Q. 2.106 - An Introduction to General, Organic, and Biological Chemistry

Question

A graduated cylinder contains $155 \mathrm{mL}$ of water. A $15.0 - E$ piece of iron and a $20.0-g$ piece of lead are added. What is the new water level, in milliliters, in the cylinder (see Table 2.8)(2.7)

Step-by-Step Solution

Answer

The new water level in the cylinder in milliliters is determined as $158.7$ mL

Introduction

The given is the water level of a graduated cylinder 

The objective is the find the new level of water in the cylinder in milliliters

Step 1

The volume of the new water level will be determined by the volume of water and the volume of the substances added.

The following is the formula for calculating density, mass, and volume:

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

Mass of iron $=15.0 \mathrm{g}$

Density of iron $=7.86 \mathrm{g}/{\mathrm{cm}}^3$

The following formula is used to compute the volume of iron:

$$\begin{gathered} 7.86 g c{m}^3=\frac{15.0 g}{volume} \\ Volume=\frac{15.0 \backslash g}{7.86 g / c{m}^3} \end{gathered}$$

$=1.91{ \mathrm{cm}}^3$

Convert ${\mathrm{cm}}^3$ to $\mathrm{mL}$.

$$\begin{gathered} 1.91 c{m}^3 \&=1.91 c{m}^3 \times \frac{1 mL}{1 c{m}^3} \\ =1.91 mL \end{gathered}$$

Step 2

Mass of lead$=20.0 \mathrm{g}$

Density of iron$=11.3 \mathrm{g}/{\mathrm{cm}}^3$

The following formula is used to compute the volume of lead:

$11.3 \mathrm{g}/{\mathrm{cm}}^3=\frac{20.0 \mathrm{g}}{\text{ Volume }}$

$Volume =\frac{20.0 g}{11.3 g / c{m}^3}$

$1.77 c{m}^3$

Convert ${\mathrm{cm}}^3$ to $\mathrm{mL}$.

$$\begin{gathered} 1.77 c{m}^3 \&=1.77 c{m}^3 \times \frac{1 mL}{1 c{m}^3} \\ =1.77 mL \end{gathered}$$

Step 3

Initially, the volume of water  $=155 \mathrm{mL}$

To calculate the new volume, multiply the volume of lead and iron by the initial volume of water.

$$\begin{gathered} New volume=155 mL+1.91 mL+1.77 mL \\ =158.7 mL \end{gathered}$$

Hence, the new water level is $158.7 mL$