Chapter 4, Problem 106QP

(a)

Interpretation Introduction

Interpretation:

The number of moles and mass of ${\text{CaCl}}_2$ are to be determined.

Explanation of Solution

The conversion of volume from milliliters to liters is given as follows:

$\text{Volume}\left(\text{L}\right)=x\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}$ …… (1)

The molarity of the solution is determined by the following expression:

$M=\frac{n}{V}$ …… (2)

Here, $n$ is the number of moles of solute, $V$ is the volume of the solution in liters, and $M$ is the molarity of the solution.

Substitute $\text{150}\text{.0 mL}$ for $x$ in equation (1) in order to determine the volume of ${\text{CaCl}}_2$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=150.0\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}\\ =0.1500\text{ L}\end{array}$

Rearrange equation (2) in terms of the number of moles.

$n=M\times V\mathrm{......}\text{ (3)}$

Substitute $0.1500\text{ L}$ for $V$ and $\text{0}{\text{.245 molL}}^{-1}$ for $M$ in equation (3) to determine the moles of ${\text{CaCl}}_2$ as follows:

$\begin{array}{c}n=\frac{\text{0}{\text{.245 mol CaCl}}_2}{\cancel{{\text{L CaCl}}_2}}\times 0.1500\cancel{{\text{L CaCl}}_2}\\ =\text{0}\text{.0368 mol }\end{array}$

The mass of solute is determined by the following expression:

$m=n\times M_m\mathrm{......}\text{ (4)}$

Here, $M_m$ is the molar mass of solute and $n$ is the number of moles of solute.

The molar mass of ${\text{CaCl}}_2$ is $\text{110}{\text{.98 gmol}}^{-1}$ .

Substitute $\text{110}{\text{.98 gmol}}^{-1}$ for $M_m$ and $\text{0}\text{.0368 mol }$ for $n$ in equation (4) in order to determine the mass of ${\text{CaCl}}_2$ as follows:

$\begin{array}{c}m=\text{0}\text{.0368 }\cancel{{\text{mol CaCl}}_2}\times \frac{\text{110}{\text{.98 g CaCl}}_2}{1\cancel{{\text{mol CaCl}}_2}}\\ =4.08\text{ g}\end{array}$

Therefore, the number of moles and mass of ${\text{CaCl}}_2$ are $\text{0}\text{.0368 mol}$ and $4.08\text{ g}$ , respectively.

(b)

Interpretation Introduction

Interpretation:

The number of moles and mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are to be determined.

Explanation of Solution

Substitute $\text{1450 mL}$ for $x$ in equation (1) in order to determine the volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=1450\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}\\ =1.45\text{ L}\end{array}$

Substitute $1.45\text{ L}$ for $V$ and $\text{0}{\text{.00187 molL}}^{-1}$ for $M$ in equation (3) to determine the moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ as follows:

$\begin{array}{c}n=\frac{\text{0}{\text{.00187 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\cancel{{\text{L H}}_{\text{2}}{\text{SO}}_{\text{4}}}}\times 1.45\cancel{{\text{L H}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ =\text{0}\text{.00271 mol }\end{array}$

The molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{98}{\text{.08 gmol}}^{-1}$ .

Substitute $\text{98}{\text{.08 gmol}}^{-1}$ for $M_m$ and $\text{0}\text{.00271 mol}$ for $n$ in equation (4) in order to determine the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ as follows:

$\begin{array}{c}m=\text{0}\text{.00271}\cancel{{\text{mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}\times \frac{\text{98}{\text{.08 H}}_{\text{2}}{\text{SO}}_{\text{4}}}{1\cancel{{\text{mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}}\\ =0.266\text{ g}\end{array}$

Therefore, the number of moles and mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are $\text{0}\text{.00271 mol}$ and $0.266\text{ g}$ , respectively.