Part 2 I. The density of a 50% water and 50% hanol mixture, by mass, is 0.913 g/
ID: 546872 • Letter: P
Question
Part 2 I. The density of a 50% water and 50% hanol mixture, by mass, is 0.913 g/mL. Note: the molar masses of water by mass, and ethanol are 18 0 g and 460g the universal gas constant R is 0.0821 L-atm/mol-K. Calculate the molality of this solution assuming that water 8 and 4çQ g the universal gas constant R is 0,0821 L-atm/molK. a. is the solvent in this case. b. Calculate the molarity of this solution assuming that water is the solvent in this case. c. Calculate the mole fractions of water and ethanol in this solution. Calculate the total vapor pressure of this solution if the vapor pressure of pure water at 25 °C is 0.03 atm and the vapor pressure of ethanol is O.0724 atm. Assuming that water is the solvent, calculate the osmotic pressure of this solution at 25 °C. e.Explanation / Answer
Solution:- (a) let's say the mass of solution is 100 g. Then mass of water would be 50 g and mass of ethanol would also be 50 g.
moles of ethanol = 50 g x (1mol/46.0 g) = 1.09 mol
molality is moles of solute/kg of solvent.
50 g of water x (1kg/1000g) = 0.05 kg of water
molality = 1.09 mol/0.05 kg = 21.8 m
where m stands for molality.
(b) molarity is moles of solute per liter of solution.
volume = mass/density = 100 g x (1mL/0.913 g) = 109.5 mL = 0.1095 L
Molarity = 1.09 mol/0.1095 L = 9.95 M
where M stands for molarity.
(c) Moles of water = 50 g x (1mol/18.0g) = 2.78 mol
Total moles = 2.78 + 1.09 = 3.87
mole fraction of water = 2.78/3.87 = 0.718
mole fraction of ethanol = 1.09/3.87 = 0.282