Answer to Question #300234 in General Chemistry for daryl

Question

1.     How many grams of BaCl₂ should be added to 300 g H₂O so that the freezing point of the solution is lowered – 8.3 ⁰C ? Assume that the BaCl₂  completely dissociates in the solution.

Answer 1

Using the freezing point depression equation ∆T_f = i x K_f x m

Where ∆T(f) is the freezing point depression = 8.3 °C

i is the van't Hoff factor = 3 (because BaCl₂ completely dissociates into Ba²⁺ + and 2Cl^-)

K_f is the molal freezing point depression constant, it's units are °C kg/mol. For water it is 1.86 °C kg/mol

m is the molality of the solution

Isolating m in the freezing point depression equation gives

m = ∆T_f / (i x K_f)

m = 8.3 °C / (3 x 1.86 °C kg/mol)

m = 1.50 mol/kg

Using the equation

Molality = moles of solute / kg of solvent

Where

Molality = 1.50 mol/kg

Given 300 grams of water must be converted to kilograms.

Since 1000 grams = 1 kg

300 grams = 300/1000 = 0.300 kg of water

1.50 mol/kg = moles of solute / 0.300 kgm

moles of solute = 1.50 mol/kg x 0.300 kg = 0.45 mol

Using the equation

Moles = mass/molar mass

Moles = 0.45 moles

Molar mass of BaCl₂ = 208.23 g/mol

From the above equation

mass of BaCl₂ = moles x molar mass = 0.45 mol x 208.23 g/mol = 93.7 grams