Freezing Point Depression - Video Tutorials & Practice Problems

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Freezing Point Depression is the phenomenon when adding a solute to a pure solvent results in decreased freezing point of the solvent.

Freezing Point Depression Calculations

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concept

Freezing Point Depression Concept 1

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the phenomenon when adding a salute to appear solvent results in the decreasing of the freezing point of the solvent. Now, with this whole idea of freezing point depression, we have what are called the normal freezing point and the freezing point of the solution. Now, the normal freezing point which I'm going to bring it is F. P. Has to do with our pure solvent. We're going to say this is the freezing point of the solvent before the addition of the salute. And the freezing point of the solution, which we're going to say is FP solution. This is the freezing point of the solvent after the addition of salyut. Remember once you had sought to solve it, it becomes a solution. Now, if we take a look at freezing point depression, let's look at some important areas. First we have the freezing point depression formula, which is delta T. F. Which stands for our changing freezing point equals I which is your mental factor times KF, which is your freezing point constant of the solvent in degrees Celsius. Over morality times, morality itself remember morality is just moles of solute over kilograms of solvent. Now that we know the components of the freezing point depression formula. Realized. How does it relate to the freezing point of a solution? Well, we're gonna say freezing point of the solution equals freezing point of the pure solvent minus delta T. F. If we take a look here, we can see that we have some common types of solvents. We have water, benzene, chloroform and methanol, each with their own normal freezing point before the addition of any salute. Each of them also has their own unique freezing point constant value. No, you don't need to memorize these. This is just for reference of some of the most common types of solvents you may see. So just remember when it comes to the freezing point depression are freezing point decreases the more salient we add to our solvent.

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example

Freezing Point Depression Example 1

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Calculate the freezing point of a solution containing 110.7 g of glucose, which is C6 H 1206 dissolved in 302.6 g of water. All right, so, we're gonna say here that are freezing point of our solution equals the freezing point of our solution equals the freezing point of our solvent minus delta T F. Are solvent. Here is water pure water freezes at 0°C. So we need to do here is figure out what DELTA T. F. Is. We're gonna say DELTA T F equals I times KF times M glucose. Here's our salute. It is not ionic, it's co violent so it doesn't break up into ions. So I is one KF. We're dealing with water as a solvent. It's freezing point constant is 1.86°C over morality. Then all we have to do now is find the moles of glucose and convert grams of water into kilograms of water kg. Just move the decimal over three. So there goes our kilograms of our solvent water. And here we're going to find out the moles of our glucose. So we have this many grams of glucose. Mhm. We're gonna sit here for every one mole of glucose. Its mass is 180 .156 gramps and that's the weight of the six carbons 12 hydrogen is and six oxygen's together grams cancel out. And now I'm gonna have moles which comes out 2. moles of glucose. So we plugged that here. So here these counselor with this modality here, What I get at the end is 3.78°C which is what I plug over here. So the new freezing point after the addition of glucose is negative three 78°C. So that will be our final answer.

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Problem

Problem

How many moles of ethylene glycol, C_{2}H_{6}O_{2}, 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

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Problem

Problem

An ethylene glycol solution contains 28.3 g of ethylene glycol, C_{2}H_{6}O_{2} in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.

A

–8.72°C

B

–0.848°C

C

–0.541°C

D

–17.4°C

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Problem

Problem

When 825 g of an unknown is dissolved in 3.45 L of water, the freezing point of the solution is decreased by 2.89ºC. Assuming that the unknown compound is a non-electrolyte, calculate its molar mass.

A

154 g/mol

B

42.4 g/mol

C

44.5 g/mol

D

159 g/mol

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