What is the freezing point in degrees Celsius of a solution prepared by dissolving 7.00 g of MgCl2 in 110 g of water? The value of Kf for water is 1.86 (∘C⋅kg)/mol, and the van't Hoff factor for MgCl2 is i=2.7.

Freezing Temperature of solution is -3.36°C

Explanation:

ΔT = Kf . m . i → Freezing point depression

ΔT = Freezing T° of pure solution - Freezing T° of solution

Kf = Cryoscopic constant → 1.86°C /m

m = molality (mol of solute/1kg of solvent)

i = Van't Hoff factor → 2.7

Let's determine molality.

Moles of solute = 7 g . 1mol/ 95.2g = 0.0735 moles

Mass of solvent (g) to kg = 110g . 1kg/1000g = 0.110kg

0.0735 mol / 0.110kg = 0.668 m → molality

We replace data given and determined:

0° - Freezing T° of solution = 1.86°C/m . 0668m . 2.7

Freezing T° of solution = - (1.86°C/m . 0668m . 2.7) → - 3.36°C

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2b2t

explanation:

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