Chapter 8.6, Problem 8.24P

Interpretation Introduction

(a)

Interpretation:

Volume of $1.5\text{ M}$ glucose solution that contains $\text{0}\text{.15 mol}$ is to be calculated in milliliters.

Concept introduction:

The formula to calculate molarity is given as follows:

  $\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}$

Rearrange the above formula to obtain the formula for volume.

  $\text{Volume}\left(\text{L}\right)=\frac{\text{number of moles}}{\text{molarity}\left(\text{mol/L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is given as follows:

  $1\text{ L}={\text{10}}^3\text{ mL}$

Interpretation Introduction

(b)

Interpretation:

Volume of $1.5\text{ M}$ glucose solution that contains $\text{0}\text{.020 mol}$ is to be calculated in milliliters.

Concept introduction:

The formula to calculate volume from molarity isgiven as follows:

  $\text{Volume}\left(\text{L}\right)=\frac{\text{number of moles}}{\text{molarity}\left(\text{mol/L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is given as follows:

  $1\text{ L}={\text{10}}^3\text{ mL}$

Interpretation Introduction

(c)

Interpretation:

Volume of $1.5\text{ M}$ glucose solution that contains $\text{0}\text{.0030 mol}$ is to be calculated in milliliters.

Concept introduction:

The formula to calculate volume from molarity isgiven as follows:

  $\text{Volume}\left(\text{L}\right)=\frac{\text{number of moles}}{\text{molarity}\left(\text{mol/L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is given as follows:

  $1\text{ L}={\text{10}}^3\text{ mL}$

Interpretation Introduction

(d)

Interpretation:

Volume of $1.5\text{ M}$ glucose solution that contains $\text{3 mol}$ is to be calculated in milliliters.

Concept introduction:

The formula to calculate volume from molarity isgiven as follows:

  $\text{Volume}\left(\text{L}\right)=\frac{\text{number of moles}}{\text{molarity}\left(\text{mol/L}\right)}$

The conversion factor to convert $\text{L}$ to $\text{mL}$ is given as follows:

  $1\text{ L}={\text{10}}^3\text{ mL}$