An ideal solution is formed from mixing six grams of the nonvolatile solute urea, CH4N2O, with 32 grams of methanol, CH3OH. The vapor pressure of pure methanol at 20 degrees celsius is 89 mmHg...?

Question

What is the vapor pressure of the solution at 20 degrees celsius? Why might the measured/ observed solution vapor pressure be lower than the calculated vapor pressure?

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Answer 1 · 2018-02-16T21:26:07

The vapour pressure is proportional to the mole fraction of the volatile component....

And so we calculate the mole fraction, #chi#, of each component of the solution....

#chi_("urea")="moles of urea"/"moles of urea + moles of methanol"#

#chi_"urea"=((6.0*g)/(60.06*g*mol^-1))/((6.0*g)/(60.06*g*mol^-1)+(32.0*g)/(32.04*g*mol^-1))=0.091#

#chi_"methanol"=((32.0*g)/(32.04*g*mol^-1))/((6.0*g)/(60.06*g*mol^-1)+(32.0*g)/(32.04*g*mol^-1))=0.909#

And #chi_"urea"+chi_"methanol"=1# is UNITY as is absolutely required for a binary solution....

And so #"Vapour pressure of solution"=chi_"methanol"xxP_"methanol"^@#...

#=0.909xx89*mm*Hg=80.9*mm*Hg#..

The vapour pressure is thus reduced with respect to that of the pure solvent.