An aqueous salt solution is formed by adding 11.67 g sodium sulfate (solute) to water (solvent). What mass (in g) of water is used if the freezing point of the solution is -12.9 oF.
Kf H2O = 1.86 oC/m
When Determining The molality Of The Dissolved Particles, Assume Complete Ionization Of Ionic Compounds - 1 Molecule Can Form 1, 2, 3, Or More Dissolved Particles.

- Lena772

- katieb

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- Lena772

I got 64 g and it was wrong. I used the delta T = Kfmolal formula and rearranged it to find that kg solv = (Kf)(Moles solute)/delta T

- Lena772

Then I figured out moles of solute, mol pieces and the change in temperature. The Kf was given. However, I feel my delta T may have been incorrect.

- Lena772

I got 7.1667 C for delta T. I calculated that their was 0.0822 mol of solute and multiplied this by 3 mol pieces to get my moles of solute. Which made that value 0.2466.

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## More answers

- taramgrant0543664

dT=iKfm
dT is the change in temperature (Tfinal-Tinitial)
i is the van't Hoff disassociation constant
Kf would be the 1.86
m is moles of solute divided by kg of solvent

- Lena772

It's sodium sulfate not sodium nitrate @Photon336 Na2SO4 -> 2 Na+ + SO4 2-

- Lena772

I understand all that @taramgrant0543664 . I just don't know where I went wrong in my individually calculations or rearranging?

- taramgrant0543664

How did you find that temperature?

- Lena772

I converted Fahrenheit to Celsius. I got that -12.9 o F = -24 o C and then I converted 0F to Celsius and that gave me like -17C so i found the delta T of that change in C and it ws like the 7.1667 C value.

- taramgrant0543664

Which one is initial and final because you could have possibly put them the wrong way which would change the answer

- Lena772

Should it have been -32 + 12.9 = |-19.1| C because the degree value for both F and C are the same?

- Lena772

Initial had to be -32F cause that's the freezing pt of water, but when the solution was added it went to -12.9

- Photon336

Forget I said that, your math is correct.
\[\Delta T = iK _{b}m \]
\[\frac{ \Delta T }{ iK _{b} } = moles/kg \]
\[\frac{ moles*i*Kb}{ \Delta T } = kg \]
I think here is your formula.
(°F - 32) x 5/9 = °C

- taramgrant0543664

So it should be -12.9-(-32) so -12.9+32 as final temperature minus initial temperature

- Lena772

@Photon would it be 24.0 g?

- Lena772

Cause that equals 19.1 @taramgrant0543664 .

- taramgrant0543664

And if you put that in to the equation do you get that same number that you already tried?

- Photon336

according to wikipedia the freezing point of water is 32 degrees F
So originally we had Tf as -12.9 F right?
(-12.9-32)*(5/9) = C
-32F = Temperature 1 = 0C
-24 = Temperature 2 tf
i think both have to be in celsius: so T1 = 0 degrees celsius
(T2-T1) = (-24-0) Celcius -24 celcius

- Photon336

but My assumption is that our final answer that the freezing point had to be 0 Degrees C and i converted that -12.9 F right? to Celsius as Temp final

- Lena772

Yes

- taramgrant0543664

The change in temperature in celsius would be different in F then it is in C so you would have to convert before hand

- Lena772

With that value I get 19.1 g, but if I use 19.1 as the change in temp I get 24 g solv.

- Lena772

do you guys get 19.1 when you calculate with that value?

- Photon336

Tara what did you get for this?
\[11.67 grams /142 grams/mol = 0.82 moles \]
\[\frac{ 0.82*3*1.86 }{24 } = 0.191 kg \]

- Photon336

one question though that Kf shouldn't that be negative as well?

- taramgrant0543664

I got 0.019
Your moles @Photon336 I think you're missing a 0 in there

- Photon336

0.019 you're right

- taramgrant0543664

I honestly can't remember the rule when Kf is negative or not it's been a little bit since I've had to do these questions let me see if I can find that rule

- taramgrant0543664

It would have been stated if it was negative but water always has a Kf that is 1.86

- Lena772

kg to g would be 19.1 though right?

- Photon336

yep

- Lena772

It says I have a rounding error of less than 10%. Should I try 19.0 or 19.2?

- taramgrant0543664

For me it comes out as 19.065 and 10% is +/-0.1 so you could try 19.2 if you want, I'm getting an answer of 19.065 so I would think it would be 19.1 when rounding

- Lena772

Well 12.9 F = -24.9444 and we used 24, so that may be the problem. When I calculate with that value I get 18.4 g

- Lena772

18.4 was correct! Hooray :)

- taramgrant0543664

Ya that could be a problem, I never went back and checked the conversion but that could effect it for sure

- taramgrant0543664

Yay!!!!

- Lena772

Thank you both !

- taramgrant0543664

No problem!

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