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3 years ago

Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

1 answer:

Rudiy273 years ago

5 0

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

ΔTf = (Kf)(m)(i)
where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg

For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius

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