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Originally Posted by skmackie
Anyone know how I should go at this problem?
A solution of an unknown substance is made by dissolving 712.0g. of the unknown solid in 150.0g. water. The vapor pressure of the solution is measured at 60 degrees C and is found to be 119.55mm Hg. If the vapor pressure of pure water is 149.44 mm Hg at 60C, calculate the molar mass of the solute.
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You need to use Raoult's Law.
Psoln= Psolv * Xsolv
Start out by rearranging the equation to
Xsolv= (Psoln)/(Psolv)
Where Psoln is the pressure of solution. Psolv is the pressure of the solvent which is water and and X is the mole fraction of water.
First calculate the mole fraction of water.
Xsolv= (119.55mm Hg) / (149.44 mm Hg)
that equals .7999
Next calculate the number of moles of water.
150.0g H20*(1 mol / 18g) = 8.333 mol H2O
Mole fraction is defined as number of moles of water over the total moles.
X= Mole water/(Total moles of solution+ Moles of water)
re arrange and we get
Total moles = Moles of water/Xsolv
Plug our numbers for moles of water and mole fraction of the solution in and we get.
8.333 mol / .7999 = 10.41 mol
now we minus the total moles from the moles of solvent.
10.41 mol - 8.333 mol
2.08 mol solute.
so grams divided by moles eqauals molar mass so we get
712.0g/ 2.08 mol solute = 341.7 g/mol is the molar mass.