Chapter 2, Problem 68A

Interpretation Introduction

Interpretation:

The density of the object having mass $\text{7}\text{.5}\text{g}$ needs to be determined if it is placed in a graduated cylinder resulting rise in the water level from $\text{25}\text{.1}\text{mL}\text{to}\text{30}\text{.1}\text{mL}$.

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 68A

The density of the object is $1.5\text{g/mL}$.

Explanation of Solution

Mass of the object = $7.5\text{g}$

Initial volume of water in the graduated cylinder = $\text{25}\text{.1}\text{mL}$

Final volume of water in the graduated cylinder after placing the object = $\text{30}\text{.1}\text{mL}$

Volume of the object = 30.1 − 25.1

= 5 mL

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

$\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ =\frac{7.5\text{g}}{5\text{mL}}\\ =1.5\text{g/mL}\end{array}$

Hence, the density of the object is $1.5\text{g/mL}$.

Conclusion

Thus, the density of the object is $1.5\text{g/mL}$.