle's Law
P1*V1 = P2*V2
760 torr * 30 mL = 1520 torr * V2
V2 = 760 torr * 30 mL / 1520 torr
( C ) is correct
It depends on the substance,but for most substance it is in the gaseous state.
