Chapter 15, Problem 118AP

Interpretation Introduction

(a)

Interpretation:

The moles and grams of the indicated solute in the given solution are to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 118AP

The moles and grams of the indicated solute in the given solution are $0.446\text{moles}$ and $33.25\text{g}$ respectively.

Explanation of Solution

The volume and molarity of $\text{KCl}$ solution is given to be $4.25\text{L}$ and $0.105\text{M}$ respectively.

The molar mass of $\text{KCl}$ is $74.55\text{g}/\text{mol}$.

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of $\text{KCl}$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=4.25\text{L}\times 0.105\text{M}\\ =0.446\text{moles}\end{array}$

The mass of $\text{KCl}$ is calculated by the formula,

$\text{Mass of KCl}=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}$

Substitute the values of molar mass and number of moles of $\text{KCl}$ in the above expression.

$\begin{array}{c}\text{Mass of KCl}=74.55\text{g}/\text{mol}\times 0.446\text{moles}\\ =33.25\text{g}\end{array}$

Therefore, the moles and grams of the indicated solute in the given solution are $0.446\text{moles}$ and $33.25\text{g}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The moles and grams of the indicated solute in the given solution are to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 118AP

The moles and grams of the indicated solute in the given solution are $0.0034\text{moles}$ and $0.289\text{g}$ respectively.

Explanation of Solution

The volume and molarity of ${\text{NaNO}}_3$ solution is given to be $15.1\text{mL}$ and $0.225\text{M}$ respectively.

The conversion of units of $15.1\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}15.1\text{mL}=\frac{15.1}{1000}\text{L}\\ =0.0151\text{L}\end{array}$

The molar mass of ${\text{NaNO}}_3$ is $84.995\text{g}/\text{mol}$.

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of ${\text{NaNO}}_3$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.0151\text{L}\times 0.225\text{M}\\ =0.0034\text{moles}\end{array}$

The mass of ${\text{NaNO}}_3$ is calculated by the formula,

$\text{Mass of}{\text{NaNO}}_3=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}$

Substitute the values of molar mass and number of moles of ${\text{NaNO}}_3$ in the above expression.

$\begin{array}{c}{\text{Mass of NaNO}}_3=84.995\text{g}/\text{mol}\times 0.0034\text{moles}\\ =0.289\text{g}\end{array}$

Therefore, the moles and grams of the indicated solute in the given solution are $0.0034\text{moles}$ and $0.289\text{g}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The moles and grams of the indicated solute in the given solution are to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 118AP

The moles and grams of the indicated solute in the given solution are $0.075\text{moles}$ and $2.73\text{g}$ respectively.

Explanation of Solution

The volume and molarity of $\text{HCl}$ solution is given to be $25\text{mL}$ and $3.0\text{M}$ respectively.

The conversion of units of $25\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}25\text{mL}=\frac{25}{1000}\text{L}\\ =0.025\text{L}\end{array}$

The molar mass of $\text{HCl}$ is $36.46\text{g}/\text{mol}$.

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of $\text{HCl}$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.025\text{L}\times 3.0\text{M}\\ =0.075\text{moles}\end{array}$

The mass of $\text{HCl}$ is calculated by the formula,

$\text{Mass of}\text{HCl}=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}$

Substitute the values of molar mass and number of moles of $\text{HCl}$ in the above expression.

$\begin{array}{c}\text{Mass of HCl}=36.46\text{g}/\text{mol}\times 0.075\text{moles}\\ =2.73\text{g}\end{array}$

Therefore, the moles and grams of the indicated solute in the given solution are $0.075\text{moles}$ and $2.73\text{g}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The moles and grams of the indicated solute in the given solution are to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 118AP

The moles and grams of the indicated solute in the given solution are $0.0505\text{moles}$ and $4.953\text{g}$ respectively.

Explanation of Solution

The volume and molarity of ${\text{H}}_2{\text{SO}}_4$ solution is given to be $100.\text{mL}$ and $0.505\text{M}$ respectively.

The conversion of units of $100.\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}100.\text{mL}=\frac{100.}{1000}\text{L}\\ =0.1\text{L}\end{array}$

The molar mass of ${\text{H}}_2{\text{SO}}_4$ is $98.08\text{g}/\text{mol}$.

The number of moles of a solute is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$

Substitute the values of volume of solution and molarity of ${\text{H}}_2{\text{SO}}_4$ solution in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.1\text{L}\times 0.505\text{M}\\ =0.0505\text{moles}\end{array}$

The mass of ${\text{H}}_2{\text{SO}}_4$ is calculated by the formula,

$\text{Mass of}{\text{H}}_2{\text{SO}}_4=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}$

Substitute the values of molar mass and number of moles of ${\text{H}}_2{\text{SO}}_4$ in the above expression.

$\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_4=98.08\text{g}/\text{mol}\times 0.0505\text{moles}\\ =4.953\text{g}\end{array}$

Therefore, the moles and grams of the indicated solute in the given solution are $0.0505\text{moles}$ and $4.953\text{g}$ respectively.