Considerations freezing point depression
- Page ID
- 424503
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consideration \(\PageIndex{1}\)
Define Freezing point
- Answer
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The temperature at a given pressure in an isolated system for which the solid and liquid phase can co-exist in equilibrium.
Consideration \(\PageIndex{2}\)
Relate the temperature change to the heat transfer between ice and the solution?
- Answer
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The solution cools as the ice melts. It takes energy to break the hydrogen bonds in the ice (ice has more hydrogen bonds than liquid water).
Consideration \(\PageIndex{3}\)
Why can't we apply the first law and say the heat lost by the ice equals the heat gained by the water?
- Answer
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We do not have an energetically isolated system and need to take into account heat transfer to the room. As both the room and the water are above the temperature of the ice, it will all melt and eventually rise to room temperature, which is being maintained by the heating and cooling system of the building
Consideration \(\PageIndex{4}\)
How could we improve our results and avoid dealing with heat transfer from the room?
- Answer
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By running the experiment in a calorimeter, like we did with the enthalpy of neutralization experiment in general chemistry 1.
Consideration \(\PageIndex{5}\)
We are making the assumption that the once the temperature stabilizes, it represents the freezing point. This implies that the rate the ice melts equals the rate heat is transferred to the room and water, and as long as there is ice, it will maintain that temperature. Yet according to the freezing point depression equation, the freezing point will rise during the experiment, why?
- Answer
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As ice melts it turns to liquid water and the mass of liquid water goes up. As \(\Delta T\) is anti-proportional to mass of solvent, there is less of a freezing point depression
Consideration \(\PageIndex{6}\)
Why do we use the temperature when the last piece of ice melts, even if it is not the lowest temperature?
- Answer
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As only then do we know the mass of liquid water, which is the total mass of water and ice originally added, and so only at that point can we calculate the van't Hoff factor.
Consideration \(\PageIndex{7}\)
What could be the mistake if the temperature never flattens out (i.e. temperature drops continuously until the ice is all melted)?
- Answer
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There was not enough ice to cool the system to the freezing point
Consideration \(\PageIndex{8}\)
Why might the temperature hit a bottom and start to rise, and why would we not want to use the lowest temperature?
- Answer
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If the temperature hits a bottom and starts to rise it means we have hit the freezing point, but as ice melts the solution becomes more dilute and the freezing point depression is reduced. So even if the lowest temperature did represent the freezing point at the concentration, we do not know that concentration, as we do not know how much ice remains unmelted. We could remove the ice with a sieve at that point in order to measure the mass of water, though.
Consideration \(\PageIndex{9}\)
What would you expect to happen to the temperature once the last piece of ice melts?
- Answer
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The temperature will now rise to room temperature
So there are a lot of flaws in this experimental design, but these flaws can give us a bearing on the nature of running an experiment.