Respuesta :
When a solute is added to a solvent, some properties are affected and these set of properties are called colligative properties. The properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. For this case, we use freezing point depression:
ΔT(freezing point)
= (Kf)m where Kf is the freezing point depression constant and m is the molality
ΔT(freezing point) = 1.86 °C kg / mol (500 g c2h6o2 ) ( 1 mol / 62.08 g ) / 0.500 kg water
ΔT(freezing point) = 29.96 °C
Tf - T = 29.96 °C
T = -29.96 °C
Therefore, the freezing point of the solution of ethylene glycol is -29.96 °C
Freezing point of solution of 500.0 g of ethylene glycol dissolved in 500.0 g of water is [tex]\boxed{29.96\text{ }^{\circ}\text{C}}[/tex]
Further Explanation:
Colligative properties:
Such properties depend only on concentration of solute particles, not on their identities.
Four colligative properties are listed below.
I. Relative lowering of vapor pressure.
II. Elevation in boiling point.
III. Depression in freezing point.
IV. Osmotic pressure.
Freezing point depression is evaluated by following formula:
[tex]\Delta\text{T}_\text{f}=\text{k}_\text{f}\text{m}[/tex] ...... (1)
Here,
[tex]\Delta\text{T}_\text{f}[/tex] is depression in freezing point.
[tex]\text{k}_\text{f}[/tex] is molal freezing point constant.
m is molality of solution.
The formula to calculate moles of ethylene glycol is as follows:
[tex]\text{Moles of ethylene glycol}=\dfrac{\text{Mass of ethylene glycol}}{\text{Molar mass of ethylene glycol}}[/tex] ...... (2)
Substitute 500.0 g for mass of ethylene glycol and 62.07 g/mol for molar mass of ethylene glycol in equation (2).
[tex]\begin{aligned}\text{Moles of ethylene glycol}&=\dfrac{500\text{ g}}{62.07\text{ g/mol}}\\&=8.055\text{ mol}\end{aligned}[/tex]
The molality of ethylene glycol solution is calculated as follows:
[tex]\begin{aligned}\text{m}&=\left(\dfrac{8.055\text{ mol}}{500.0\text{ g}}\right)\left(\dfrac{1\text{ g}}{10^{-3}\text{ kg}}\right)\\&=16.11\text{ m}\end{aligned}[/tex]
Substitute 16.11 m for m and [tex]1.86\text{ }^{\circ}\text{C/m}[/tex]for [tex]\text{k}_\text{f}[/tex] in equation (1).
[tex]\begin{aligned}\Delta\text{T}_\text{f}&=\left(1.86\text{ }^{\circ}\text{C/m}\right)\left(16.11\text{ m}\right)\\&=29.96\text{ }^{\circ}\text{C/m}\end{aligned}[/tex]
The formula to calculate the change in freezing point is as follows:
[tex]\Delta\text{T}_\text{f}=\text{T}_\text{f}_\left(\text{water}\right)}-\text{T}_\text{f}_\left(\text{solution}\right)[/tex] ...... (3)
Here,
[tex]\text{T}_\text{f}_\left(\text{water}\right)}[/tex] is temperature of water.
[tex]\text{T}_\text{f}_\left(\text{solution}\right)[/tex] is temperature of solution.
Rearrange equation (3) to calculate the temperature of the solution.
[tex]\text{T}_\text{f}_\left(\text{solution}\right)}=\text{T}_\text{f}_\left(\text{water}\right)}-\Delta\text{T}_\text{f}[/tex] ...... (4)
Substitute [tex]0\text{ }^{\circ}\text{C}[/tex] for [tex]\text{T}_\text{f}_\left(\text{water}\right)}[/tex] and [tex]29.96\text{ }^{\circ}\text{C}[/tex] for[tex]\Delta\text{T}_\text{f}[/tex] in equation (4).
[tex]\begin{aligned}\text{T}_\text{f}_\left(\text{solution}\right)}&=0\text{ }^{\circ}\text{C}-29.96\text{ }^{\circ}\text{C}\\&=-29.96\text{ }^{\circ}\text{C}\end{aligned}[/tex]
Learn more:
1. Choose the solvent that would produce the greatest boiling point elevation: https://brainly.com/question/8600416
2. What is the molarity of the stock solution? https://brainly.com/question/2814870
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Colligative properties
Keywords: colligative properties, freezing point, ethylene glycol, m, 16.11 m, mass of ethylene glycol, molar mass of ethylene glycol, 8.055 mol, change in freezing point.
