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Chapter 12: Problem 72

Arrange the following aqueous solutions in order of decreasing freezing point, and explain your reasoning: \(0.50 \mathrm{~m} \mathrm{HCl}, 0.50 \mathrm{~m}\) glucose, \(0.50 \mathrm{~m}\) acetic acid.

Short Answer

Expert verified
The order of solutions, from highest to lowest freezing point, is 0.50m Glucose = 0.50m Acetic Acid > 0.50m HCl.

Step by step solution

01

Analyze the number of particles in each solution

First, identify how many disjoint ions, or particles, are produced when each of the given substances dissolve in their solutions. HCl is a strong acid and completely dissociates into two ions (\(H^+\) and \(Cl^-\)). Glucose is a covalent compound and does not dissociate at all, contributing only 1 particle to the solution. Acetic acid is a weak acid and only slightly ionizes, so it can be considered as primarily contributing 1 particle to the solution.
02

Order the freezing points comparing the number of particles

The more particles a solute contributes, the lower the freezing point of the solution. From step 1, the order of number of particles per amount of solute is HCl > Glucose = Acetic Acid. Therefore, the order of the freezing points of the solutions is Glucose = Acetic Acid > HCl.

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Most popular questions from this chapter

The density of an aqueous solution containing \(10 . \overline{0}\) percent of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) by mass is \(0.984 \mathrm{~g} / \mathrm{mL} .\) (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.125 mole of ethanol?

The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) present. Calculate the number of grams of alcohol present in \(1.00 \mathrm{~L}\) of 75 -proof gin. The density of ethanol is \(0.798 \mathrm{~g} / \mathrm{mL}\)

The molar mass of benzoic acid ( \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) ) determined by measuring the freezing-point depression in benzene is twice what we would expect for the molecular formula, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{2}\). Explain this apparent anomaly.

Explain the variations in solubility in water of the alcohols listed here: $$\begin{array}{lc}\hline & \text { Solubility in Water } \\\\\text { Compound } & (\mathrm{g} / 100 \mathrm{~g}) \text { at } 20^{\circ} \mathrm{C} \\\\\hline \mathrm{CH}_{3} \mathrm{OH} & \infty \\\\\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & \infty \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & \infty \\\\\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 9 \\\\\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 2.7 \\\\\hline\end{array}$$

The Henry's law constant of oxygen in water at \(25^{\circ} \mathrm{C}\) is \(1.3 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm. Calculate the molarity of oxygen in water under 1 atmosphere of air. Comment on the prospect for our survival without hemoglobin molecules. (Recall from previous problems that the total volume of blood in an adult human is about \(5 \mathrm{~L}\).)
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