Login
Copyright

Using Colligative Properties to Determine Molar Mass

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Colligative Properties and Raoult's Law

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:01 Colligative Properties
  • 1:15 Freezing Point Depression
  • 5:39 Finding Molar Mass
  • 8:33 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Login or Sign up

Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!
Create An Account

Recommended Lessons and Courses for You

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.

The equation for calculating the freezing point depression
Freesing Point Depression Equation

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 Δ 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:

  • ΔT = -2 * 1.86 °C / m * 3.42 m = -12.7 °C

Wow, that is some decrease! Instead of freezing at 0 degrees, this salt water won't freeze until it reaches almost -13 degrees. No wonder salt works so well on the roads.

Finding Molar Mass

Now that you know how to use this equation to determine the change in freezing point, you can turn the equation around and use given freezing point depression data to calculate the molar mass of an unknown substance. Let's do that now:

  • You have a 500 g water solution that contains 10 g of an unknown non-ionic solute. Its freezing point is -0.5 °C. What is the molar mass of the unknown substance? (The Kf of water is 1.86 °C/m)

To unlock this lesson you must be a Study.com Member.
Create your account

Register for a free trial

Are you a student or a teacher?
I am a teacher
What is your educational goal?
 Back

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back

Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it free for 5 days!
Create An Account
Support