Finding Optimal Water and Ammonium Nitrate Ratio

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danago
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Hi. Recently, we conducted an experiment where we were supposed to determine the optimal 'water:ammonium nitrate crystal' ratio to keep the water at a minimum temperature for a maximum length of time.

What we did was take different samples of ammonium nitrate (1g, 2g, 3g...6g) and dissolve what we could into 5mL of water. We then recorded how long each solution remained under 1 degree celsius.

Before the experiment, it would have been obvious to me that the best results would have been given by the 6g sample (enough for a saturated solution). However, our best result was given by 4g, but i can't understand why. Was it just experimental error? Or will excess ammonium nitrate cause less heat to be absorbed?

All help greatly appreciated,
Thanks,
Dan.
 
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When things don't go as you expect, allways ask why you expected them to go differently. "Did i make a false assumption or over simplify?" "Did i overlook some factor?" or "Did i just screw up my 2Hr lab session by using the wrong materials?"

What makes you think saturation would be acheived at 6g?

Would 100g be any better than 2g?

Was there any undisolved solute left over at the bottom of your sample and at what stage did you notice this?
 
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I thought saturation would be achieved at 6g, because i did some research, and apparently the solubility of ammonium nitrate in water is ~119g/100mL at 0 degrees. I had 5mL of water, therefore, ~6g should have dissolved. However, at zero degrees, even at higher temperatures, not all of it would dissolve.
 
Can you tell us how the experiment was performed? How did you ensure the starting temperature was 0C (if it was), and how did you measure temperatures subsequently (and what instrument did you use)? Also, at which weight did you reach saturation?

PS: Have you looked at "enthalpy of (dis)solution" tables before?
 
Well even if your amonium Nitrate wasnt dried it should dissolve?

As a rough calculation:

Amonimum Nitrate: [tex]NH_{4}NO_{3}[/tex] is 80g/mol or [tex]NH_{4}NO_{3}.3H_{2}O[/tex] is 134g/mol

If you take 6g of the hydrated form, you will actually only have:

[tex]\frac{6g}{134g mol^{-1}}=0.045mol[/tex]

[tex]0.045mol \times 80g mol^{-1} = 3.6g[/tex]

So it isn't that...
 
Basically, what we did was take different masses of ammonium nitrate. Each quantity was dissolved into 5mL of water, at approximately 13 degrees. We then recorded the minimum temperature the solution reached, and also recorded how long the solution remained under 1 degree.
 
Okay, I missed the mention of ammonium nitrate. Nevertheless, I can't see what the point of this experiment is.

The recording of the minimum temperature reached is more likely to be a useful number than the time to reach 1C (assuming the freezing point is sufficiently depressed that freezing never occurs ... or did it occur?).

But as for your question of why the maximum time did not occur at 6 gms, you might consider that your initial guess did not incorporate the increased heat capacity of the solution from the increasing mass of ammonium nitrate in it. Moreover, the enthalpy of solution comes only from that portion of the solute that does dissolve. For whatever reason, if some of the solute did not dissolve, that solute adds heat capacity without adding heat removal (and of course [itex]Q = C' \Delta T[/itex], where C' is the effective heat capacity. So, increasing the mass of nitrate beyond saturation clearly increases C' without increasing Q, and hence results in decreasing [itex]\Delta T[/itex].
 
Ahhh i never thought of that gokul. Thanks :)