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Chapter 11: Problem 11

If a solution shows positive deviations from Raoult’s law, would you expect the solution to have a higher or lower boiling point than if it were ideal? Explain.

Short Answer

Expert verified
If a solution shows positive deviations from Raoult's Law, it would have a lower boiling point than an ideal solution. This is because the intermolecular forces between the components in the mixture are weaker than those in pure components, leading to a higher observed vapor pressure and lower temperature required to reach the boiling point.

Step by step solution

01

Understand Raoult's Law

Raoult's Law states that for a mixture of two volatile components, the partial vapor pressure of each component in the mixture is proportional to its mole fraction. In simpler terms, it describes the relationship between the vapor pressure of the individual components and their concentrations in the solution. An ideal solution follows Raoult's Law. In such a solution, the intermolecular forces between the different components are similar to those within the pure components, and there is no significant change in the energy or volume when the components are mixed.
02

Understand Positive Deviations from Raoult's Law

Positive deviations from Raoult's Law occur when the observed vapor pressure of a solution is greater than what would be predicted by the law. This can happen when the intermolecular forces between the two components in a mixture are weaker than the forces within the pure components. As a result, the molecules in the solution can escape more easily into the vapor phase, leading to a higher observed vapor pressure.
03

Compare Boiling Points of Ideal and Non-Ideal Solutions

The boiling point of a solution is defined as the temperature at which its vapor pressure equals the atmospheric pressure. Since the boiling point is directly related to vapor pressure, we can compare the boiling points of an ideal (follows Raoult's Law) and a non-ideal solution (with positive deviation) based on their vapor pressures. For a solution showing positive deviations from Raoult's Law, the vapor pressure is higher than predicted. This means that the boiling point of such a solution would be lower compared to an ideal solution, as it requires less heat to achieve the same vapor pressure as the ideal solution.
04

Conclusion

If a solution shows positive deviations from Raoult's Law, we can expect it to have a lower boiling point than if it were an ideal solution. This is because the weaker intermolecular forces in the mixture lead to a higher observed vapor pressure, and thus a lower temperature is needed to reach the boiling point.

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Most popular questions from this chapter

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{lll}{\mathrm{CaCl}_{2}(s)} & {-2247 \mathrm{k} / \mathrm{mol}} & {-46 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Cal}_{2}(s)} & {-2059 \mathrm{k} / \mathrm{mol}} & {-104 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-},\) is more strongly attracted to water?

A 2.00 -g sample of a large biomolecule was dissolved in 15.0 \(\mathrm{g}\) carbon tetrachloride. The boiling point of this solution was determined to be \(77.85^{\circ} \mathrm{C}\) . Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is $5.03^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol},$ and the boiling point of pure carbon tetrachloride is \(76.50^{\circ} \mathrm{C} .\)

An aqueous solution is 1.00\(\% \mathrm{NaCl}\) by mass and has a density of 1.071 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) . The observed osmotic pressure of this solution is 7.83 atm at \(25^{\circ} \mathrm{C}\) . a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution.

Rubbing alcohol contains 585 g isopropanol $\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)$ per liter (aqueous solution). Calculate the molarity.
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