Question

In: Chemistry

pre-lab 1. Name at least three common substances applied to road surfaces to prevent or reduce...

pre-lab

1. Name at least three common substances applied to road surfaces to prevent or reduce freezing during cold weather. Cite the sources for your answer and justify their reliability.

2. Discuss advantages and disadvantages for each of the substances you reported in question 1 in relation to environmental implications. Cite the sources for your answer and justify their reliability.

3. Compare and contrast ionic and covalent substances as they relate to freezing point depression of aqueous solutions and the extent of the depression. Cite the sources for your answer and justify their reliability.

4. What is the freezing point of a water solution containing 0.45 m solute? 5. If a water solution had a freezing point of –1.50 oC, what is the molality of the solution?

Expert Solution

1. Name at least three common substances applied to road surfaces to prevent or reduce freezing during cold weather.

we use --> CaCl2, calcium chloride, NaCl --> Sodum chloride, and even KCl, Potassium chlorde

2. Discuss advantages and disadvantages for each of the substances you reported in question 1 in relation to environmental implications.

CaCl2 --> has the most ions per mol, i.e. Ca+2 and 2Cl- ions, will depress it the most, and it is readily achievable,m since it is a product of the chlor-alkali industr ( byproduct)

NaCl, KCl --> those are expensive, will not depress enough point han CaCl2

3. Compare and contrast ionic and covalent substances as they relate to freezing point depression of aqueous solutions and the extent of the depression.

ionic species = will form ions in solution

covalent --> will not likely form ions, remain molecular

depression points for ionic species --> larger depressions are achieved

covalent --> not likely to form large depressions

4. What is the freezing point of a water solution containing 0.45 m solute?

for water:

dTf = -Kf*m

assume i = 1, since no data is given

dTf = -1.86*0.45 = -0.837°c

5. If a water solution had a freezing point of –1.50 oC, what is the molality of the solution?

dTf = -Kf*m

m = dTf/kf = 1.5/1.86 = 0.806

queen_honey_blossom answered 2 years ago

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