Multiple Choice
When 825 g of an unknown is dissolved in 3.45 L of water, the freezing point of the solution is decreased by 2.89ºC. Assuming that the unknown compound is a non-electrolyte, calculate its molar mass.
13. Solutions
Freezing Point Depression
Multiple Choice
An ethylene glycol solution contains 28.3 g of ethylene glycol, C2H6O2 in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.
A
–8.72°C
B
–0.848°C
C
–0.541°C
D
–17.4°C
Verified step by step guidance1
First, calculate the molality of the ethylene glycol solution. Molality (m) is defined as the moles of solute per kilogram of solvent. Start by finding the moles of ethylene glycol using its molar mass. The molar mass of ethylene glycol (C2H6O2) is calculated as follows: (2 * 12.01 g/mol for carbon) + (6 * 1.008 g/mol for hydrogen) + (2 * 16.00 g/mol for oxygen).
Next, convert the mass of water from grams to kilograms. Since the density of water is 1.00 g/mL, the mass of 97.2 mL of water is 97.2 g. Convert this mass to kilograms by dividing by 1000.
Calculate the molality of the solution using the formula: m = moles of solute / kilograms of solvent. Use the moles of ethylene glycol calculated in step 1 and the kilograms of water from step 2.
Use the freezing point depression formula to find the change in freezing point: ΔTf = i * Kf * m, where i is the van't Hoff factor (which is 1 for ethylene glycol, a non-electrolyte), Kf is the freezing point depression constant for water (1.86°C kg/mol), and m is the molality calculated in step 3.
Subtract the change in freezing point (ΔTf) from the normal freezing point of water (0°C) to find the freezing point of the solution. This will give you the final answer.
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