Determining Molar Mass of Unknown using Freezing Point Depression (Colligative Properties)

[@gabbyl3856]

Amazing! You saved my butt on this lab report!

[@xxxsteviexxx]

Thank you!!! So much clearer than how I was being taught in class!

[@TheScienceClassroom]

I'm so glad this video helped!

[@johndelmarkpagulong5348]

Thank you so much for this! Been stuck with a question for 2 hours now thank you!

[@TheScienceClassroom]

I'm glad this video helped you! 🎉

[@jirdalys]

This helped so much, thank you!

[@mrrisky1m184]

Hi

[@naimamohomoud3401]

This video is 2016
And now 2023 Iam watching this
Thank you a lot it's clearer than how I thought in the class💗

[@TheScienceClassroom]

Fantastic! I'm glad it helped! 😃

[@arunmoses2197]

Thanks for the video, it was exactly what I needed!

[@1314Fher]

Yeeees! This made so much sense. Thank you so much! 😃😃

[@keneanbekele1143]

saved my life on lab homework!!!!!!!!!!!! Thank you so much!!!!!!!!!!

[@victoria9396]

What if the problem also gives the van't hoff factor? Would I need it?

[@johnx5k]

DIOS MÍO !!!!!! Thank you !!!!! You are a live saver…. OMG !!!!! Finally 🎉

[@TheScienceClassroom]

Hurray! I'm glad this video helped you! 💯😀

[@saihyss7646]

Supper explanation by this videos

[@cameroncoyle2487]

Question: when you're converting molality to mols, why do you multiply by H2O's mass? Why not the unknown substance's or the solution's total mass?

[@cameroncoyle2487]

I think i answered my own question. Because molality = mols solute/ kg SOLVENT, you have to multiply that by the given mass of solvent in order for it to cancel.

[@anthonyakale4776]

Your a life saver

[@TheScienceClassroom]

😀👍

[@magznificent7845]

so would our molality be 2.74 or is it 2.06?

[@arunmoses2197]

The molality was 2.74. The moles of solute was 0.206.

[@michaelmartinez848]

Awesome, thanks!

[@josh.e7702]

Owen Wilson?

[@isaacvuelvas4619]

What if the freezing point constant isn't given?

[@TheScienceClassroom]

You may have to look it up in your textbook or online. Freezing constants are widely available, so just search for "freezing point constant of....(whatever the solvent is)...."

[@codygruberg7294]

i finally get it

[@moonchironYT]

thank you so much!

[@cathierainwater]

Thank you!!

[@TheScienceClassroom]

You're welcome 😁

[@julianacassandravillaflor9742]

Thank you sir😊😊😊

[@jezreellucena5194]

Thankyou very much

[@1829Emily]

Thank you 😊

[@seanpetersen8146]

Thank you so much

[@merpuffedy]

Thank you. I now refer to you as Jesus Christ

[@graceamazing7349]

I doubt the last calculation for molar mass is stoichiometrically incorrect

[@graceamazing7349]

Should it be (mol x g/mol) = .206mol x 26.4g/mol ~ 5.44g ?

[@cuteag2540]

How about the formula ∆Tf=Kfm, why didn't you use it?😅

[@TheScienceClassroom]

That's what I used

[@usa2250]

Thank you so much!

[@TheScienceClassroom]

😁