Scelta multipla
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.
2803
views
1
rank
13. Solutions
Freezing Point Depression
Scelta multipla
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?
A
334 moles
B
5.4 moles
C
3.2 moles
D
200 moles
Guida verificata passo dopo passo1
Understand the concept of freezing point depression, which is a colligative property. It states that the freezing point of a solvent decreases when a solute is added.
Use the formula for freezing point depression: ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for ethylene glycol as it does not dissociate), Kf is the freezing point depression constant for water (1.86 °C kg/mol), and m is the molality of the solution.
Calculate the change in freezing point (ΔTf) by subtracting the desired freezing point (-10°C) from the normal freezing point of water (0°C), which gives ΔTf = 10°C.
Rearrange the formula to solve for molality (m): m = ΔTf / (i * Kf). Substitute the values: m = 10°C / (1 * 1.86 °C kg/mol).
Convert the molality to moles of solute. Molality (m) is defined as moles of solute per kilogram of solvent. Multiply the molality by the mass of the solvent (1 kg of water) to find the number of moles of ethylene glycol needed.
Video correlati
Pratica correlata
