Chapter 14, Problem 92AE

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ present in $45.8\text{ mL}$ of solution have to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Expression to calculate molarity is as follows:

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

Explanation of Solution

Expression to calculate molarity of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{Volume }\left(\text{L}\right){\text{ of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}}$        (1)

Rearrange equation (1) for moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

  ${\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\begin{array}{c}\left({\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\\ \left({\text{Volume of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}\right)\end{array}\right]$        (2)

Substitute $0.25M$ for molarity and $\text{45}\text{.8 mL}$ for volume of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution in equation (2).

  $\begin{array}{c}{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(0.25M\right)\left(\text{45}\text{.8 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.01145\text{ mol}\end{array}$

Hence, moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ produced are $0.01145\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

Grams of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ present in $750\text{ mL}$ of solution have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression to calculate molarity of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{Volume }\left(\text{L}\right){\text{ of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}}$        (1)

Rearrange equation (1) for moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

  ${\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\begin{array}{c}\left({\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\\ \left({\text{Volume of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}\right)\end{array}\right]$        (2)

Substitute $0.25M$ for molarity and $750\text{ mL}$ for volume of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution in equation (2).

  $\begin{array}{c}{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(0.25M\right)\left(750\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.1875\text{ mol}\end{array}$

Expression for mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Mass of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\begin{array}{c}\left({\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\\ \left({\text{Molar mass of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\end{array}\right]$        (3)

Substitute $0.1875\text{ mol}$ for moles and $\text{73}\text{.89 g/mol}$ for molar mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (3).

  $\begin{array}{c}{\text{Mass of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(0.1875\text{ mol}\right)\left(\text{73}\text{.89 g/mol}\right)\\ =13.85\text{ g}\end{array}$

Hence, mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ produced is $13.85\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

Volume $\left(\text{mL}\right)$ of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution that supply $6.0\text{ g}$ of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression to calculate molarity of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{Volume }\left(\text{L}\right){\text{ of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}}$        (1)

Rearrange equation (1) for volume of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

  ${\text{Volume of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}=\frac{{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{Molarity of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}$        (3)

Expression for moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{{\text{Mass of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{Molar mass of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}$        (4)

Substitute $6.0\text{ g}$ for mass and $\text{73}\text{.89 g/mol}$ for molar mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (4).

  $\begin{array}{c}{\text{Moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{\text{6}\text{.0 g}}{\text{73}\text{.89 g/mol}}\\ =0.0812\text{ mol}\end{array}$

Substitute $0.0812\text{ mol}$ for moles and $0.25M$ for molarity of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution in equation (3).

  $\begin{array}{c}{\text{Volume of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}=\left(\frac{0.0812\text{ mol}}{\text{0}\text{.25 }M}\right)\left(\frac{{10}^3\text{ mL}}{1\text{ L}}\right)\\ =324.8\text{ mL}\end{array}$

Hence, volume of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $324.8\text{ mL}$.

(d)

Interpretation Introduction

Interpretation:

Mass percent has to be determined if density of solution is $1.22\text{ g/mL}$.

Concept Introduction:

Mass percent is concentration term used to determine concentration of solution in terms of solute percent in particular mass of solution. Expression for mass percent is as follows:

  $\text{Mass percent}=\left(\frac{\text{Mass of solute}}{\text{Mass of solution}}\right)\left(100\right)$

Explanation of Solution

Expression for mass percent of ${\text{Li}}_2{\text{CO}}_3$ is as follows:

  ${\text{Mass percent of Li}}_2{\text{CO}}_3=\left(\frac{{\text{Mass of Li}}_2{\text{CO}}_3}{\text{Mass of solution}}\right)\left(100\right)$        (5)

Expression for mass of solution is as follows:

  $\text{Mass of solution}=\left[\begin{array}{c}\left(\text{Density of solution}\right)\\ \left(\text{Volume of solution}\right)\end{array}\right]$        (6)

Substitute $1.22\text{ g/mL}$ for density and $324.8\text{ mL}$ for volume of solution in equation (6).

  $\begin{array}{c}\text{Mass of solution}=\left(\text{1}\text{.22 g/mL}\right)\left(324.8\text{ mL}\right)\\ =396.256\text{ g}\end{array}$

Substitute $\text{6}\text{.0 g}$ for mass of ${\text{Li}}_2{\text{CO}}_3$ and $396.256\text{ g}$ for mass of solution in equation (5).

  $\begin{array}{c}{\text{Mass percent of Li}}_2{\text{CO}}_3=\left(\frac{\text{6}\text{.0 g}}{396.256\text{ g}}\right)\left(100\right)\\ =1.51\text{ \%}\end{array}$

Hence, mass percent of ${\text{Li}}_2{\text{CO}}_3$ is $\text{1}\text{.51 \%}$.