A glucose solution contains 55.8 g of glucose (C6 H12 O6) in 455 g of water. Determine the fre

A glucose solution contains 55.8 g of glucose (C6 H12 O6)...

Key Concepts

Van't Hoff Factor

The van't Hoff factor (i) is a measure of the number of particles a solute yields in solution compared to the number of formula units dissolved. It is important in colligative property calculations because it quantifies the effect of solute dissociation or association on properties like freezing point depression and boiling point elevation.

Boiling Point Elevation

This is the process by which the boiling point of a solvent increases when a non-volatile solute is added. It is quantitatively described by the equation ?Tb = i × Kb × m, where i, the van't Hoff factor, accounts for the number of particles produced in solution, Kb is the ebullioscopic constant, and m is the molality.

Molality

Molality is a measure of the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. It is especially useful in colligative property calculations because it is independent of temperature changes.

Colligative Properties

These are properties of solutions that depend on the number of solute particles present rather than the type of particles. They include changes in boiling point and freezing point, osmotic pressure, and vapor pressure lowering, and are crucial for understanding how solute addition affects solvent behavior.

Freezing Point Depression

This phenomenon occurs when a solute is dissolved in a solvent, leading to a lower freezing point of the solution compared to the pure solvent. The extent of freezing point depression is determined using the equation ?Tf = i × Kf × m, where i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality.

Transcript

00:01 So when we're thinking about how solutions, boiling, and freezing point have changed, they give us an equation for that.

00:11 The change in temperature is equal to some constant that is known, in this case a constant for water, that's freezing temperature and its boiling temperature.

00:24 There are these numbers 1 .86 for the freezing, 0 .512 for the boiling.

00:32 We have to multiply each one of these against malality.

00:38 And then if the compound breaks up, say if it's an ionic compound, then this i called the van hoff factor will come into play.

00:49 But because we're talking about glucose, it doesn't break up in solution.

00:56 We can pretty much assume i is going to equal just one in this case and pretty much disregard this.

01:03 And so we already have the constants for the boiling and freezing.

01:08 What we really need is to malau in this case.

01:12 And if you recall malau is moles of the solute over the kilograms of the same.

01:20 Solvent this is a solvent and this is going to be the solute right so we need to figure that part first so we're looking at it they already give us that the solute was 55 .8 grams and we just need to multiply it against the molecular weight and that's about 180 .156.

02:03 I'm going to do something a little bit different here.

02:06 We're also, remember, it's moles of the solute of the kilograms solutions.

02:12 We actually put the solute on top, but i know 55 .8 is a solute.

02:17 We know there's 455 grams of the solvent.

02:22 Now we just got to do some conversions.

02:25 Now this one's going to convert.

02:27 Inverts solute from grams into moles.

02:32 I'm going to add another factor here where we take, you know, for every 1 ,000 grams of the solvent, you get one kilogram of the solvent.

02:47 And it strikes this one out and this one out.

02:51 From there, all i do is put in a calculator.

02:53 Went through which you get 0 .681.

03:03 Then go off and figure out, let's do the freezing point first.