Calculate the freezing point and boiling point of each aqueous solution, assuming complete disso

Calculate the freezing point and boiling point of each...

Key Concepts

Complete Dissociation

This assumption implies that an ionic solute fully separates into its constituent ions when dissolved in a solvent. It is fundamental to colligative property calculations because it ensures that the total number of solute particles in the solution corresponds directly to the expected number based on the van't Hoff factor.

Molality

Molality is a concentration unit defined as moles of solute per kilogram of solvent. It is particularly useful in colligative property calculations because it is independent of temperature, making it more reliable than molarity when temperature changes may occur during phase transitions.

Van't Hoff Factor

This factor (denoted as i) represents the number of particles into which a solute dissociates in solution. For complete dissociation, electrolytes break apart into multiple ions, so i is greater than 1. It plays a critical role in calculating the magnitude of colligative effects, as it amplifies the changes in freezing and boiling points based on the number of solute particles.

Freezing Point Depression

This is a colligative property where the freezing point of a solvent is lowered by the addition of a solute. It is quantified by the equation ?Tf = i × Kf × m, where ?Tf is the change in freezing point, i is the van't Hoff factor, Kf is the molal freezing point depression constant specific to the solvent, and m is the molality of the solution.

Boiling Point Elevation

Boiling point elevation is another colligative property observed when a solute is dissolved in a solvent; the boiling point of the solvent increases. This is described by the equation ?Tb = i × Kb × m, where ?Tb is the elevation in boiling point, i is the van't Hoff factor, Kb is the molal boiling point elevation constant, and m is the molality. The presence of solute particles causes a reduction in the vapor pressure of the solvent, necessitating a higher temperature to reach the boiling point.

Colligative Properties

These are properties of solutions that depend solely on the number of solute particles present, rather than their chemical identity. Examples include boiling point elevation and freezing point depression, which occur when a solute is added to a solvent, altering its physical properties by affecting the solvent’s phase transitions.

Transcript

00:01 To calculate the freezing and boiling points of each solution, we need to know the molality of each solution.

00:06 The molality is provided for the first one.

00:09 It's 0 .1 molal.

00:12 However, you'll notice that it will separate into two potassium ions and one sulfide anion.

00:18 So the vaughfactor will be three.

00:21 So it'll be three ions for every potassium sulfide.

00:27 So we'll multiply the concentration by three.

00:30 Then we'll multiply it by the freezing point depression constant for water of 1 .86.

00:35 To get the change in the freezing point of the water when the solute is added, we get 0 .558 degrees celsius.

00:44 Because it's freezing point depression, that's a decrease off of the normal freezing point of water of 0 to negative 0 .558 degrees celsius.

00:53 Then for the boiling point elevation, tb will be equal to, again, the vant haft factor 3, multiplied by the molality, multiplied by the boiling point elevation constant of water, of 0 .512, giving us 0 .154 degrees celsius.

01:13 This will be the change in boiling point off of 100 degrees celsius, and this is a boiling point elevation.

01:19 Boiling point increases, freezing point decreases.

01:22 So we'll add .154 to 100, and we get the new boiling point of 100 .154 degrees celsius.

01:32 For the next one, we don't know the molality.

01:34 We'll have to calculate the molality.

01:36 We have 21 .5 grams of copper two chloride.

01:40 So we'll convert the copper two chloride mass into moles by dividing by the molar mass of copper two chloride.

01:46 Then to get molality, we need to divide by the kilograms of just the solvent, just the water...

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