Question

1. When a 6.31 g sample of CaCl2 is dissolved in water, the freezing point of the solution is -2.63 degrees C. The Kf for the water is 1.86 degrees Clm. Calculate the mass of water used.

2. Suppose a salt (molar mass=58.74g/mol) containing two ions has a solubility of 5.1 mg/L in water at 8.84 degrees C. What is the osmotic pressure in atmospheres of a saturated solution of the salt at 8.84 degrees C?

Answer 1

1.We know that ΔT _f = iK_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0 oC - (-2.36)

        = 2.36

K _f = depression in freezing constant = 1.86 oC/m

i= vanthoff’s factor = 3 ( CaCl₂ ---> Ca²⁺ + 2Cl^- )

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

   = ( 6.31g/111(g/mol)) / W

= 0.057/W

Plug the values we get 2.36 = 3 x 1.86 x (0.057/W)

                                    W = 0.134 kg

                                        = 134 g

2. Given solubility , S = 5.1 mg/L

We know that osmotic pressure = iCRT

Where

T = Temperature = 8.84 oC = 8.84+273 = 281.84 K

C = concentration = (mass/molar mass)x ( 1000 / volume of solution in mL)

   = (5.1mgx10^-3g/mg)/58.74 (g/mol) x ( 1000 / 1000 mL)

   = 8.68x10^-5 mol/L

i = Vant hoff's factor = 2

R = gas constant = 0.0821 L atm / mol - K

Plug the values we get = 2x 8.68x10^-5 x0.0821 x 281.84 = 4.02x10^-3 atm