How many grams of NaCl (58 g/mol) would produce the same freezing point depression of 1 L of water a

[baywatch123]

After going through this problem over and over, I do understand how they got 14.5gram of NaCl. But I'd like to see how other's would have tackled the problem?

I basically set an equation like so:

-imKf = -imKf

where the negatives cancel out and the freezing point constant Kf cancels out as well. i = dissociation number and m = molals.

So you're left with
im (of NaCl) = im (of SiO2)

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[baywatch123]

Here's the question:

How many grams of NaCl (58 g/mol) would produce the same freezing point depression of 1 L of water as 30 grams of SiO2 (60 g/mol)?

 

[Promethean]

I would have set it up very similarly.

If you understand that you need as many moles of ions on each side of the equation, it becomes pretty trivial to figure out how grams you need from there.

 

[Queen Beach]

Yep, that's it
x/58 * 2 = 30/60 * 1
Thus x= 14.5 g

 

[WheatLom]

1/4 of a mol.

 

[London22]

Hey I know it's late to bump this thread but could someone explain to me how you know to multiple the left side of the equation by 2 and the right side by 1?
I know it's probably something easy that I'm skipping over. When I got this question wrong on a practice exam I had ended up with 29 as my answer.

 

[baywatch123]

It's been a while for me now, but I think it has to do with NaCl being able to dissolve into "two entities" making it 2, and SiO2 does not break up.

NaCl is a salt, whereas SiO2 is a molecule that's not a "salt". I think that's why.

Go back to the review lecture notes on freezing point, i think it was under like solutions for gen chem. Chad's videos goes over it nicely and his examples - that's what i used and repeated like crazy to get the point

 

[DAT Destroyer]

NaCl is an electrolyte, and therefore i=2,,,this is the Vant Hoff Factor. SiO2 has i = 1 since it is a network solid. Join my FaceBook Study Group and I will be happy to discuss this further. I had some wonderful colligative property questions yesterday that may interest you.

Dr. Jim Romano

 

[Bersabeh]

After going through this problem over and over, I do understand how they got 14.5gram of NaCl. But I'd like to see how other's would have tackled the problem?

I basically set an equation like so:

-imKf = -imKf

where the negatives cancel out and the freezing point constant Kf cancels out as well. i = dissociation number and m = molals.

So you're left with
im (of NaCl) = im (of SiO2)

Chad is super in explaining this look at his explanation, chad for chemistries!