Freezing point depression of 1 % fluorescein sodium is 0.18^∘ C. Calculate the volume of an iso-

step 1 So now we're going to work on problem 118 from chapter 13. In this problem, we're asked about a

Freezing point depression of 1 % fluorescein sodium is...

Key Concepts

Iso-osmotic Solutions

An iso-osmotic (or isotonic) solution is one that has the same osmotic pressure as another solution, typically body fluids. Preparing such a solution involves adjusting the solute concentration so that the effective osmolarity matches a target value, ensuring that no net water movement occurs across semipermeable membranes when the solution is in contact with cells.

Molality and Osmolarity

Molality is a concentration unit defined as the number of moles of solute per kilogram of solvent and is used in calculating colligative properties like freezing point depression. Osmolarity, on the other hand, is defined as the number of osmoles of solute per liter of solution and is crucial when preparing iso?osmotic (or isotonic) solutions that have the same osmotic pressure as a reference, such as bodily fluids.

Van't Hoff Factor

The van’t Hoff factor (i) indicates the number of particles into which a solute dissociates in solution. For salts like fluorescein sodium, which dissociate into two ions, the factor is important for accurately determining the effective particle concentration, which in turn influences the freezing point depression and the osmolarity of the solution.

Colligative Properties

These are properties of solutions that depend only on the number of solute particles relative to the solvent, not on the identity of the solute. In the context of this problem, freezing point depression is a colligative property, meaning that the depression is directly related to the total number of dissolved particles from the fluorescein salt present in the solution.

Freezing Point Depression

This phenomenon occurs when a solute is added to a solvent, lowering the freezing point of the resultant solution compared to that of the pure solvent. The effect is quantified by the equation ?Tf = Kf · m · i, where ?Tf is the change in freezing point, Kf is the cryoscopic constant, m is the molality, and i is the van’t Hoff factor, reflecting the number of particles generated per formula unit of solute.

Transcript

00:01 So now we're going to work on problem 118 from chapter 13.

00:08 In this problem, we're asked about a glucose solution, which is 10 % by mass.

00:14 We want to calculate its freezing point, boiling point, and osmotic pressure.

00:21 So first, let's talk about this glucose solution.

00:24 If it's 10 % mass glucose, what that means is that it's, let's assume 10 grams of glucose.

00:36 If there's 10 grams of glucose, there's going to be 90 grams of water.

00:43 So the percent mass of glucose is calculated by taking the mass of glucose over the total mass, and multiplying by 100.

00:51 So that would give us 10%.

00:55 So we're going to need to know the moles of glucose.

00:59 So we go ahead and multiply or divide the mass, which is 10 grams, by the molar mass, which is 180 .156 grams.

01:14 And we get a value of 0 .055 moles of glucose.

01:22 So we can directly calculate molality, small m, by dividing these moles by the mass of the water.

01:37 So we wrote up there that there's 90 grams of water.

01:41 If we convert that to kilograms, it's 0 .090 kilograms.

01:44 Kilograms.

01:48 So we do this division and we get a value of 0 .617 molality.

01:57 In, so we can calculate now the freezing point.

02:02 So freezing point will be equal to this molality.

02:08 We ignore the band off factor since it's glucose and will not dissociate.

02:13 So we multiply the malality by the freezing point constant and we get a value of negative 1 .15 degrees celsius and so the freezing point is actually equal to the same thing since we're subtracting it from zero which is the normal freezing point of water.

02:38 So just a small correction here this should be dt since we're finding the freezing point...

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