Answer:
The mass of ethyleneglycol that is been required to reduce the vapor pressure of H2O from 1.00 atm to 0.800 atm at 100 °C, is 193,6 g
Explanation:
ΔP = P° - X st
where ΔP is P° - P' (Pressure pure sv - Pressure in sl)
and X st is molar fraction from solute
We replace the values:
1,00 atm - 0,800 atm = 1 atm . Xst
0,2 atm = 1 atm . Xst
0,2 = X st (remember, molar fraction has no units)
X st which is molar fraction from solute means: moles from solute / moles from solute + moles from solvent so
0,2 = moles st / moles st + moles sv
You have mass from solvent so let's get the moles:
225 g / 18,02 g/m = 12,48 moles
0,2 = moles st / moles st + 12, 48 moles
0,2 (moles st + 12,48 moles sv) = moles st
0,2 moles st + 2,496 moles sv = moles st
2,496 moles = moles st - 0,2 moles st
2,496 moles = 0,8 moles st
2,496 moles /0,8 moles = 3,12 moles
So now let's get the mass with the molar mass
3,12 moles x 62,07 g/moles = 193,6 g
