Problem 30 - Chemistry: The Central Science

Question

(a) A cube of osmium metal \(1.500 \mathrm{~cm}\) on a side has a mass of \(76.31 \mathrm{~g}\) at \(25^{\circ} \mathrm{C}\). What is its density in \(\mathrm{g} / \mathrm{cm}^{3}\) at this temperature? (b) The density of titanium metal is \(4.51 \mathrm{~g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\). What mass of titanium displaces \(125.0 \mathrm{~mL}\) of water at \(25^{\circ} \mathrm{C}\) ? (c) The density of benzene at \(15^{\circ} \mathrm{C}\) is \(0.8787 \mathrm{~g} / \mathrm{mL}\). Calculate the mass of \(0.1500 \mathrm{~L}\) of benzene at this temperature.

Step-by-Step Solution

Answer

(a) The density of osmium metal at 25°C is \(22.6 \frac{\mathrm{g}}{\mathrm{cm}^3}\). (b) The mass of titanium that displaces 125.0 mL of water at 25°C is 564 g. (c) The mass of 0.1500 L of benzene at 15°C is 132 g.

Step 1: Problem (a)

We're given the size of the cube (1.500 cm on a side) and its mass (76.31 g) and asked to find its density. First, we need to find the volume of the cube. Recall that the volume of a cube (V) is calculated using the formula: V = side^3. \(V = (1.500 \mathrm{~cm})^3 = 3.375 \mathrm{~cm}^3 \) Next, use the formula for density: Density = Mass / Volume. \(\rho = \frac{Mass}{Volume} = \frac{76.31 \mathrm{~g}}{3.375 \mathrm{~cm}^3} = 22.6 \frac{\mathrm{g}}{\mathrm{cm}^3} \) So, the density of osmium metal at 25°C is 22.6 g/cm³.

Step 2: Problem (b)

The density of titanium metal is given as 4.51 g/cm³, and we need to find the mass of titanium that displaces 125.0 mL of water. First, convert the volume of water to cm³ (1 mL = 1 cm³). Volume of water = 125.0 mL = 125.0 cm³ Now, use the formula for density: Density = Mass / Volume. We need to find the mass, so we can rearrange the formula as follows: Mass = Density * Volume. \(Mass = \rho * Volume = 4.51 \frac{\mathrm{g}}{\mathrm{cm}^3} * 125.0 \mathrm{~cm}^3 = 564 \mathrm{~g} \) So, the mass of titanium that displaces 125.0 mL of water at 25°C is 564 g.

Step 3: Problem (c)

The density of benzene at 15°C is 0.8787 g/mL. We need to find the mass of 0.1500 L of benzene at this temperature. First, convert the volume of benzene to mL (1 L = 1000 mL). Volume of benzene = 0.1500 L = 150.0 mL Now, use the formula for density: Density = Mass / Volume. Rearrange the formula to find the mass: Mass = Density * Volume. \(Mass = \rho * Volume = 0.8787 \frac{\mathrm{g}}{\mathrm{mL}} * 150.0 \mathrm{~mL} = 132 \mathrm{~g} \) So, the mass of 0.1500 L of benzene at 15°C is 132 g.

Key Concepts

Understanding the Mass-Volume Relationship

The mass-volume relationship is a fundamental concept in understanding the physical properties of materials. It simply refers to how much space (volume) a certain amount of substance (mass) occupies. This is crucial when determining the density of a substance, which is a measure of how tightly matter is packed together. In the context of the textbook exercise, for the osmium metal cube, we know both the mass (76.31 g) and the volume, which is calculated from the cube's dimensions (1.500 cm on each side).

The volume calculation involves cubing the side of the cube, leading to a volume of 3.375 cm³. The relationship between mass and volume is expressed through the density formula, which is essentially the ratio of mass to volume. The cube's mass and volume are directly used in the formula to compute its density.

The Density Formula Explained

Density, symbolically represented as \( \rho \), is a characteristic property of a material. The density formula, \( \rho = \frac{Mass}{Volume} \), relates to how much mass is contained in a given volume. This formula is essential for solving the problems in our textbook exercise. For instance, with the osmium metal cube, after finding the volume, we apply the density formula to find the density at a specific temperature, 25°C.

In the context of problem (b) and (c), we know the density of the materials and are asked to find the mass corresponding to a certain volume. By rearranging the density formula, we can solve for mass as \( Mass = \rho * Volume \). Knowing how to manipulate the density formula allows us to solve a variety of practical problems, such as determining the mass of a material that would displace a certain volume of liquid, as seen with the titanium example.

Conversion of Units for Accurate Calculations

Converting units is a vital step in many scientific calculations. This step ensures that all aspects of a calculation are expressed in compatible units, leading to accurate and meaningful results. For density calculations, it's common to convert volumes between mL (milliliters) and cm³ (cubic centimeters) since 1 mL is equivalent to 1 cm³.

In problem (b) of the exercise, the volume of water (125.0 mL) is converted to cm³ to align with the given density units. Similarly, in problem (c), the volume of benzene (0.1500 L) is converted to mL before applying the density formula. Remember to convert liters to milliliters by multiplying by 1000, as 1 L equals 1000 mL. Mastering unit conversions is crucial when working with various measurements, avoiding errors, and ensuring that your results correctly reflect the physical quantities being measured.