Question

Forgetful Freddie placed \(25.0 \mathrm{~mL}\) of a liquid in a graduated cylinder with a mass of \(89.450 \mathrm{~g}\) when empty. When Freddie placed a metal slug with a mass of \(15.454\) g into the cylinder, the volume rose to \(30.7 \mathrm{~mL}\). Freddie was asked to calculate the density of the liquid and of the metal slug from his data, but he forgot to obtain the mass of the liquid. He was told that if he found the mass of the cylinder containing the liquid and the slug, he would have enough data for the calculations. He did so and found its mass to be \(125.934 \mathrm{~g}\). Calculate the density of the liquid and of the metal slug.

Short answer

The density of the liquid is 0.8412 g/mL, and the density of the metal slug is 2.7112 g/mL.

Step-by-step solution

- Find the mass of the liquid

We know the mass of the cylinder when it is empty and when it contains both the liquid and the metal slug. We can calculate the mass of the liquid and the slug together by subtracting the mass of the empty cylinder from the total mass.Mass of liquid and slug = 125.934 g - 89.450 g = 36.484 g.

- Find the mass of the liquid alone

The mass of the metal slug is given as 15.454 g. To find the mass of the liquid alone, subtract the mass of the slug from the combined mass of the liquid and slug.Mass of liquid = 36.484 g - 15.454 g = 21.03 g.

- Calculate the density of the liquid

Density is defined as mass divided by volume. The volume of the liquid is given as 25.0 mL, and we have calculated the mass of the liquid as 21.03 g.Density of liquid = \( \frac{21.03 \text{ g}}{25.0 \text{ mL}} \) = 0.8412 g/mL.

- Calculate the volume of the metal slug

When the slug is placed in the graduated cylinder, the total volume changes from 25.0 mL to 30.7 mL. The volume increase is due to the slug. Therefore, the volume of the slug isVolume of slug = 30.7 mL - 25.0 mL = 5.7 mL.

- Calculate the density of the metal slug

Using the mass and volume of the metal slug, we can calculate its density. The mass of the slug is given as 15.454 g and its volume is 5.7 mL.Density of slug = \( \frac{15.454 \text{ g}}{5.7 \text{ mL}} \) = 2.7112 g/mL.

Key concepts

mass and volume

To solve density problems, you must understand the basic concepts of mass and volume.
Mass is a measure of how much matter an object contains and is usually measured in grams (g).
Volume, on the other hand, refers to the amount of space an object occupies and is commonly measured in milliliters (mL) or cubic centimeters (cm³).

In the given exercise, we are provided the mass of the empty graduated cylinder (89.450 g), the volume of the initial liquid (25.0 mL), and the masses involved after adding different objects.
These pieces of data serve as the foundation for our calculations, allowing us to find the densities of both the liquid and the metal slug.

density formula

To calculate density, you need to use the density formula:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \] This formula shows that density (\text{\rho}) is the mass (m) of a substance divided by its volume (V).
This relationship is crucial for understanding how compact or sparse matter is within an object.

For example, in the exercise, we first determined the mass of the liquid and the metal slug, and then their respective volumes.
Using the density formula, we calculated:
\[ \text{Density of liquid} = \frac{21.03 \text{ g}}{25.0 \text{ mL}} = 0.8412 \text{ g/mL} \] and
\[ \text{Density of slug} = \frac{15.454 \text{ g}}{5.7 \text{ mL}} = 2.7112 \text{ g/mL} \] This confirms how density can be used to identify the nature of the substances involved.

graduated cylinder

A graduated cylinder is vital in measuring the volume of liquids accurately.
It typically features marked lines indicating volume increments (mL), allowing for precise measurements.
In the provided exercise, a graduated cylinder is used to measure the volume changes when the liquid and the metal slug are placed within it.

First, the initial volume of the liquid is noted (\text{25.0 mL}).
After adding the metal slug, the volume increases to \text{30.7 mL}.
This rise in volume indicates the space occupied by the slug, which can be calculated:
\[ \text{Volume of slug} = 30.7 \text{ mL} - 25.0 \text{ mL} = 5.7 \text{ mL} \] Through this change in volume, the graduated cylinder helps determine the slug's displacement, an essential step in calculating its density.

step-by-step solution

Breaking down a complex problem into manageable steps simplifies calculations.

In the exercise, we followed a step-by-step approach:

• Step 1 - Find the mass of the liquid and metal slug together:

\[ 125.934 \text{ g} - 89.450 \text{ g} = 36.484 \text{ g} \]
• Step 2 - Obtain the mass of the liquid alone:

\[ 36.484 \text{ g} - 15.454 \text{ g} = 21.03 \text{ g} \]
• Step 3 - Calculate the density of the liquid:

\[ \frac{21.03 \text{ g}}{25.0 \text{ mL}} = 0.8412 \text{ g/mL} \]
• Step 4 - Determine the volume of the metal slug:

\[ 30.7 \text{ mL} - 25.0 \text{ mL} = 5.7 \text{ mL} \]
• Step 5 - Calculate the density of the metal slug:

\[ \frac{15.454 \text{ g}}{5.7 \text{ mL}} = 2.7112 \text{ g/mL} \]

This methodical approach ensures accuracy and helps in understanding each component of the problem separately.