Answer:
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The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.
There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.
B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.
The basis of ionic compounds are ions and the basis of polar compounds are dipoles.
The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.
Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.
When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.
Then, the answer is dispersion interaction.
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
The masses of the objects and the distance between them
-hope it helps
