We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

Freezing Point Deperession

essay
The whole doc is available only for registered users
  • Pages: 2
  • Word count: 354
  • Category: Chemistry

A limited time offer! Get a custom sample essay written according to your requirements urgent 3h delivery guaranteed

Order Now

Conclusion: My Lab unsuccessfully found the molar mass of the unknown given to us in the lab. We were tasked with finding the molar mass of an unknown compound using the colligative property of Freezing Point Depression. The way we did this was first finding the temperature at which ice crystals form for just regular BHT. Luckily thanks to our TA we were given temperature probes so we could pinpoint exact temperatures. We found that the temperature at which crystallization of BHT was 68.9 Celsius in our lab. After finding temperature of BHT we were tasked with adding the solute cetyl acid to our BHT mixture. We then heated and melted the two substances than found the freezing point of the two substances. We found the temperature at which crystals formed of the two to be 63.8 degrees Celsius. Now the main reason we mixed these two substances was to find the freezing point of BHT. Now to do this we need to take the grams of cetyl acid and find moles of solute and divide it by kilograms of our solvent.

This gave the molality of our solution which turned out to be .515 mol/kg. After finding molality we could find kfp or the freezing point. To do this you would use the formula ∆Tfp = kfpm, so we take the change in temperature which is 5.1 degrees Celsius equal to kfp times .515 mol/kg all which turns out to be 9.90 oc/molal. After we find this crucial piece of information we can add the unknown to our BHT melt the two down and find the ∆T of it which comes out to be 3.3. With all this information given we can finally determine the molar mass of our unknown. To do this we would use the equation MM (unknown solute) = kfp x g(solute)/ kg(solvent) x ∆Tfp now when we plugged all our information into said formula and crunched the numbers we found our molar mass to be 386 g/mol. We found out that the correct molar mass was 284 g/mol and the unknown was stearic acid.

Related Topics

We can write a custom essay

According to Your Specific Requirements

Order an essay
icon
300+
Materials Daily
icon
100,000+ Subjects
2000+ Topics
icon
Free Plagiarism
Checker
icon
All Materials
are Cataloged Well

Sorry, but copying text is forbidden on this website. If you need this or any other sample, we can send it to you via email.

By clicking "SEND", you agree to our terms of service and privacy policy. We'll occasionally send you account related and promo emails.
Sorry, but only registered users have full access

How about getting this access
immediately?

Your Answer Is Very Helpful For Us
Thank You A Lot!

logo

Emma Taylor

online

Hi there!
Would you like to get such a paper?
How about getting a customized one?

Can't find What you were Looking for?

Get access to our huge, continuously updated knowledge base

The next update will be in:
14 : 59 : 59