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You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

Short Answer

Expert verified
In short, adding a solute to a solvent causes boiling point elevation and freezing point depression due to colligative properties. The solute particles interfere with the solvent particles' ability to escape into vapor, requiring more energy to boil the solution. Similarly, the solute disrupts the ordered arrangement of solvent particles during freezing, necessitating more energy to be removed for the solution to freeze. Your friend's explanation was partially correct but misunderstood the reason behind the boiling point elevation.

Step by step solution

01

Introduce Colligative Properties

Colligative properties are properties of solutions that depend on the ratio of the solute particles to the solvent particles. These properties include boiling point elevation and freezing point depression. It is important to note that these properties are independent of the nature of the solute, meaning that it doesn't matter if it's salt, sugar, etc.
02

Boiling Point Elevation

When a solute is added to the solvent, it increases the boiling point of the solution. This is because the solute particles interfere with the solvent particles' ability to escape into the vapor phase. Due to the solute particles presence, there are fewer solvent particles that can easily escape. Consequently, more energy (higher temperature) is needed for the solvent particles to overcome the attractive forces between them and become vapor. This increases the boiling point of the solution compared to the pure solvent.
03

Freezing Point Depression

Similarly, adding a solute to the solvent can decrease the freezing point of the solution. When a solute is added to a solvent, it causes disorder between the solvent particles. When the solution freezes, the solute particles get in the way of the solvent particles and disrupt the perfect arrangement of particles needed to form a solid. Therefore, more energy needs to be removed from the system (lower temperature) to overcome the disorder caused by the solute and allow the solvent particles to arrange themselves into a solid (frozen) state.
04

Address Your Friend's Explanation

Your friend's explanation is partially correct. It's true that the solute particles, like salt, get in the way of the water molecules forming a solid (freezing point depression); however, the part about the salt acting like a strong bond holding the water molecules together is not accurate. Instead, the salt (or other solutes) interferes with the water molecules ability to escape the solution, meaning that more energy is needed to boil the solution (boiling point elevation).

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Most popular questions from this chapter

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}_{3} \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \(\left[143\right.\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.)

You have a solution of two volatile liquids, \(\mathrm{A}\) and \(\mathrm{B}\) (assume ideal behavior). Pure liquid A has a vapor pressure of \(350.0\) torr and pure liquid \(\mathrm{B}\) has a vapor pressure of \(100.0\) torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. What is the mole fraction of liquid \(\mathrm{A}\) in the solution?

The solubility of nitrogen in water is \(8.21 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) when the \(\mathrm{N}_{2}\) pressure above water is \(0.790 \mathrm{~atm} .\) Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of \(\mathrm{mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0{ }^{\circ} \mathrm{C}\).

A solution of sodium chloride in water has a vapor pressure of \(19.6\) torr at \(25^{\circ} \mathrm{C}\). What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is \(23.8\) torr at \(25^{\circ} \mathrm{C}\) and \(71.9\) torr at \(45^{\circ} \mathrm{C}\) and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of \(31.57 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). The molar mass is determined by measuring the freezing- point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving \(10.56 \mathrm{~g}\) of the compound in \(25.0 \mathrm{~g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

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