A solution is made by mixing 42.0 mL of ethanol, C2H6O, and 58.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 C in torr?

Question

Values at 20C
ethanol density: 0.789 g/mL & vapor pressure (torr): 43.9 torr
water: denisty: 0.998 g/mL & 17.5 torr

Any steps would be GREATLY appreciated!

Answer 1

Calculate moles H2O.
Calculate moles EtOH.
Calculate XH2O
Calculate XEtOH.
Calculate pH2O
Calculate pEtOH.
Total P = pH2O + pEtOH

Answer 2

Hi Dr Bob, is this right so far?
mass h2o = (d)(v) = (0.998 g/mL)(58 mL) = 57.884 g
57.884 g H2O/18g H2O = 3.2157 mol H2O
mass ethanol = (0.789 g/mL)(42 mL) = 33.138 g
33.138 g C2H6O/46.0687 g C2H6O = 0.719 mol C2H6O

molality (m) = 0.71937 mol C2H6O/0.057884 kg H2O = 12.43 m

xH2O = 12.45/(12.43+3.2157) = 0.794
xEthanol = 12.43/(12.43 + 0.719) = 0.91
?

Answer 3

Hi Dr Bob, is this right so far?
mass h2o = (d)(v) = (0.998 g/mL)(58 mL) = 57.884 g
57.884 g H2O/18g H2O = 3.2157 mol H2O
mass ethanol = (0.789 g/mL)(42 mL) = 33.138 g
33.138 g C2H6O/46.0687 g C2H6O = 0.719 mol C2H6O

You are ok to here with moles.
You don't need molality. You want mole fraction. mole fraction EtOH = XEtOH = moles EtOH/total moles. mole fraction H2O = XH2O = moles H2O/total moles
Then pH2O = XH2O*PoH2O and
pEtOH = XEtOH*PoEtOH.
Total P = pH2O + pEtOH.

molality (m) = 0.71937 mol C2H6O/0.057884 kg H2O = 12.43 m

xH2O = 12.45/(12.43+3.2157) = 0.794
xEthanol = 12.43/(12.43 + 0.719) = 0.91
?

Answer 4

i got it! thanks Dr Bob