What is the density (grams/mL) of a 48.0-gram sample of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added?

A. 6.0 g/mL
B. 1.92 g/mL
C. 8 g/mL
D. 1.45 g/mL



Answer :

To find the density of the metal, we need to follow these steps:

1. Determine the volume of the metal:
- First, we find the initial volume of water in the graduated cylinder, which is 25.0 mL.
- After the metal is added, the water level rises to 33.0 mL.
- The volume of the metal is therefore the difference between the final and initial water levels:
[tex]\[ \text{Volume of the metal} = \text{Final water level} - \text{Initial water level} = 33.0\ \text{mL} - 25.0\ \text{mL} = 8.0\ \text{mL} \][/tex]

2. Calculate the density of the metal:
- Density is defined as mass per unit volume. The mass of the metal is given as 48.0 grams and we have already found the volume to be 8.0 mL.
- Using the formula for density:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
- Substituting the known values:
[tex]\[ \text{Density} = \frac{48.0\ \text{g}}{8.0\ \text{mL}} = 6.0\ \text{g/mL} \][/tex]

Therefore, the density of the metal is [tex]\(6.0\ \text{g/mL}\)[/tex].

The correct answer is:
[tex]\[ 6.0\ \text{g/mL} \][/tex]