Question

The vapor pressure of water at 25 degrees Celsius is 23.76 mm of Hg; Kf for water is 1.86 (degree Celsius kg/mol). The normal boiling point of chloroform is 61.0 degrees Celsius and Kb is 3.63 (degree Celsius kg/mol) R=0.08206 (L atm/mol.k).
1. What is the vapor pressure of a solution on 10.0g of NaNO3 in 25.0g of water?
2. What is the freezing point of a 0.20 m
aqueous solution of sucrose at 1 atm pressure?
3. What is the osmotic pressure, pi, of a 0.100 M aqueous solution of a CaCl2 at 25 degrees Celsius

Answer 1

1) We know from Raoult's law

P⁰-P/P⁰ = i x mole fraction of solute

P⁰ = 23.76 mm and i = 2 for NaNO3(two ions)

mole fraction of NaNO3 = (10/85) /[(10/85) + 25/18]

= 0.1176/1.50

Thus 23.76-P /23.76 = 2 x [0.1176/1.5]

P = 20.03mm

2) depression in freezing point is calculated

Delta Tf = i xKf x m

= 1x 1.86 x 0.2

= 0.372

Thus the freezing point of solution = 0C -0.372

= -0.372C

3) We can calculate osmotic pressure using

pi = i x CRT where C = molar concentration

R = gas constant

T= temperature in K

pi = 3 x 0.100mol/L x 0.08206 L.atm/K.mol x 298K

= 7.34atm