Question

A chemist mixes two liquids \(\mathrm{A}\) and \(\mathrm{B}\) to form a homogeneous mixture. The densities of the liquids are \(2.0514 \mathrm{~g} / \mathrm{mL}\) for \(\mathrm{A}\) and \(2.6678 \mathrm{~g} / \mathrm{mL}\) for \(\mathrm{B}\). When she drops a small object into the mixture, she finds that the object becomes suspended in the liquid; that is, it neither sinks nor floats. If the mixture is made of 41.37 percent \(\mathrm{A}\) and 58.63 percent \(\mathrm{B}\) by volume, what is the density of the object? Can this procedure be used in general to determine the densities of solids? What assumptions must be made in applying this method?

Short answer

The object's density is 2.4133 g/mL. Yes, it's generally usable with assumptions of ideal mixing and no interactions.

Step-by-step solution

Understanding Density of Mixtures

When two liquids are mixed, the density of the resulting mixture is the weighted average of the densities of the two liquids, based on their volume proportions in the mixture.

Formula for Mixture Density

The formula to calculate the density of the mixture is:\[\rho_m = \left( \frac{V_A \cdot \rho_A + V_B \cdot \rho_B}{V_A + V_B} \right)\]where \( \rho_m \) is the density of the mixture, \( V_A \) and \( V_B \) are the volumes, and \( \rho_A \) and \( \rho_B \) are the densities of liquids A and B respectively.

Converting Percentage to Volume Fraction

Convert the given percentages into a fraction of total volume. For liquid A, it is 41.37% or 0.4137, and for liquid B, it is 58.63% or 0.5863. Assume a total volume, say 1 mL, for simplicity: \( V_A = 0.4137 \) mL, \( V_B = 0.5863 \) mL.

Calculate the Density of the Mixture

Substitute the known values into the formula for mixture density:\[\rho_m = \left( \frac{0.4137 \times 2.0514 + 0.5863 \times 2.6678}{1} \right)\]Calculate each term:- \(0.4137 \times 2.0514 = 0.8489\)- \(0.5863 \times 2.6678 = 1.5644\)Add these:\(\rho_m = 0.8489 + 1.5644 = 2.4133 \text{ g/mL}\)This is the density of the mixture.

Making Conclusions About the Object

Since the object is suspended in the mixture, it has the same density as the mixture. Thus, the density of the object is \(2.4133 \text{ g/mL}\).

General Use of the Procedure and Assumptions

This method of using a suspension in a mixture can be used generally to find the density of objects as long as the mixture is homogeneous. The method assumes no chemical interaction between the liquids and the object, and that the liquids mix ideally following the volume percentages precisely.

Key concepts

Weighted Average Density

When dealing with mixtures, particularly those involving liquids, calculating the density requires using the concept of weighted average density. This concept is essential because it accounts for both the density and volume fraction of each component in the mixture. The mixture's density is not merely an average, but rather it is calculated by multiplying each liquid's density by its respective volume fraction, and then summing these products. This is expressed mathematically as:\[\rho_m = \frac{V_A \cdot \rho_A + V_B \cdot \rho_B}{V_A + V_B}\]

• \( \rho_m \): Density of the mixture.
• \( V_A \) and \( V_B \): Volumes of liquids A and B.
• \( \rho_A \) and \( \rho_B \): Densities of liquids A and B.

This approach ensures that the contribution of each liquid to the overall density is proportional to its amount in the total mixture.

Volume Percentage

The concept of volume percentage is integral when calculating mixture densities, as it determines how much of each component is present. In our exercise, liquid A makes up 41.37%, and liquid B makes up 58.63%. To utilize these percentages in density calculations, you must convert them into volume fractions by dividing by 100. As such:

• The volume fraction for liquid A becomes 0.4137 (equal to 41.37%).
• The volume fraction for liquid B becomes 0.5863 (equal to 58.63%).

Assume a total volume to simplify calculations; for example, assuming 1 mL total volume, the volume of liquid A is 0.4137 mL and for liquid B it is 0.5863 mL. This technique makes it easier to plug values into the weighted density formula.

Suspension in Liquid

Understanding suspension in a liquid is crucial for density determination. When an object becomes suspended in a liquid mixture, it neither sinks to the bottom nor rises to the top. This state indicates that the object's density is equal to the density of the mixture. In our context, the object is suspended in a mixture whose density we've calculated to be 2.4133 g/mL. Thus, the object's density is the same as the mixture.
Using this method, once you achieve suspension, you can conclude that the object's density matches that of the mixed liquid, making it a useful technique for measuring densities if one can create a homogeneous mixture that balances the object's density.

Assumptions in Density Calculation

To effectively use the method described, certain assumptions must be made. These assumptions ensure accuracy and reliability when determining densities through suspension:

• Homogeneity: The mixture needs to be thoroughly mixed with no phase separation to ensure uniform density throughout.
• No Chemical Reactions: It’s assumed there is no chemical interaction affecting the densities of the liquids or the object involved. Such interactions can alter the results by changing volume or density.
• Ideal Mixture: The liquids must ideally mix according to volume percentages without deviation.

These conditions help preserve the integrity of the calculation and allow the suspended object's density to match the mixture accurately.