Problem 5: we use Raoult's Law y=K*x where y is the vapor mole fraction and x is
the liquid mole fraction. We can calculate the x for each compound from the
concentration and then solve for y. So for Glycerol we have a concentration of 5.5g
Glycerol/100 g water. To get a mole fraction we must divide the top by the molecular
weight of Glycerol (92.1 g/mol) and divide the bottom by the molecular weight of water
(18 g/mol). So our concentration then becomes .05972 mol Glycerol/5.556 mol water
or .01075 mol Glycerol/1 mol water. If we assume we have 1 mol of water we would
then have .01075 mol Glycerol, giving a mole fraction of .01075/(.01017+1)=.0106. We
plug this number into Raoult's law to get a vapor mole fraction, y, of 1.276*10^-9. We
can then use this number to get back to a concentration by doing the opposite of what we
did before. We say 1.276*10^-9= x mol Glycerol/(x+1) and find there to be 1.276*10^-9
mol Glycerol/mol water. Multiplying again by the molecular weights gives a final
answer of 1.175*10^-7 g Glycerol/18 g water. The same thing can be done for MEK
and Phenol.
Problem 7: Extent of reaction describes the amount of reaction that has happened. The
equation we need is that extent of reation (e)= (moles at end - moles
initially)/stoichiometric coefficient. so for this equation we can write this for each
compound: e= (CO2,f-CO2,i)/-1, e=(H,f-H,i)/-4, e= (CH2O,f-CH2O,i)/1, & e=(H2O,fH2O,i)/1. We know how much CH2O we started with (none) and how much we ended
with (.7g) so we will use that equation. We change into moles CH2O by dividing by the
molecular weight (30g/mol) to get that we end with .02333 moles. Plugging into our
equation we get e=(.02333-0)/1 so e=.02333 moles. For part be we use this extent of
reaction in the CO2 equation. We started with 1.81 g CO2 which is .04114 moles CO2,
so our equation is .02333=(moles CO2,f-.04114)/-1. This comes out to give us .0178
moles CO2 remaining. Multiplying by the molecular weight gives us .783 g CO2 left.