Chapter 7, Problem 7.45E

Interpretation Introduction

(a)

Interpretation:

The number of moles of $\text{NaI}$ in $\text{50}\text{.0}\text{mL}$ of a $\text{0}\text{.400}\text{M}$ solution is to be predicted.

Concept Introduction:

The Molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 7.45E

The number of moles of $\text{NaI}$ in $\text{50}\text{.0}\text{mL}$ of a $\text{0}\text{.400}\text{M}$ solution is $0.02\text{moles}$.

Explanation of Solution

The number of moles of $\text{NaI}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

The given volume and molarity is $\text{50}\text{.0}\text{mL}$ and $\text{0}\text{.400}\text{M}$ respectively.

Substitute the volume and molarity in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{\text{0}\text{.400}\text{M}\times 50.0\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.02\text{moles}\end{array}$

Thus, the number of moles of $\text{NaI}$ is $0.02\text{moles}$.

Conclusion

The number of moles of $\text{NaI}$ in $\text{50}\text{.0}\text{mL}$ of a $\text{0}\text{.400}\text{M}$ solution is $0.02\text{moles}$.

Interpretation Introduction

(b)

Interpretation:

The number of grams of $\text{KBr}$ in $\text{120}\text{.0}\text{mL}$ of a $\text{0}\text{.720}\text{M}$ solution is to be predicted.

Concept Introduction:

The number of moles is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 7.45E

Thenumber of grams of $\text{KBr}$ in $\text{120}\text{.0}\text{mL}$ of a $\text{0}\text{.720}\text{M}$ solution is $10.28\text{g}$.

Explanation of Solution

The number of moles of $\text{KBr}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

The volume and molarity is $\text{120}\text{.0}\text{mL}$ and $\text{0}\text{.720}\text{M}$ respectively.

Substitute the volume and molarity in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{\text{0}\text{.720}\text{M}\times 120.0\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.0864\text{moles}\end{array}$

Thus, the number of moles of $\text{KBr}$ is $0.0864\text{moles}$.

The amount of $\text{KBr}$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

The molar mass of $\text{KBr}$ is $119.0\text{g}/\text{mol}$.

Substitute the value of molar mass in the given formula.

$\begin{array}{rll}0.0864\text{moles}& =& \frac{\text{Amount}\text{of}\text{KBr}}{\text{119}\text{.0}\text{g}/\text{mol}}\\ \text{Amount}\text{of}\text{KBr}& =& 0.0864\text{moles}\times \text{119}\text{.0}\text{g}/\text{mol}\\ & =& 10.28\text{g}\end{array}$

Thus, the number of grams of $\text{KBr}$ in $\text{120}\text{.0}\text{mL}$ of a $\text{0}\text{.720}\text{M}$ solution is $10.28\text{g}$.

Conclusion

Thenumber of grams of $\text{KBr}$ in $\text{120}\text{.0}\text{mL}$ of a $\text{0}\text{.720}\text{M}$ solution is $10.28\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The number of grams of $\text{NaCl}$ in $\text{20}\text{.0}\text{mL}$ of a $1.20\%\left(\text{w}/\text{v}\right)$$\text{NaCl}$ solution is to be predicted.

Concept Introduction:

The concentration of the solution in $\%\left(\text{w}/\text{v}\right)$ is given by the formula,

$\%\left(\text{w}/\text{v}\right)=\frac{\text{grams}\text{of}\text{solute}}{\text{milliliters}\text{of}\text{solution}}\times 100$

Answer to Problem 7.45E

The number of grams of $\text{NaCl}$ in $\text{20}\text{.0}\text{mL}$ of a $1.20\%\left(\text{w}/\text{v}\right)$$\text{NaCl}$ solution is $0.24\text{g}$.

Explanation of Solution

The concentration of the solution in $\%\left(\text{w}/\text{v}\right)$ is given by the formula,

$\%\left(\text{w}/\text{v}\right)=\frac{\text{grams}\text{of}\text{solute}}{\text{milliliters}\text{of}\text{solution}}\times 100$ …(1)

The given value of $\%\left(\text{w}/\text{v}\right)$ is $1.20\%\left(\text{w}/\text{v}\right)$. The volume of $\text{NaCl}$ solution is $\text{20}\text{.0}\text{mL}$.

Substitute the value of $\%\left(\text{w}/\text{v}\right)$ and volume in equation (1).

$\begin{array}{rll}1.20& =& \frac{\text{grams}\text{of}\text{solute}}{\text{20}\text{.0}\text{L}}\times 100\\ \text{grams}\text{of}\text{solute}& =& \frac{1.20\times \text{20}\text{.0}\text{L}}{100}\\ & =& 0.24\text{g}\end{array}$

Hence, the number of grams of $\text{NaCl}$ in $\text{20}\text{.0}\text{mL}$ of a $1.20\%\left(\text{w}/\text{v}\right)$$\text{NaCl}$ solution is $0.24\text{g}$.

Conclusion

The number of grams of $\text{NaCl}$ in $\text{20}\text{.0}\text{mL}$ of a $1.20\%\left(\text{w}/\text{v}\right)$$\text{NaCl}$ solution is $0.24\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $20.0\%\left(\text{v}/\text{v}\right)$ solution is to be predicted.

Concept Introduction:

The concentration of the solution in $\%\left(\text{v}/\text{v}\right)$ is given by the formula,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{solute}\text{volume}}{\text{solultion}\text{volume}}\times 100$

Answer to Problem 7.45E

The number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $20.0\%\left(\text{v}/\text{v}\right)$ solution is $50\text{mL}$.

Explanation of Solution

The concentration of the solution in $\%\left(\text{v}/\text{v}\right)$ is given by the formula,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{solute}\text{volume}}{\text{solultion}\text{volume}}\times 100$ …(1)

The given value of $\%\left(\text{v}/\text{v}\right)$ is $20.0\%\left(\text{v}/\text{v}\right)$. The volume of solution is $\text{250}\text{.}\text{mL}$.

Substitute the value of $\%\left(\text{v}/\text{v}\right)$ and solution volume in equation (1).

$\begin{array}{rll}20.0& =& \frac{\text{volume}\text{of}\text{alcohol}}{250\text{mL}}\times 100\\ \text{volume}\text{of}\text{alcohol }& =& \frac{250\text{mL}\times 20.0}{100}\\ & =& 50\text{mL}\end{array}$

Hence, the number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $20.0\%\left(\text{v}/\text{v}\right)$ solution is $50\text{mL}$.

Conclusion

The number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $20.0\%\left(\text{v}/\text{v}\right)$ solution is $50\text{mL}$.