Chapter 1, Problem 1.97E

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced. Find the volume and density of the following:

a. An irregular piece of the mineral quartz is found to weigh $12.4\text{g}$. It is then placed into a graduated cylinder that contains some water. The quartz does not float. The water in the cylinder was at a level of $25.2\text{mL}$ before the quartz was added and at $29.9\text{mL}$ afterward.

b. The volume of a sample of lead shot is determined using a graduated cylinder, as in part (a). The cylinder readings are $16.3\text{mL}$ before the shot is added and $21.7\text{mL}$ after. The sample of shot weighs $61.0\text{g}$.

c. A sample of coarse rock salt is found to have a mass of $11.7\text{g}$. The volume of the sample is determined by the graduated-cylinder method described in (part a), but kerosene is substituted for water because the salt will not dissolve in kerosene. The cylinder readings are $20.7\text{mL}$ before adding the salt and $26.1\text{mL}$ after.

Interpretation Introduction

(a)

Interpretation:

The volume and density of an irregular piece of the mineral quartz weighing $12.4\text{g}$ and having volume change from $25.2\text{mL}$ to $29.9\text{mL}$ in a graduated cylinder are to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of an irregular piece of the mineral quartz weighing $12.4\text{g}$ and having volume change from $25.2\text{mL}$ to $29.9\text{mL}$ in graduated cylinder are $\text{4}\text{.7}\text{mL}$ and $2.64\text{g}/\text{mL}$ respectively.

Explanation of Solution

The volume displaced by immersing the mineral quartz in a graduated cylinder that contains water is calculated as given below.

$V=V_{\text{final}}-V_{\text{initial}}$

Where,

• $V_{\text{final}}$ is the volume in the graduated cylinder after immersing solid

• $V_{\text{initial}}$ is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

$\begin{array}{rll}V& =& V_{\text{final}}-V_{\text{initial}}\\ & =& 29.9\text{mL}-25.2\text{mL}\\ & =& \text{4}\text{.7}\text{mL}\end{array}$

The density is calculated as given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

Substitute the values in the above expression as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ & =& \frac{12.4\text{g}}{\text{4}\text{.7}\text{mL}}\\ & =& 2.64\text{g}/\text{mL}\end{array}$

Thus, the density of mineral quartz is found to be $2.64\text{g}/\text{mL}$.

Conclusion

The volume and density of an irregular piece of the mineral quartz weighing $12.4\text{g}$ and having volume change from $25.2\text{mL}$ to $29.9\text{mL}$ in graduated cylinder are calculated as $\text{4}\text{.7}\text{mL}$ and $2.64\text{g}/\text{mL}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The volume and density of a lead shot weighing $61.0\text{g}$ and having volume change from $16.3\text{mL}$ to $21.7\text{mL}$ in graduated cylinder are to becalculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of a lead shot weighing $61.0\text{g}$ and having volume change from $16.3\text{mL}$ to $21.7\text{mL}$ in graduated cylinder are $5.\text{4}\text{mL}$ and $11.3\text{g}/\text{mL}$ respectively.

Explanation of Solution

The volume displaced by immersing the lead shot in a graduated cylinder that contains water is calculated as given below.

$V=V_{\text{final}}-V_{\text{initial}}$

Where,

• $V_{\text{final}}$ is the volume in the graduated cylinder after immersing solid

• $V_{\text{initial}}$ is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

$\begin{array}{rll}V& =& V_{\text{final}}-V_{\text{initial}}\\ & =& 21.7\text{mL}-16.3\text{mL}\\ & =& 5.\text{4}\text{mL}\end{array}$

The density is calculated as given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

Substitute the values in the above expression as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ & =& \frac{61\text{g}}{5.4\text{mL}}\\ & =& 11.3\text{g}/\text{mL}\end{array}$

Thus, the density of lead shot is found to be $11.3\text{g}/\text{mL}$.

Conclusion

The volume and density of a lead shot weighing $61.0\text{g}$ and having volume change from $16.3\text{mL}$ to $21.7\text{mL}$ in graduated cylinder are calculated as $5.\text{4}\text{mL}$ and $11.3\text{g}/\text{mL}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The volume and density of rock salt weighing $11.7\text{g}$ and having volume change from $20.7\text{mL}$ to $26.1\text{mL}$ in graduated cylinder are to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of its mass to the volume. The expression for density is given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

The volume of an irregularly shaped solid can be determined by immersing the solid in a liquid and measuring the volume of liquid displaced.

Answer to Problem 1.97E

The volume and density of rock salt weighing $11.7\text{g}$ and having volume change from $20.7\text{mL}$ to $26.1\text{mL}$ in graduated cylinder are $5.\text{4}\text{mL}$ and $2.16\text{g}/\text{mL}$ respectively.

Explanation of Solution

The volume displaced by immersing rock salt in a graduated cylinder that contains kerosene is calculated as given below.

$V=V_{\text{final}}-V_{\text{initial}}$

Where,

• $V_{\text{final}}$ is the volume in the graduated cylinder after immersing solid

• $V_{\text{initial}}$ is the volume initially taken in the graduated cylinder

Substitute the values in the above expression as follows.

$\begin{array}{rll}V& =& V_{\text{final}}-V_{\text{initial}}\\ & =& 26.1\text{mL}-20.7\text{mL}\\ & =& 5.\text{4}\text{mL}\end{array}$

The density is calculated as given below.

$d=\frac{m}{V}$

Where,

• $d$ is the density

• $m$ is the mass

• $V$ is the volume

Substitute the values in the above expression as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ & =& \frac{11.7\text{g}}{5.4\text{mL}}\\ & =& 2.16\text{g}/\text{mL}\end{array}$

Thus, the density of rock salt is found to be $2.16\text{g}/\text{mL}$.

Conclusion

The volume and density of rock salt weighing $11.7\text{g}$ and having volume change from $20.7\text{mL}$ to $26.1\text{mL}$ in graduated cylinder are calculated as $5.\text{4}\text{mL}$ and $2.16\text{g}/\text{mL}$ respectively.