Lena772
  • Lena772
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.
Chemistry
katieb
  • katieb
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Lena772
  • 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
  • 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
  • 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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taramgrant0543664
  • 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
  • Lena772
It's sodium sulfate not sodium nitrate @Photon336 Na2SO4 -> 2 Na+ + SO4 2-
Lena772
  • Lena772
I understand all that @taramgrant0543664 . I just don't know where I went wrong in my individually calculations or rearranging?
taramgrant0543664
  • taramgrant0543664
How did you find that temperature?
Lena772
  • 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
  • taramgrant0543664
Which one is initial and final because you could have possibly put them the wrong way which would change the answer
Lena772
  • Lena772
Should it have been -32 + 12.9 = |-19.1| C because the degree value for both F and C are the same?
Lena772
  • 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
  • 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
  • taramgrant0543664
So it should be -12.9-(-32) so -12.9+32 as final temperature minus initial temperature
Lena772
  • Lena772
@Photon would it be 24.0 g?
Lena772
  • Lena772
Cause that equals 19.1 @taramgrant0543664 .
taramgrant0543664
  • taramgrant0543664
And if you put that in to the equation do you get that same number that you already tried?
Photon336
  • 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
  • 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
  • Lena772
Yes
taramgrant0543664
  • 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
  • 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
  • Lena772
do you guys get 19.1 when you calculate with that value?
Photon336
  • 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
  • Photon336
one question though that Kf shouldn't that be negative as well?
taramgrant0543664
  • taramgrant0543664
I got 0.019 Your moles @Photon336 I think you're missing a 0 in there
Photon336
  • Photon336
0.019 you're right
taramgrant0543664
  • 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
  • taramgrant0543664
It would have been stated if it was negative but water always has a Kf that is 1.86
Lena772
  • Lena772
kg to g would be 19.1 though right?
Photon336
  • Photon336
yep
Lena772
  • Lena772
It says I have a rounding error of less than 10%. Should I try 19.0 or 19.2?
taramgrant0543664
  • 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
  • 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
  • Lena772
18.4 was correct! Hooray :)
taramgrant0543664
  • taramgrant0543664
Ya that could be a problem, I never went back and checked the conversion but that could effect it for sure
taramgrant0543664
  • taramgrant0543664
Yay!!!!
Lena772
  • Lena772
Thank you both !
taramgrant0543664
  • taramgrant0543664
No problem!

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