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List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m}\) \(\mathrm{KBr}, 0.030 \mathrm{~m}\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

Short Answer

Expert verified
The order of decreasing freezing points for the given aqueous solutions is Phenol > Glycerin = KBr.

Step by step solution

01

1. Recall the formula for freezing point depression.

The formula for freezing point depression (\(\Delta T_{f}\)) is given by: \[\Delta T_{f} = K_f \cdot i \cdot m \] Where \(K_f\) is a constant called the cryoscopic constant (which is the same for all solutes in a given solvent), \(i\) is the van't Hoff factor, and \(m\) is the molality of the solute in the solution.
02

2. Determine the van't Hoff factor for each solute.

The van't Hoff factor (\(i\)) is the number of dissolved particles produced from a single molecule of solute. For non-electrolytes like glycerin and phenol, the van't Hoff factor is 1. For electrolytes like potassium bromide, \(i\) has to be calculated based on the number of ions produced during the dissociation. In this case, for KBr: \[ \mathrm{KBr} \rightarrow \mathrm{K}^{+} + \mathrm{Br}^{-} \] Thus, the van't Hoff factor for KBr is 2.
03

3. Calculate the freezing point depression for each solution.

Use the calculated van't Hoff factors and the given molalities to find the freezing point depressions: \[\Delta T_{f \, glycerin} = K_f \cdot 1 \cdot 0.040 \] \[\Delta T_{f \, KBr} = K_f \cdot 2 \cdot 0.020 \] \[\Delta T_{f \, phenol} = K_f \cdot 1 \cdot 0.030 \] As \(K_f\) is common for all solutes in the same solvent, we directly compare the numerical values: \[\Delta T_{f \, glycerin} : \Delta T_{f \, KBr} : \Delta T_{f \, phenol} = (1 \cdot 0.040) : (2 \cdot 0.020) : (1 \cdot 0.030) = 0.040 : 0.040 : 0.030\]
04

4. List the solutions in order of decreasing freezing point.

Since the freezing point is inversely proportional to freezing point depression, the solution with the lowest freezing point depression will have the highest freezing point. From step 3, we find the following order of freezing points; Phenol > Glycerin = KBr So, the order of decreasing freezing points for these aqueous solutions is Phenol > Glycerin = KBr.

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Most popular questions from this chapter

List four properties of a solution that depend on the total concentration but not the type of particle or particles present as solute. Write the mathematical expression that describes how each of these properties depends on concentration.

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The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is \(208 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(15^{\circ} \mathrm{C}\). A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving \(324 \mathrm{~g}\) in \(100 \mathrm{~g}\) water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C}\), no precipitate forms. (a) What term describes this solution? (b) What action might you take to initiate crystallization? Use molecular- level processes to explain how your suggested procedure works.

Fish need at least 4 ppm dissolved \(\mathrm{O}_{2}\) for survival. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above the water is needed to obtain this concentration at \(10{ }^{\circ} \mathrm{C}\) ? (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(\left.1.71 \times 10^{-3} \mathrm{~mol} / \mathrm{L}-\mathrm{atm} .\right)\)

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