3. Mixtures a.) The vapor pressure of 2-propanol is 50.00 kPa at 338.8K, but it

Question

3. Mixtures a.) The vapor pressure of 2-propanol is 50.00 kPa at 338.8K, but it falls to 49.62 kPa as 8.69 g of a non-volatile organic compound is dissolved in 250 g of 2-propanol. Calculate the molar mass of the solute, assuming the solution is ideal. (3 credits) b) The vapor pressure of pure liquid water is p*(H2O) = 0.02308 atm. Above an aqueous solution with 122 g of a non-volatile solute (molar mass 241 g/mol) in 920 g H20 has a vapor pressure of 0.02239 atm. Calculate the activity and the activity coefficient of water in the solution. (2 credits)

Explanation / Answer

a) we use partial pressure equation.

partial pressure of 2-propanol = mol fraction x total pressure

mol fraction = moles of 2-propanol / (total number of moles of both component )

pA = xA.pA* = nApA* / (nA + nB) :- { A = 2-propanol and B = non volatile organic compound }

pA = (pA* mA / MA ) /(mA / MA + mB / MB)

:- { mA = mass of 2- propanol, and mB = mass of non..com. MA = molar mass of A and MB = molar mass of B }

49.62 = (50.0 KPa x 250 g / 60.1 g/mol ) / (8.69 g / MB + 250 / 60.1g/mol )

MB = 272.78 g/mol

B) activity of water = vapor pressure of solution / vapor pressure of pure water

aA = pA/pA* = 0.02239 atm / 0.02308 atm = 0.9701

activity coefficient of water = activity / mol fraction

gA = aA / xA = aA (nA + nB) / nA

gA = 0.9701 (920 g / 18 g/mol + 122 g / 241 g/mol) / (920 g / 18 g/mol)

gA = 0.9797

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