Multiple Choice
How many moles of ethylene glycol, C2H6O2, must be added to 1,000 g of water to form a solution that has a freezing point of – 10ºC?
2029
views
1
rank
13. Solutions
Freezing Point Depression
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.
A
154 g/mol
B
42.4 g/mol
C
44.5 g/mol
D
159 g/mol
Verified step by step guidance1
Start by understanding the concept of freezing point depression, which is a colligative property. It depends on the number of solute particles in a solution, not their identity.
Use the formula for freezing point depression: ΔT_f = i * K_f * m, where ΔT_f is the change in freezing point, i is the van't Hoff factor (which is 1 for non-electrolytes), K_f is the freezing point depression constant for the solvent (water in this case), and m is the molality of the solution.
Calculate the molality (m) of the solution using the formula: m = moles of solute / kilograms of solvent. First, convert the volume of water (3.45 L) to mass, assuming the density of water is approximately 1 g/mL, so 3.45 L of water is 3450 g or 3.45 kg.
Rearrange the freezing point depression formula to solve for molality: m = ΔT_f / K_f. Use the given ΔT_f = 2.89ºC and the known K_f for water, which is 1.86ºC kg/mol.
Once you have the molality, calculate the moles of solute using the formula: moles = molality * kilograms of solvent. Then, find the molar mass of the unknown compound by dividing the mass of the solute (825 g) by the moles of solute calculated.
Related Videos
Related Practice
