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Chapter 11: Problem 6

You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\). Explain what happens and why.

Short Answer

Expert verified
When an ice cube made of pure water is dropped into a saltwater solution at \(0^{\circ}\mathrm{C}\), it begins to melt due to the surrounding saltwater solution having a lower freezing point than pure water. This is because dissolved salt ions interfere with the formation of the crystalline ice structure, making it harder for the water molecules to freeze. As the ice cube melts, it dilutes the saltwater solution, causing its freezing point to rise closer to that of pure water. Eventually, thermal equilibrium is reached and the ice cube may dissolve completely, unless the ice cube is too large or the saltwater solution is already at its maximum freezing point depression.

Step by step solution

01

Understanding the Components

Before we proceed, let's understand the two components involved in this scenario. First, we have an ice cube made of pure water, meaning it is frozen water without any contaminants. Second, we have a saltwater solution at 0 degrees Celsius. This means the solution has salt (typically NaCl) dissolved in water, and it is at the freezing point of pure water.
02

Melting Point Depression

In a saltwater solution, the presence of salt particles interferes with the formation of solid water structures and lowers the freezing point compared to that of pure water. This phenomenon is known as "melting point depression" or "freezing point depression". When salt is dissolved in water, the freezing point of the solution decreases because the dissolved salt ions interfere with the ability of the water molecules to form crystalline ice structures, making it harder for the water to freeze.
03

What Happens to the Ice Cube

When the ice cube is dropped into the saltwater solution, the surrounding liquid is at 0 degrees Celsius, which is the freezing point of pure water but not the freezing point of the saltwater solution. As a result, the ice cube begins to melt since it is now in contact with a solution that is warmer than its own freezing point.
04

What Happens to the Saltwater Solution

As the ice cube melts, the water from the ice cube will mix with the saltwater solution. This addition of pure water will dilute the saltwater solution slightly, causing its freezing point to rise closer to that of pure water. However, as long as the salt concentration remains higher than zero, the freezing point of the saltwater solution will still be lower than 0 degrees Celsius.
05

Equilibrium and Final Scenario

The melting of the ice cube will continue until thermal equilibrium is reached between the ice and the saltwater solution. At this point, the temperature of the saltwater solution may drop slightly due to the melting of the ice and the dilution of the salt concentration. In most cases, the ice cube will eventually dissolve completely in the saltwater solution, unless the ice cube is too large or the saltwater solution is already at its maximum freezing point depression.

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Most popular questions from this chapter

Erythrocytes are red blood cells containing hemoglobin. In a saline solution they shrivel when the salt concentration is high and swell when the salt concentration is low. In a \(25^{\circ} \mathrm{C}\) aqueous solution of \(\mathrm{NaCl}\), whose freezing point is \(-0.406^{\circ} \mathrm{C}\), erythrocytes neither swell nor shrink. If we want to calculate the osmotic pressure of the solution inside the erythrocytes under these conditions, what do we need to assume? Why? Estimate how good (or poor) of an assumption this is. Make this assumption and calculate the osmotic pressure of the solution inside the erythrocytes.

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

What mass of sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is needed to prepare \(0.250 \mathrm{~L}\) of a \(0.100 M\) solution?

A solution is prepared by mixing \(25 \mathrm{~mL}\) pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}, d=\right.\) \(\left.0.63 \mathrm{~g} / \mathrm{cm}^{3}\right)\) with \(45 \mathrm{~mL}\) hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}, d=0.66 \mathrm{~g} / \mathrm{cm}^{3}\right)\). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.

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?
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