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Chapter 13: Problem 72

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m} \mathrm{KBr}$,

Short Answer

Expert verified
The order of decreasing freezing point for the given aqueous solutions is: Sodium acetate \((0.010 \mathrm{~m}\)) > Glycerin \((0.040 \mathrm{~m})\) = KBr \((0.020 \mathrm{~m})\).

Step by step solution

01

Calculate the van't Hoff factor for each solute

The van't Hoff factor (i) for glycerin, a non-electrolyte, is 1 because it does not dissociate into ions in solution. For KBr, a strong electrolyte, the van't Hoff factor is 2, as it dissociates into one K+ and one Br- ion. For sodium acetate, a strong electrolyte, the van't Hoff factor is 2, as it dissociates into one Na+ and one \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\) ion.
02

Calculate freezing point depression for each solution

We can use the formula ΔTf = Kf * m * i to find the freezing point depression for each of the solutions. For simplicity, we'll assume Kf is the same for all three solutions since we are only comparing the relative decrease in freezing point. For glycerin: ΔTf = Kf * 0.040 m * 1 = 0.040 Kf For KBr: ΔTf = Kf * 0.020 m * 2 = 0.040 Kf For sodium acetate: ΔTf = Kf * 0.010 m * 2 = 0.020 Kf
03

List the solutions in order of decreasing freezing point

The solution with the highest freezing point depression will have the lowest freezing point. Based on the calculations above, - Sodium acetate has the lowest ΔTf (0.020 Kf), so it has the highest freezing point. - Glycerin and KBr have the same ΔTf (0.040 Kf), so they would have the same freezing point, which is lower than sodium acetate. So, the order of decreasing freezing point is sodium acetate > glycerin = KBr.

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Most popular questions from this chapter

Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a \(0.2 \mathrm{M}\) solution of some solute and is submerged in a 0.1 \(M\) solution of the same solute: Initially, the volume of solution in the balloon is \(0.25 \mathrm{~L}\). Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5]

Indicate the type of solute-solvent interaction (Section 11.2) that should be most important in each of the following solutions: $(\mathbf{a}) \mathrm{CCl}_{4}\( in benzene \)\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),(\mathbf{b})\( methanol \)\left(\mathrm{CH}_{3} \mathrm{OH}\right)\( in water, \)(\mathbf{c}) \mathrm{KBr}$ in water, (d) HCl in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right).\)

Indicate whether each statement is true or false: (a) A solute will dissolve in a solvent if solute-solute interactions are stronger than solute-solvent interactions. (b) In making a solution, the enthalpy of mixing is always a positive number. (c) An increase in entropy favors mixing.

Aerosols are important components of the atmosphere. Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth's surface, compared to an "aerosol-free" atmosphere? Explain your reasoning.

Indicate whether each statement is true or false: \((\) a) \(\mathrm{NaCl}\) dissolves in water but not in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) because benzene is denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.
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