Chapter 15, Problem 56QAP

Interpretation Introduction

(a)

Interpretation:

The new molarity after the addition of water is 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,

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

Answer to Problem 56QAP

The new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.0837\text{M}$.

Explanation of Solution

The initial volume and molarity of $\text{HCl}$ solution is given to be $125\text{mL}$ and $0.251\text{M}$ respectively.

The conversion of units of $125\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}125\text{mL}=\frac{1250}{1000}\text{L}\\ =0.125\text{L}\end{array}$

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

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$        (1)

Substitute the values of initial volume and molarity of $\text{HCl}$ solution in the equation (1).

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.125\text{L}\times 0.251\text{M}\\ =0.0314\text{moles}\end{array}$

It is given that $250.0\text{mL}$ of water is added in the given solution. During the dilution, the amount of the solute particles in initial and final solution remains same, only the amount of solvent gets changed.

The conversion of units of $250.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}250.0\text{mL}=\frac{250.0}{1000}\text{L}\\ =0.250\text{L}\end{array}$

Thus, the total volume is calculated by the formula,

$\text{Final}\text{volume}=\text{Initial}\text{volume}+\text{Volume}\text{of}\text{water}\text{added}$        (2)

Substitute the values of initial volume and volume of water added in the equation (2).

$\begin{array}{c}\text{Final}\text{volume}=0.125\text{L}+0.250\text{L}\\ =0.375\text{L}\end{array}$

The new molarity of the solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$        (3)

Substitute the values of number of moles of solute and final volume in the equation (3).

$\begin{array}{c}\text{Molarity}=\frac{0.0314\text{moles}}{0.375\text{L}}\\ =0.0837\text{M}\end{array}$

Therefore, the new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.0837\text{M}$.

Interpretation Introduction

(b)

Interpretation:

The new molarity after the addition of water is 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,

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

Answer to Problem 56QAP

The new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.319\text{M}$.

Explanation of Solution

The initial volume and molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is given to be $445\text{mL}$ and $0.499\text{M}$ respectively.

The conversion of units of $445\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}445\text{mL}=\frac{445}{1000}\text{L}\\ =0.445\text{L}\end{array}$

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

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$        (1)

Substitute the values of initial volume and molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution in the equation (1).

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.445\text{L}\times 0.499\text{M}\\ =0.222\text{moles}\end{array}$

It is given that $250.0\text{mL}$ of water is added in the given solution. During the dilution, the amount of the solute particles in initial and final solution remains same, only the amount of solvent get changes.

The conversion of units of $250.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}250.0\text{mL}=\frac{250.0}{1000}\text{L}\\ =0.250\text{L}\end{array}$

Thus, the total volume is calculated by the formula,

$\text{Final}\text{volume}=\text{Initial}\text{volume}+\text{Volume}\text{of}\text{water}\text{added}$        (2)

Substitute the values of initial volume and volume of water added in the equation (2).

$\begin{array}{c}\text{Final}\text{volume}=0.445\text{L}+0.250\text{L}\\ =0.695\text{L}\end{array}$

The new molarity of the solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$        (3)

Substitute the values of number of moles of solute and final volume in the equation (3).

$\begin{array}{c}\text{Molarity}=\frac{0.222\text{moles}}{0.695\text{L}}\\ =0.319\text{M}\end{array}$

Therefore, the new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.319\text{M}$.

Interpretation Introduction

(c)

Interpretation:

The new molarity after the addition of water is 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,

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

Answer to Problem 56QAP

The new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.0964\text{M}$.

Explanation of Solution

The initial volume and molarity of ${\text{HNO}}_3$ solution is given to be $\text{5}\text{.25}\text{L}$ and $0.101\text{M}$ respectively.

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

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$        (1)

Substitute the values of initial volume and molarity of ${\text{HNO}}_3$ solution in the equation (1).

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{5}\text{.25}\text{L}\times 0.101\text{M}\\ =0.530\text{moles}\end{array}$

It is given that $250.0\text{mL}$ of water is added in the given solution. During the dilution, the amount of the solute particles in initial and final solution remains same, only the amount of solvent gets changed.

The conversion of units of $250.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}250.0\text{mL}=\frac{250.0}{1000}\text{L}\\ =0.250\text{L}\end{array}$

Thus, the total volume is calculated by the formula,

$\text{Final}\text{volume}=\text{Initial}\text{volume}+\text{Volume}\text{of}\text{water}\text{added}$        (2)

Substitute the values of initial volume and volume of water added in the equation (2).

$\begin{array}{c}\text{Final}\text{volume}=\text{5}\text{.25}\text{L}+0.250\text{L}\\ =5.5\text{L}\end{array}$

The new molarity of the solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$        (3)

Substitute the values of number of moles of solute and final volume in the equation (3).

$\begin{array}{c}\text{Molarity}=\frac{0.530\text{moles}}{5.5\text{L}}\\ =0.0964\text{M}\end{array}$

Therefore, the new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.0964\text{M}$.

Interpretation Introduction

(d)

Interpretation:

The new molarity after the addition of water is 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,

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

Answer to Problem 56QAP

The new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.6217\text{M}$.

Explanation of Solution

The initial volume and molarity of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution is given to be $11.2\text{mL}$ and $14.5\text{M}$ respectively.

The conversion of units of $11.2\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}11.2\text{mL}=\frac{11.2}{1000}\text{L}\\ =0.0112\text{L}\end{array}$

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

$\text{Number}\text{of}\text{moles}\text{of}\text{solute}=\text{Volume}\text{of}\text{solution}\times \text{Molarity}$        (1)

Substitute the values of initial volume and molarity of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution in the equation (1).

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{solute}=0.0112\text{L}\times 14.5\text{M}\\ =0.1624\text{moles}\end{array}$

It is given that $250.0\text{mL}$ of water is added in the given solution. During the dilution, the amount of the solute particles in initial and final solution remains same, only the amount of solvent get changes.

The conversion of units of $250.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}250.0\text{mL}=\frac{250.0}{1000}\text{L}\\ =0.250\text{L}\end{array}$

Thus, the total volume is calculated by the formula,

$\text{Final}\text{volume}=\text{Initial}\text{volume}+\text{Volume}\text{of}\text{water}\text{added}$        (2)

Substitute the values of initial volume and volume of water added in the equation (2).

$\begin{array}{c}\text{Final}\text{volume}=0.0112\text{L}+0.250\text{L}\\ =0.2612\text{L}\end{array}$

The new molarity of the solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$        (3)

Substitute the values of number of moles of solute and final volume in the equation (3).

$\begin{array}{c}\text{Molarity}=\frac{0.1624\text{moles}}{0.2612\text{L}}\\ =0.6217\text{M}\end{array}$

Therefore, the new molarity that results when $250.0\text{mL}$ of water is added to the given solution is $0.6217\text{M}$.