Raoult's Law - How To Calculate The Vapor Pressure of a Solution
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- เผยแพร่เมื่อ 22 ม.ค. 2021
- This chemistry video tutorial provides a basic introduction into Raoult's law which says that the vapor pressure of a solution is the product of the mole fraction of the solvent and the partial pressure of the pure solvent at a given temperature. This video explains how to calculate the vapor pressure of a solution with a nonvolatile solute. It also explains how to calculate the mass of solute / glucose needed to reduce the vapor pressure of the solution to a certain value.
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FANTASTIC concise explanation and the practice questions you chose were perfect at covering this topic. Thank you!
I would like to correct or point out a slight mistake in the last problem. Both sides should be divided by 0.9765. The answer would be 0.3338 moles.
Yeah that confused me for a bit
he converted it into gram. that 0.3338 is a mol
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Why is the timing so perfect, my prof have just finished teaching this topic.
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Maybe your professor was also learning from here and then teaching you
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can someone explain why at 5:38 he calculates the molarity, but not of the compound. Instead the # of ions??
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Why did you use the mols ions on the second one and not the first
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Slight error with the numbers sir. just 0.9765 and not 0.9675 but other than that thank you.
You’re my life savor. Yet at 13’10”, I believe the denominator is supposed to be 0.9765. I made lots of similar mistakes.
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Why moles of ions? Can't we take no
Of moles of cacl2 directly?
how can you tell if an ionic solute is given in a question? like the difference between the 1st and 2nd question?
When there are metal and nonmetal atoms bonded together, it forms an ionic compound.
is the last example here correct? i computed for the mol of glucose and it turned out 576.64 mol (although i doubt this because the solvent should be greater). When i inserted this in the Raoult's equation it resulted to 54 Torr. Correct me if I'm wrong tho thank you
2 months old but just here to tell you, he read his mol wrong, so I’m guessing what you got is the correct answer.
@@Micahiscoo lmao thanks! It's already vacation but I appreciate the reply man God bless on your studies
Ok so you mentioned at 6:30 that you need to mutiply by 3 ions and you explained how CaCl2 forms one calcium ion and 2 chlorine ions. How were we supposed to know that? That shoudnt be just something we have to memorize Im sure right. Would that be something your teacher would have to give you kind of like the periodic table of elements. How are we supposed to know how many ions form from a solute dissolving? Please explain this
How would it be if we're to find the vapour pressure of only one substance and not the entire solution...??
Can you explain when you're trying to find the mole of the compound you put the molar mass as 'g' instead of gmol^-1.Also why are you dividing the the number of moles by the molar mass instead of using the mole formula = mass/molar mass to find the mole of the compound. Thanks
why did we have to convert the moles of cacl2 to ions for the second problem but not the first problem?
Because glucose is not an ionic compound, CaCl2 is
@@engertejada6930 how are we supposed to know CaCl2 is an ionic compound. No way you have to just memorize every single compound formula right? Is this something your teacher should give you?
why did u not change mL to L
at the last example, i didn't understand why he multiple the moles by 180.15.. from where did he get that number?
180.15 is the molar mass of glucose. To convert moles to grams you have to multiply the number of moles( here he got 0.337 moles) with the molar mass. (in this case, multiply it with the molar mass of glucose)
@@lenamathew5713 Oh i see! Thank you so much!
at 0:55 , mind if i ask why should the answer be less than 23.8 torr?
dont know if you need it anymore or not but the bases of raoult's law says that the vapour pressure of solvent decreases if you mix a Solute
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how if the question dont have torr
It does have torr
I'm among the first😎
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1
Lie lol
i dont get it
Not me watching this 1 houre before exam😂
1mL=1g you're welcome
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You put 0.9675 instead of 0.9765
First.
True
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First
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