17. What mass of ethylene glycol, when mixed with 225 g H2O, will reduce the equilibrium vapor pressure of H2O from 1.00 atm to 0.800 atm at 100 °C? The molar masses of water and ethylene glycol are 18.02 g/mol and 62.07 g/mol, respectively. Assume ideal behavior for the solution. a. 15.6 g

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