How many grams of ethylene glycol (c2h6o2) must be added to 1.00 kg of water to produce a solution that freezes at -5.00oc? (kf = 1.86 oc/m)?

167 g

Explanation:

1) The depression of the freezing point of a solution is a colligative property ruled by this equation:

ΔTf = i × m × Kf

Where:

ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.

i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1

Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C

2) Calculate the molality (m) of the solution

ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m

3) Calculate the number of moles from the molality definition

m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent

moles of solute = 2.69 m × 1.00 kg = 2.69 moles

4) Convert moles to grams using the molar mass

molar mass of C₂H₆O₂ = 62.07 g/mol

mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g

h

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