Chapter 8, Problem 8.74P

Interpretation Introduction

(a)

Interpretation:

The molarity of $\text{NaOH}$ solution when $\text{2}\text{.4 mol}$ is present in $1.50\text{ L}$ is to be determined.

Concept introduction:

The formula to calculate molarity is given as follows:

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

Answer to Problem 8.74P

The molarity of $\text{NaOH}$ solution when $\text{2}\text{.4 mol}$ is present in $1.50\text{ L}$ is $1.6\text{ mol/L}$.

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 solution in liters is given as $1.50\text{ L}$.

The number of moles of $\text{NaOH}$ is $\text{2}\text{.4 mol}$.

Substitute the values in above formula to calculate the molarity of $\text{NaOH}$.

  $\begin{array}{c}\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ =\frac{\text{2}\text{.4 mol}}{1.5\text{ L}}\\ =1.6\text{ mol/L}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The molarity of ${\text{KNO}}_3$ solution when $\text{0}\text{.48 mol}$ is present in $\text{750 mL}$ aqueous solution is to be determined.

Concept introduction:

The formula to calculate molarity is given as follows:

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

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

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

Answer to Problem 8.74P

The molarity of ${\text{KNO}}_3$ solution when $\text{0}\text{.48 mol}$ is present in $\text{750 mL}$ aqueous solution is $0.64\text{ mol/L}$.

Explanation of Solution

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

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

Hence convert volume from $\text{mL }$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}\left(\text{L}\right)=\left(\text{750 }\cancel{\text{mL}}\right)\left(\frac{1\text{ L}}{{10}^3\cancel{\text{mL}}}\right)\\ =\text{ 0}\text{.75 L}\end{array}$

The volume of solution is given as $\text{0}\text{.855 L}$.

The number of moles of ${\text{KNO}}_3$ is $\text{0}\text{.48 mol}$.

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

  $\begin{array}{c}\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ =\frac{\text{0}\text{.48 mol}}{\text{0}\text{.75 L}}\\ =0.64\text{ mol/L }\end{array}$

Interpretation Introduction

(c)

Interpretation:

The molarity of $\text{KCl}$ solution when $25\text{ g}$ is present in $\text{650 mL}$ aqueous solution is to be calculated.

Concept introduction:

The formula to calculate molarity is given as follows:

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

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

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

Answer to Problem 8.74P

The molarity of $\text{KCl}$ solution when $25\text{ g}$ is present in $\text{650 mL}$ aqueous solution is $0.5158\text{ mol/L}$.

Explanation of Solution

The formula to calculate the number of moles of $\text{KCl}$ is given as follows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

Mass of $\text{KCl}$ is $25\text{ g}$.

Molar mass of $\text{KCl}$ is $\text{74}\text{.55 g/mol}$.

Substitute the values in above formula to calculate the number of moles of $\text{KCl}$.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}\\ =\frac{\text{25 g}}{\text{74}\text{.55 g/mol}}\\ =0.3353\text{ mol}\end{array}$

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

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

Hence, convert volume from $\text{mL }$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}\left(\text{L}\right)=\left(\text{650 }\cancel{\text{mL}}\right)\left(\frac{1\text{ L}}{{10}^3\cancel{\text{mL}}}\right)\\ =\text{ 0}\text{.65 L}\end{array}$

The volume of solution is given as $\text{0}\text{.65 L}$.

The number of moles of $\text{KCl}$ is $0.3353\text{ mol}$.

Substitute the values in above formula to calculate the molarity of $\text{KCl}$.

  $\begin{array}{c}\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ =\frac{0.3353\text{ mol}}{0.65\text{ L}}\\ =0.5158\text{ mol/L }\end{array}$

Interpretation Introduction

(d)

Interpretation:

The molarity of ${\text{Na}}_2{\text{CO}}_3$ solution when $10\text{ g}$ is present in $3.8\text{ L}$ aqueous solution is to be calculated.

Concept introduction:

The formula to calculate molarity is given as follows:

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

The formula to calculate number of moles from given mass is as follows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

Answer to Problem 8.74P

The molarity of ${\text{Na}}_2{\text{CO}}_3$ solution when $10\text{ g}$ is present in $\text{250 mL}$ aqueous solution is $0.02489\text{ mol/L }$.

Explanation of Solution

The formula to calculate the number of moles of ${\text{Na}}_2{\text{CO}}_3$ is given as follows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

Mass of ${\text{Na}}_2{\text{CO}}_3$ is given as $10\text{ g}$.

Molar mass of ${\text{Na}}_2{\text{CO}}_3$ is $\text{105}\text{.98 g/mol}$.

Substitute the values in above formula to calculate the number of moles of ${\text{Na}}_2{\text{CO}}_3$.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ =\frac{10\text{ g}}{\text{105}\text{.98 g/mol}}\\ =0.09435\text{ mol}\end{array}$

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 solution is given as $\text{3}\text{.8 L}$.

The number of moles of ${\text{Na}}_2{\text{CO}}_3$ is $0.09435\text{ mol}$.

Substitute the values in above formula to calculate the molarity of ${\text{Na}}_2{\text{CO}}_3$.

  $\begin{array}{c}\text{Molarity}\left(\text{mol/L}\right)=\frac{\text{number of moles}}{\text{volume}\left(\text{L}\right)}\\ =\frac{0.09435\text{ mol}}{\text{3}\text{.8 L}}\\ =0.02489\text{ mol/L }\end{array}$