Chapter 7, Problem 7.46E

Calculate the following:

a. The number of grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.}\text{mL}$ of $\text{1}\text{.75}\text{M}$ ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution

b. The number of moles of ${\text{NH}}_3$ in $\text{200}\text{.}\text{mL}$ of $\text{3}\text{.50}\text{M}$ ${\text{NH}}_3$ solution

c. The number of $\text{mL}$ of alcohol in $\text{250}\text{.}\text{mL}$ of $12.5\%\left(\text{v}/\text{v}\right)$ solution

d. The number of grams of ${\text{CaCl}}_2$ in $\text{50}\text{.0}\text{mL}$ of $4.20\%\left(\text{w}/\text{v}\right)$ ${\text{CaCl}}_2$ solution

Interpretation Introduction

(a)

Interpretation:

The number of grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.}\text{mL}$ of $\text{1}\text{.75}\text{M}$${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ 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 7.46E

The number of grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.}\text{mL}$ of $\text{1}\text{.75}\text{M}$${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $\text{32}\text{.3}\text{g}$.

Explanation of Solution

The number of moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ 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 $\text{250}\text{.}\text{mL}$ and $\text{1}\text{.75}\text{M}$ respectively.

Substitute the volume and molarity in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{1.75\text{M}\times 250.0\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.4375\text{moles}\end{array}$

Thus, the number of moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $0.4375\text{moles}$.

The amount of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated by the formula,

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

The molar mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $73.8\text{g}/\text{mol}$.

Substitute the value of molar mass in the given formula.

$\begin{array}{rll}0.4375\text{moles}& =& \frac{\text{Amount}\text{of}{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{73.8\text{g}/\text{mol}}\\ \text{Amount}\text{of}{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}& =& 0.4375\text{moles}\times 73.8\text{g}/\text{mol}\\ & =& 32.3\text{g}\end{array}$

Thus, the number of grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.}\text{mL}$ of $\text{1}\text{.75}\text{M}$${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $\text{32}\text{.3}\text{g}$.

Conclusion

The number of grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.}\text{mL}$ of $\text{1}\text{.75}\text{M}$${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $\text{32}\text{.3}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The number of moles of ${\text{NH}}_3$ in $\text{200}\text{.}\text{mL}$ of $\text{3}\text{.50}\text{M}$${\text{NH}}_3$ 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 7.46E

The number of moles of ${\text{NH}}_3$ in $\text{200}\text{.}\text{mL}$ of $\text{3}\text{.50}\text{M}$${\text{NH}}_3$ solution is $0.700\text{moles}$.

Explanation of Solution

The number of grams of ${\text{NH}}_3$ 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 $\text{200}\text{.}\text{mL}$ and $\text{3}\text{.50}\text{M}$ respectively.

Substitute the volume and molarity in the given formula.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{\text{3}\text{.50}\text{M}\times 200.\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.700\text{moles}\end{array}$

Thus, the number of moles of ${\text{NH}}_3$ is $0.700\text{moles}$.

Conclusion

The number of moles of ${\text{NH}}_3$ in $\text{200}\text{.}\text{mL}$ of $\text{3}\text{.50}\text{M}$${\text{NH}}_3$ solution is $0.700\text{moles}$.

Interpretation Introduction

(c)

Interpretation:

The number of mL of alcohol in $\text{250}\text{.}\text{mL}$ of $12.5\%\left(\text{v}/\text{v}\right)$ solution is to be predicted.

Concept Introduction:

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

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

Answer to Problem 7.46E

The number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $12.5\%\left(\text{v}/\text{v}\right)$ solution is $31.3\text{mL}$.

Explanation of Solution

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

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

The given value of $\%\left(\text{v}/\text{v}\right)$ is $12.5\%\left(\text{v}/\text{v}\right)$. 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).

$\begin{array}{rll}12.5& =& \frac{\text{volume}\text{of}\text{alcohol}}{250\text{mL}}\times 100\\ \text{volume}\text{of}\text{alcohol }& =& \frac{250\text{mL}\times 12.5}{100}\\ & =& 31.3\text{mL}\end{array}$

Hence, the number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $12.5\%\left(\text{v}/\text{v}\right)$ solution is $31.3\text{mL}$.

Conclusion

The number of milliliters of alcohol in $\text{250}\text{.}\text{mL}$ of a $12.5\%\left(\text{v}/\text{v}\right)$ solution is $31.3\text{mL}$.

Interpretation Introduction

(d)

Interpretation:

The number of grams of ${\text{CaCl}}_2$ in $\text{50}\text{.}\text{mL}$ of $4.20\%\left(\text{w}/\text{v}\right)$${\text{CaCl}}_2$ 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 7.46E

The number of grams of ${\text{CaCl}}_2$ in $\text{50}\text{.}\text{mL}$ of $4.20\%\left(\text{w}/\text{v}\right)$${\text{CaCl}}_2$ solution is $2.10\text{g}$.

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 $4.20\%\left(\text{w}/\text{v}\right)$. The volume of ${\text{CaCl}}_2$ solution is $\text{50}\text{.}\text{mL}$.

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

$\begin{array}{rll}4.20& =& \frac{\text{grams}\text{of}\text{solute}}{\text{50}\text{.}\text{L}}\times 100\\ \text{grams}\text{of}\text{solute}& =& \frac{4.20\times \text{50}\text{.}\text{L}}{100}\\ & =& 2.10\text{g}\end{array}$

Hence, the number of grams of ${\text{CaCl}}_2$ in $\text{50}\text{.}\text{mL}$ of $4.20\%\left(\text{w}/\text{v}\right)$${\text{CaCl}}_2$ solution is $2.10\text{g}$.

Conclusion

The number of grams of ${\text{CaCl}}_2$ in $\text{50}\text{.}\text{mL}$ of $4.20\%\left(\text{w}/\text{v}\right)$${\text{CaCl}}_2$ solution is $2.10\text{g}$.