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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

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

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

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

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

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

6. Lena772

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

7. taramgrant0543664

How did you find that temperature?

8. 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.

9. taramgrant0543664

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

10. Lena772

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

11. 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

12. 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

13. taramgrant0543664

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

14. Lena772

@Photon would it be 24.0 g?

15. Lena772

Cause that equals 19.1 @taramgrant0543664 .

16. taramgrant0543664

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

17. 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

18. 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

19. Lena772

Yes

20. 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

21. 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.

22. Lena772

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

23. Photon336

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

24. Photon336

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

25. taramgrant0543664

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

26. Photon336

0.019 you're right

27. 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

28. taramgrant0543664

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

29. Lena772

kg to g would be 19.1 though right?

30. Photon336

yep

31. Lena772

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

32. 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

33. 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

34. Lena772

18.4 was correct! Hooray :)

35. taramgrant0543664

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

36. taramgrant0543664

Yay!!!!

37. Lena772

Thank you both !

38. taramgrant0543664

No problem!

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