Question

A graduated cylinder contains \(155 \mathrm{~mL}\) of water. A \(15.0-\mathrm{g}\) piece of iron (density \(\left.=7.86 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and a \(20.0-\mathrm{g}\) piece of lead (density \(\left.=11.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) are added. What is the new water level, in milliliters, in the cylinder?

Short answer

158.68 mL

Step-by-step solution

- Find the volume of the iron piece

The volume of the iron piece can be calculated using the formula for density: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \] Given: \[ \text{Mass} = 15.0 \text{ g} \] \[ \text{Density} = 7.86 \text{ g/cm}^3 \] So, \[ \text{Volume} = \frac{15.0 \text{ g}}{7.86 \text{ g/cm}^3} = 1.91 \text{ cm}^3 \]

- Find the volume of the lead piece

The volume of the lead piece can be calculated using the same formula: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \] Given: \[ \text{Mass} = 20.0 \text{ g} \] \[ \text{Density} = 11.3 \text{ g/cm}^3 \] So, \[ \text{Volume} = \frac{20.0 \text{ g}}{11.3 \text{ g/cm}^3} = 1.77 \text{ cm}^3 \]

- Calculate the total volume added to the cylinder

Add the volumes of the iron and lead pieces: \[ \text{Total Volume} = 1.91 \text{ cm}^3 + 1.77 \text{ cm}^3 = 3.68 \text{ cm}^3 \]

- Convert total volume to milliliters

Since 1 cm\textsuperscript{3} = 1 mL, \[ \text{Total Volume} = 3.68 \text{ cm}^3 = 3.68 \text{ mL} \]

- Calculate the new water level in the cylinder

Add the total volume added to the initial water volume: \[ \text{New Water Level} = 155 \text{ mL} + 3.68 \text{ mL} = 158.68 \text{ mL} \]

Key concepts

Density

Density is a fundamental concept in chemistry that relates mass to volume. In simple terms, it's how much stuff is packed into a given space. Density is calculated using the formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). This tells us how heavy a material is for a given volume. Densities are commonly given in units like g/cm³ (grams per cubic centimeter).
For example, the density of iron is 7.86 g/cm³, meaning a cubic centimeter of iron weighs 7.86 grams. Understanding density is crucial for converting between mass and volume, as seen in our exercise.

Mass-Volume Relationship

The relationship between mass and volume is closely linked to density. If you know two of these properties, you can find the third using the formula: \( \text{Mass} = \text{Density} \times \text{Volume} \). This relationship helps us solve various problems in chemistry.
Let's consider our exercise: We know the mass and density of iron and lead, allowing us to find their volumes via \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). For the iron piece with a mass of 15.0 g and a density of 7.86 g/cm³, the volume is calculated as 1.91 cm³. For the lead piece with a mass of 20.0 g and density of 11.3 g/cm³, the volume is 1.77 cm³.

Volume Displacement

Volume displacement is a method used to measure the volume of an irregularly shaped object. By submerging the object in a liquid, such as water, the volume of liquid displaced is equal to the volume of the object. This is based on Archimedes' principle.
In our exercise, adding pieces of iron and lead to a graduated cylinder of water displaces the water, causing the water level to rise. By calculating the volumes of iron and lead (1.91 cm³ + 1.77 cm³), we find that their combined volume of 3.68 cm³ displaces an equal volume of water. Consequently, the water level in the cylinder increases by 3.68 mL, as 1 cm³ equals 1 mL.

Graduated Cylinder Measurement

A graduated cylinder is a common lab instrument used to measure liquid volumes. It features marked lines that indicate volume in milliliters (mL). This precision helps in accurate measurement of liquid volumes.
When measuring the water level in a graduated cylinder, ensure it's on a flat surface, and your eye is level with the meniscus (the curved surface of the liquid). This reduces parallax error. In our example, the initial water volume is 155 mL. When the iron and lead pieces are added, they displace water, causing the new water level to be 158.68 mL. Understanding how to accurately read the measurements is key to obtaining precise results.