Chapter 8, Problem 64P

Interpretation Introduction

(a)

Interpretation:

The number of grams of solute present in $\text{0}\text{.55 M}{\text{NaNO}}_3$ is to be calculated.

Concept introduction:

Several concentration units may be used to express the concentration. It is often useful to express the concentration in terms of percentage. This can be expressed as the amount of solute in grams dissolved in $100\text{ mL}$ of solution that is termed as weight by volume percentage. This is useful to express the concentration of solid solutes. The other way is to express the number of milliliters of solute per $100\text{ mL}$ of the solution. This is called the volume /volume percentage that is useful to express the concentrations of liquid solutes.

The formula to calculate molarity is given as follows:

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

The formula to determine the mass from number of moles is as follows:

  $\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)$

Answer to Problem 64P

The number of grams of ${\text{NaNO}}_3$ present in $\text{0}\text{.55 M}$ of ${\text{NaNO}}_3$ is $11.69\text{ g}$.

Explanation of Solution

The formula to calculate molarity is given as follows:

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

Since the units of molarity are $\text{mol/L}$ so $\text{0}\text{.55 M}$ of ${\text{NaNO}}_3$ represents $\text{0}\text{.55 mol}$ of ${\text{NaNO}}_3$ present in $1\text{ L}$.

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

  $1\text{ L}=\text{1000 mL}$

Therefore, $\text{0}\text{.55 mol}$ of ${\text{NaNO}}_3$ are present in $\text{1000 mL}$.

Therefore, the number of moles of ${\text{NaNO}}_3$ required to prepare a solution of $250\text{}\text{mL}$ of $\text{0}\text{.55 M}{\text{NaNO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Number of moles of NaNO}}_3\text{=}\left(250\text{ mL}\right)\left(\frac{\text{0}\text{.55 mol}}{1000\text{ mL}}\right)\\ =0.1375\text{ mol}\end{array}$

The formula to determine the mass from number of moles is as follows:

  $\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)$

The molar mass of ${\text{NaNO}}_3$ is $85\text{ g/mol}$.

The number of moles of ${\text{NaNO}}_3$ is $0.1375\text{ mol}$.

Substitute the values in above formula to calculate the mass of ${\text{NaNO}}_3$.

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)\\ =\left(0.1375\text{ mol}\right)\left(\text{85 g/mol}\right)\\ =11.69\text{ g}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The number of grams of solute present in $145\text{ mL}$ of $\text{4 M}{\text{HNO}}_3$ is to be calculated.

Concept introduction:

Several concentration units may be used to express the concentration. It is often useful to express the concentration in terms of percentage. This can be expressed as the amount of solute in grams dissolved in $100\text{ mL}$ of solution that is termed as weight by volume percentage. This is useful to express the concentration of solid solutes. The other way is to express the number of milliliters of solute per $100\text{ mL}$ of the solution. This is called the volume /volume percentage that is useful to express the concentrations of liquid solutes.

The formula to calculate molarity is given as follows:

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

The formula to determine the mass from number of moles is as follows:

  $\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)$

Answer to Problem 64P

The number of grams of ${\text{HNO}}_3$ present in $145\text{ mL}$ of $\text{4 M}$

${\text{HNO}}_3$ is $36.5458\text{ g}$.

Explanation of Solution

The formula to calculate molarity is given as follows:

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

Since the units of molarity are $\text{mol/L}$ so, $\text{4 M}$

${\text{HNO}}_3$ represents $\text{4 mol}$ of ${\text{HNO}}_3$ present in $1\text{ L}$.

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

  $1\text{ L}=\text{1000 mL}$

Therefore, $\text{4 mol}$ of ${\text{HNO}}_3$ are present in $\text{1000 mL}$.

So, the number of moles of ${\text{HNO}}_3$ required to prepare a solution of $\text{4 M}$

  ${\text{HNO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Number of moles =}\left(145\text{ mL}\right)\left(\frac{\text{4 mol}}{1000\text{ mL}}\right)\\ =0.58\text{ mol}\end{array}$

The formula to determine the mass from number of moles is as follows:

  $\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)$

The molar mass of ${\text{HNO}}_3$ is $\text{63}\text{.01 g/mol}$.

The number of moles of ${\text{HNO}}_3$ is $0.58\text{ mol}$.

Substitute the values in above formula to calculate the mass of ${\text{HNO}}_3$.

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)\\ =\left(0.58\text{ mol}\right)\left(\text{63}\text{.01 g/mol}\right)\\ =36.5458\text{ g}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The number of grams of solute present in $6.5\text{ L}$ of $\text{2}\text{.5 M}\text{HCl}$ is to be calculated.

Concept introduction:

Several concentration units may be used to express the concentration. It is often useful to express the concentration in terms of percentage. This can be expressed as the amount of solute in grams dissolved in $100\text{ mL}$ of solution that is termed as weight by volume percentage. This is useful to express the concentration of solid solutes. The other way is to express the number of milliliters of solute per $100\text{ mL}$ of the solution. This is called the volume /volume percentage that is useful to express the concentrations of liquid solutes.

The formula to calculate molarity is given as follows:

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

Answer to Problem 64P

The number of grams of $\text{HCl}$ present in $6.5\text{ L}$ of $\text{2}\text{.5 M}$

$\text{HCl}$ is $\text{592}\text{.475 g}$.

Explanation of Solution

The formula to calculate molarity is given as follows:

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

The volume of $\text{HCl}$ solution to be prepared is $6.5\text{ L}$.

The molarity of $\text{HCl}$ solution to be prepared is $\text{2}\text{.5 M}$.

Substitute the values in above formula.

  $\begin{array}{c}\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ \text{2}\text{.5 M}=\frac{\text{number of moles}}{6.5\text{ L}}\end{array}$

Rearrange the above expression to calculate the number of moles of $\text{HCl}$.

  $\begin{array}{c}\text{Number of moles}=\left(6.5\text{ L}\right)\left(\frac{\text{2}\text{.5 mol}}{1\text{ L}}\right)\\ =16.25\text{ mol}\end{array}$

The formula to determine the mass from number of moles is as follows:

  $\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)$

The molar mass of $\text{HCl}$ is $\text{36}\text{.46 g/mol}$.

The number of moles of $\text{HCl}$ is $16.25\text{ mol}$.

Substitute the values in above formula to calculate the mass of $\text{HCl}$.

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)=\left(\text{number of moles}\left(\text{mol}\right)\right)\left(\text{molar mass}\left(\text{g/mol}\right)\right)\\ =\left(16.25\text{ mol}\right)\left(\text{36}\text{.46 g/mol}\right)\\ =\text{592}\text{.475 g}\end{array}$