# Using Colligative Properties to Determine Molar Mass Video

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• 0:01 Colligative Properties
• 1:15 Freezing Point Depression
• 5:39 Finding Molar Mass
• 8:33 Lesson Summary
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Lesson Transcript
Instructor: Nicola McDougal

Nicky has taught a variety of chemistry courses at college level. Nicky has a PhD in Physical Chemistry.

In this lesson, we will explore the effect of colligative properties on a solution. We will learn how to calculate freezing point depression and see how it can be used to calculate the molar mass of an unknown substance.

## Colligative Properties

Anyone who lives in a cold place, like I do, knows the danger of icy roads and sidewalks. To stop cars and people from slipping around, the roads are often treated with a salt. The salt melts the ice and makes the roads and sidewalks much safer. Have you ever wondered why?

The salt has dissolved in the ice and created a solution of icy slush. The salt is a solute that has affected the property of the pure solvent. This property is a colligative property. A colligative property is a property that depends on the number of solute particles present, but not on the type of particle. In other words, the more solute particles in the solution, the greater the effect. Don't worry if this doesn't make much sense now; we will look at this more in the next part of the lesson.

Colligative properties include:

• Vapor pressure lowering
• Boiling point elevation
• Freezing point depression
• Osmotic Pressure

In this lesson, we will focus on freezing point depression and see how it can be used to calculate the molar mass of an unknown substance.

## Freezing Point Depression

Now, we know that under normal conditions, water freezes at 0 °Celsius, but when we add some salt or other type of solute, the freezing temperature goes down. Ice is formed below the normal freezing point, and it remains liquid to a colder temperature.

This is called the freezing point depression, and it is the decrease in the freezing point of a solvent due to the presence of solute particles. When a solute is dissolved in a solvent, the change in the freezing temperature is easily calculated using the following equation:

Let's go through this equation. The first thing to notice is the negative sign; this is here because the freezing point goes down, so the change in temperature, or Î” T, must be a negative number.

The symbol i refers to the number of molecules or ions the solute splits into when it dissolves. So, hopefully, you can see the more particles it breaks down into, the more the effect on the change in temperature.

Kf is the freezing point depression constant, and each solvent has its own value of Kf . (And you will be pleased to know, you will never have to remember these values; you can always look them up.)

And the final symbol is m, the molality of the solution. This is the number of moles of solute per kilogram of solvent.

So, let's look at an example:

• When 20 g of NaCl are added to 100 g of water, what is the change in water's freezing point? (Kf = 1.86 °C/m)

Okay, to tackle this, let us bring back out freezing point depression equation. So, we are given Kf and we need to find out m and i.

Firstly, the molality: remember that this is the number of moles of solute per kilogram of solvent. It is the measure of concentration. (Do not muddle this up with the more commonly used molarity.)

To calculate molality, we firstly need to figure out the number of moles of NaCl using the mass and molar mass. The molar mass of a substance is usually given as the grams of a substance per mole. NaCl has a molar mass of 58.5 g per mole. We can multiply the mass of NaCl in our solution by one over molar mass to to get the number of moles.

• So, 20 g of NaCl * 1 mole NaCl / 58.5 g NaCl = 0.342 moles NaCl.

Then, we calculate the mass of solvent in kilograms:

• 100 g water * 1 kg / 1000 g = 0.1000 kg water
• m = 0.342 moles NaCl / 0.1000 kg water = 3.42 m

To figure out i, we need to remind ourselves how NaCl dissolves. NaCl is ionic and breaks down into ions in a solution.

• For each molecule of NaCl, one Na+ and one Cl- ions are formed.
• So, i = 2.

Now, we can just put the information back into our freezing point depression equation. So, we have:

• Î”T = -2 * 1.86 °C / m * 3.42 m = -12.7 °C

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