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Lena772

  • one year ago

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.

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  1. Lena772
    • one year ago
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    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

  2. Lena772
    • one year ago
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    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.

  3. Lena772
    • one year ago
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    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.

  4. taramgrant0543664
    • one year ago
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    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

  5. Lena772
    • one year ago
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    It's sodium sulfate not sodium nitrate @Photon336 Na2SO4 -> 2 Na+ + SO4 2-

  6. Lena772
    • one year ago
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    I understand all that @taramgrant0543664 . I just don't know where I went wrong in my individually calculations or rearranging?

  7. taramgrant0543664
    • one year ago
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    How did you find that temperature?

  8. Lena772
    • one year ago
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    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.

  9. taramgrant0543664
    • one year ago
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    Which one is initial and final because you could have possibly put them the wrong way which would change the answer

  10. Lena772
    • one year ago
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    Should it have been -32 + 12.9 = |-19.1| C because the degree value for both F and C are the same?

  11. Lena772
    • one year ago
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    Initial had to be -32F cause that's the freezing pt of water, but when the solution was added it went to -12.9

  12. Photon336
    • one year ago
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    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

  13. taramgrant0543664
    • one year ago
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    So it should be -12.9-(-32) so -12.9+32 as final temperature minus initial temperature

  14. Lena772
    • one year ago
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    @Photon would it be 24.0 g?

  15. Lena772
    • one year ago
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    Cause that equals 19.1 @taramgrant0543664 .

  16. taramgrant0543664
    • one year ago
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    And if you put that in to the equation do you get that same number that you already tried?

  17. Photon336
    • one year ago
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    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

  18. Photon336
    • one year ago
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    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

  19. Lena772
    • one year ago
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    Yes

  20. taramgrant0543664
    • one year ago
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    The change in temperature in celsius would be different in F then it is in C so you would have to convert before hand

  21. Lena772
    • one year ago
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    With that value I get 19.1 g, but if I use 19.1 as the change in temp I get 24 g solv.

  22. Lena772
    • one year ago
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    do you guys get 19.1 when you calculate with that value?

  23. Photon336
    • one year ago
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    Tara what did you get for this? \[11.67 grams /142 grams/mol = 0.82 moles \] \[\frac{ 0.82*3*1.86 }{24 } = 0.191 kg \]

  24. Photon336
    • one year ago
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    one question though that Kf shouldn't that be negative as well?

  25. taramgrant0543664
    • one year ago
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    I got 0.019 Your moles @Photon336 I think you're missing a 0 in there

  26. Photon336
    • one year ago
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    0.019 you're right

  27. taramgrant0543664
    • one year ago
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    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

  28. taramgrant0543664
    • one year ago
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    It would have been stated if it was negative but water always has a Kf that is 1.86

  29. Lena772
    • one year ago
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    kg to g would be 19.1 though right?

  30. Photon336
    • one year ago
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    yep

  31. Lena772
    • one year ago
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    It says I have a rounding error of less than 10%. Should I try 19.0 or 19.2?

  32. taramgrant0543664
    • one year ago
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    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

  33. Lena772
    • one year ago
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    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

  34. Lena772
    • one year ago
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    18.4 was correct! Hooray :)

  35. taramgrant0543664
    • one year ago
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    Ya that could be a problem, I never went back and checked the conversion but that could effect it for sure

  36. taramgrant0543664
    • one year ago
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    Yay!!!!

  37. Lena772
    • one year ago
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    Thank you both !

  38. taramgrant0543664
    • one year ago
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    No problem!

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