Problem 68 - General Chemistry: Principles and Modern Applications

Question

For a solution containing \(6.38 \%\) para-diclorobenzene by mass in benzene, the density of the solution as a function of temperature ( \(t\) ) in the temperature range 15 to \(65^{\circ} \mathrm{C}\) is given by the equation \(d(\mathrm{g} / \mathrm{mL})=1.5794-1.836 \times 10^{-3}(t-15)\) At what temperature will the solution have a density of \(1.543 \mathrm{g} / \mathrm{mL} ?\)

Step-by-Step Solution

Answer

The solution will have a density of 1.543 g/mL at a temperature of \(35^{\circ} C\).

Step 1: Analyzing the given function

The function for density provided in the problem is \(d(\mathrm{g} / \mathrm{mL})=1.5794-1.836 \times 10^{-3}(t-15)\). This function describes the relationship between the temperature (\(t\)) and the density of solution (\(d)\). The target is to find the temperature when density is \(1.543 g/mL\).

Step 2: Solve the equation for \(t\)

Substitute \(d\) in the equation with \(1.543 g/mL\) to find the temperature at this density: \(1.543 = 1.5794 - 1.836 \times 10^{-3}(t-15)\).

Step 3: Isolate the term with \(t\)

The next step is to isolate the term with the variable \(t\). To do that, subtract \(1.5794\) from both sides of the equation. Now the equation looks like this: \(-0.0364 = - 1.836 \times 10^{-3}(t-15)\).

Step 4: Simplify the equation

Further simplify the equation to find \(t\). By multiplying both sides of the equation by \(-1\), we get \(0.0364 = 1.836 \times 10^{-3}(t-15)\). Divide both sides of the equation by \(1.836 \times 10^{-3}\) to solve for \(t - 15\). The result is \(t - 15 = 0.0364 / 1.836 \times 10^{-3}\). Calculate the result to get \(t - 15 = 19.8255\).

Step 5: Solve for \(t\)

Finally, solve for \(t\) by adding 15 to both sides of the equation. So, \(t = 15 + 19.8255\). Calculate the result to get the final temperature of \(t = 34.8255 ^{\circ} C\). The solution should be rounded to the nearest integer.

Step 6: Round the final result

Round to the nearest integer. So, the final temperature is \(t = 35 ^{\circ} C\).

Key Concepts

Understanding the Temperature-Density Relationship

The temperature-density relationship of a solution is critical in solution chemistry. It illustrates how changes in temperature can directly influence the density of a solution. Density, which is mass per unit volume, typically decreases as temperature increases in many substances. This happens because the particles in a substance move more vigorously and tend to occupy more space when heated, reducing the density. In the given exercise, the equation provided, \( d(g/mL) = 1.5794 - 1.836 \times 10^{-3}(t-15) \), describes how temperature \( (t) \) affects the density \( (d) \) of a para-diclorobenzene solution in benzene. Here, the coefficient \( -1.836 \times 10^{-3} \) indicates the rate of decrease in density per degree Celsius increase in temperature. Understanding such relationships can help predict a solution's behavior under different thermal conditions. This knowledge is particularly useful in fields like material sciences and chemical engineering, where the stability and flow of materials are essential.

Exploring Solution Chemistry

Solution chemistry is a broad area that investigates the characteristics and behaviors of solutes when mixed into solvents to form solutions. A solution consists of a solute (such as para-diclorobenzene in our example) that is dissolved evenly within a solvent (like benzene). - **Solubility**: The solubility of a substance is key to forming a stable solution. It depends on various factors, including temperature, pressure, and the nature of the solvent and solute. - **Concentration**: This refers to the measure of how much solute is present in a given amount of solvent or solution. In our case, the solution has a mass concentration of **6.38%** para-diclorobenzene by mass. It means that every 100 grams of solution contains 6.38 grams of para-diclorobenzene. These factors affect not just how well solutes dissolve, but also the resulting physical properties of the solutions, like their boiling and freezing points, and densities. Solution chemistry is vital in creating products such as pharmaceuticals, food products, and more.

Understanding Mass Percentage

Mass percentage is a way of expressing the concentration of a component in a mixture or a solution. It tells us how much of the total mass of the solution is made up by a specific component. To calculate mass percentage, you use the formula:\[\text{Mass Percentage} = \frac{\text{mass of component}}{\text{total mass of solution}} \times 100\%\]In the context of our exercise, a **6.38%** mass percentage of para-diclorobenzene implies that in 100 grams of solution, there are 6.38 grams of para-diclorobenzene. Knowing the mass percentage is crucial when preparing solutions for experiments in the laboratory, as it determines the reactivity and behavior of a solution. Using this parameter also aids in quality control in various industries, ensuring that products meet specific concentration requirements.