Chapter 2, Problem 105A

Interpretation Introduction

Interpretation:

The mass of the lead (density $11.4{\text{g/cm}}^3$ ) having a volume equal to the $\text{15}\text{.0}\text{g}$ of mercury (density $13.6{\text{g/cm}}^3$ ) 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 105A

Mass of the lead is $\text{12}\text{.5}\text{g}$.

Explanation of Solution

Mass of the mercury = $15\text{g}$

Density of the mercury = $13.6{\text{g/cm}}^3$

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

$\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ 13.6{\text{g/cm}}^3=\frac{15\text{g}}{\text{Volume}}\\ \text{Volume}=\frac{15\text{g}}{13.6{\text{g/cm}}^3}\\ =1.10{\text{cm}}^3\end{array}$

Volume of the mercury = volume of the lead = $1.10{\text{cm}}^3$

Now, the mass of the lead can be calculated by multiplying the volume with density.

$\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ \text{Mass}=\text{Volume}\times \text{Density}\\ =\left(1.10{\text{cm}}^3\right)\times \left(\text{11}\text{.4}{\text{g/cm}}^{\text{3}}\right)\\ =\text{12}\text{.5}\text{g}\end{array}$

Hence, the mass of lead is 12.5 g.

Conclusion

Thus, mass of the lead is $\text{12}\text{.5}\text{g}$.