Chapter 2, Problem 67A

Interpretation Introduction

Interpretation:

The volume of $8.1\text{g}$ aluminum having density $\text{2}\text{.7}\text{g/mL}$ needs to be determined.

Concept introduction:

Density of a sample of matter or an object is equal to the mass of the sample divided by the volume of the sample.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 67A

Volume of the aluminum is $3\text{mL}$.

Explanation of Solution

Mass of the aluminum = $8.1\text{g}$

Density of the aluminum = $\text{2}\text{.7}\text{g/mL}$

The volume of the aluminum can be calculated by dividing the mass by density.

$\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ \text{Volume}=\frac{\text{Mass}}{\text{Density}}\\ =\frac{8.1\text{g}}{2.7\text{g/mL}}\\ =3\text{mL}\end{array}$

Hence, the volume of the water sample is $3\text{mL}$.

Conclusion

Volume of the aluminum is $3\text{mL}$.