How to Calculate Osmotic Pressure and Freezing Point from density and gas pressure

Question

We are given that:

The vapour pressure of water at 100C is 1atm and the vapour pressure of an aceous solution of C2H6O2, non volatile, at 100C is .0.83atm. Taking into account the density of the solution, 1.013g/ml, calculate:

1. Its osmotic pressure at 100C.
2. The solutions freezing point.

Any answers really appreciated

ANSWER:

Got it!:

Pa = 1atm

Pb =0.83atm

T =100C = 373.15K

d= 1.013g/ml amu

H20 = 18 amu

C2H6O2 = 62amu

 osmotic pressure = pi (for purposes of this excercise)

pi=M*R*T

Pa=Xj*Pa=Xj=Pb/Pa = .83/1

Xj = Moles disolution/total moles = .83/1

Mole fraction = .83H2O to 0.17 C2H6O2

1-.83 = .17 moles solute (C2H6O2)

ms=0.17*62 = 10.54gC2H6O2

mj=0.83*18 = 14.94g (H2O)

g total = 25.48g

1.013g/ml = 25.48/V

V = 25.153ml =0 0.025153 Litres

pi =M*R*T

Gas constant = 0.082

= 0 .17 / 0.025153 * 0.082 *373.15 = 206.803

Osmotic Pressure = 206.803

Freezing point = Tc=Tco-Kc*m

Kc H2O = 1.86C/Kgmol

Tc=Tco - Kc*m*s/Kg Dissolution

= 0-1.86*(0.17/14.94/1000) = -21.10465

Freezing point = -21.10465C

Answer 1

Osmotic pressure of a solution is proportional to the molar concentration of a solute particles in the solution

Given that proportional relationship we can write the equation:

Po = (nRT) / V

solving for P (in this case osmosis, from our gas law equation (PV = nRT)

Variable Definitions

Po - osmosis pressure

R - ideal gas constant ( 0.0821 L * atm / mol *K)

T - temperature in KELVIN (K)

n - number of moles in solute

V - volume of solution

remember n/V is the molar concentration

MM is the molar mas

In this case we are given density so we can apply RAOULT'S LAW - equation Pa = XaP*a

(Pa is the product of the mole fraction of solvent in Solution Xa times the vapor pressure of the purse solvent)

Given C2H6O2 is nonvolatile with the density 1.013 g/ml we can find the moles using dimensional analysis with arbitrary 100 mL of solution

Moles C3H6O2 = (100 mL C3H6O2) *(1.013 g C3H6O2)/(1 mL C3H6O2) * (1 mol C3H6O2)/ (74.08 g C2H6O2) = 1.3673 mol

where 74.08 is the molar mass of C3H6O2

Moles H2O assuming 100 mL

(100 mL H2O) * (1.00 gm H20) / (1 mL H2O)* (1 mol H2O) / (18.0 g H2O) = 5.556 mol

Xc3h6o2 = mol C3H6O2 / (mol H20 + mol C3H6O2) = 1.3673 mol / ( 5.556 mol +1.3673 mol) = 0.1975

Raoult's Law to find vapor pressure of water

Pc3h6o2 = Xc3h6o2*P*c3h6o2 = 0.1975 * 0.083 atm = 0.016393 atm

Usually you are giving what concentration of what and ethylene glycol. Are you not?

IF so you may be able to use the ideal gas law as well. To find the freezing point you need to use the equation changeTf =mKf

Given that proportional relationship we can write the equation:

Po = (nRT) / V

solving for P (in this case osmosis, from our gas law equation (PV = nRT)

Variable Definitions

Po - osmosis pressure

R - ideal gas constant ( 0.0821 L * atm / mol *K)

T - temperature in KELVIN (K)

n - number of moles in solute

V - volume of solution

remember n/V is the molar concentration

MM is the molar mass

Perhaps if you could give me some more direction on your problem we can talk and walk through this in a clearer way. I do not want to set you off in the wrong direction. Please contact me with further questions.