Chapter 11, Problem 11.68QP

Interpretation Introduction

Interpretation: The concentration of different ions that are present in river water is given. The molality of these six different ions is to be calculated.

Concept introduction: Molality is defined as the number of moles of solute that is present in $1\text{kg}$ of solvent.

To determine: The molality of given six different ions that are present in the river water.

Answer to Problem 11.68QP

Solution

The concentration of ${\text{Al}}^{3+}$ is $\underset{\_}{0.19\times 10^{-5}mol/kg}$.

The concentration of ${\text{Fe}}^{3+}$ is $\underset{\_}{0.07\times 10^{-5}mol/kg}$.

The concentration of ${\text{Ca}}^{2+}$ is $\underset{\_}{3.3\times 10^{-4}mol/kg}$.

The concentration of ${\text{Na}}^{+}$ is $\underset{\_}{2.3\times 10^{-4}mol/kg}$.

The concentration of ${\text{K}}^{+}$ is $\underset{\_}{3.3\times 10^{-5}mol/kg}$.

The concentration of ${\text{Mg}}^{2+}$ is $\underset{\_}{1.4\times 10^{-4}mol/kg}$.

Explanation of Solution

Explanation

Given

The concentration of ${\text{Al}}^{3+}$ is $0.050\text{mg}/\text{kg}$.

The concentration of ${\text{Fe}}^{3+}$ is $0.040\text{mg}/\text{kg}$.

The concentration of ${\text{Ca}}^{2+}$ is $13.4\text{mg}/\text{kg}$.

The concentration of ${\text{Na}}^{+}$ is $5.2\text{mg}/\text{kg}$.

The concentration of ${\text{K}}^{+}$ is $1.3\text{mg}/\text{kg}$.

The concentration of ${\text{Mg}}^{2+}$ is $3.4\text{mg}/\text{kg}$.

The mass of solvent is $1.0\text{kg}$.

The conversion of $\text{mg}$ into $\text{g}$ is done as,

$1\text{mg}=0.001\text{g}$

Hence, the conversion of $0.050\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}0.050\text{mg}=0.050\times 0.001\text{g}\\ =5.0\times {10}^{-5}\text{g}\end{array}$

Similarly, the conversion of $0.040\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}0.040\text{mg}=0.040\times 0.001\text{g}\\ =4.0\times {10}^{-5}\text{g}\end{array}$

Similarly, the conversion of $13.4\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}13.4\text{mg}=13.4\times 0.001\text{g}\\ =0.0134\text{g}\end{array}$

Similarly, the conversion of $5.2\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}5.2\text{mg}=5.2\times 0.001\text{g}\\ =0.0052\text{g}\end{array}$

Similarly, the conversion of $1.3\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}1.3\text{mg}=1.3\times 0.001\text{g}\\ =0.0013\text{g}\end{array}$

Similarly, the conversion of $3.4\text{mg}$ into $\text{g}$ is done as,

$\begin{array}{c}3.4\text{mg}=3.4\times 0.001\text{g}\\ =0.0034\text{g}\end{array}$

The molar mass of Aluminum is $26.98\text{g}/\text{mol}$.

The molar mass of Iron is $55.85\text{g}/\text{mol}$.

The molar mass of Calcium is $40.08\text{g}/\text{mol}$.

The molar mass of Sodium is $22.99\text{g}/\text{mol}$.

The molar mass of Potassium is $39.10\text{g}/\text{mol}$.

The molar mass of Magnesium is $24.31\text{g}/\text{mol}$.

The number of moles of a substance is calculated as,

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Substitute the value of given mass and molar mass of ${\text{Al}}^{3+}$ in the above equation as,

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{5.0\times {10}^{-5}\text{g}}{26.98\frac{\text{g}}{\text{mol}}}\\ =0.19\times {10}^{-5}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{Fe}}^{3+}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{4.0\times {10}^{-5}\text{g}}{55.85\frac{\text{g}}{\text{mol}}}\\ =0.07\times {10}^{-5}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{Ca}}^{2+}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.0134\text{g}}{40.08\frac{\text{g}}{\text{mol}}}\\ =3.3\times {10}^{-4}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{Na}}^{+}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.0052\text{g}}{22.99\frac{\text{g}}{\text{mol}}}\\ =2.3\times {10}^{-4}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{K}}^{+}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.0013\text{g}}{39.10\frac{\text{g}}{\text{mol}}}\\ =3.3\times {10}^{-5}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{Mg}}^{2+}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.0034\text{g}}{24.31\frac{\text{g}}{\text{mol}}}\\ =1.4\times {10}^{-4}\text{mol}\end{array}$

The molality of the solution is calculated as,

$\text{Molality}=\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$ (2)

Substitute the number of moles of ${\text{Al}}^{3+}$ and mass of solvent in the above equation,

$\begin{array}{c}\text{Molality}=\frac{0.19\times {10}^{-5}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{0.19\times 10^{-5}mol/kg}\end{array}$

The concentration of ${\text{Al}}^{3+}$ is $\underset{\_}{0.19\times 10^{-5}mol/kg}$.

Similarly, substitute the number of moles of ${\text{Fe}}^{3+}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{0.07\times {10}^{-5}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{0.07\times 10^{-5}mol/kg}\end{array}$

The concentration of ${\text{Fe}}^{3+}$ is $\underset{\_}{0.07\times 10^{-5}mol/kg}$.

Similarly, substitute the number of moles of ${\text{Ca}}^{2+}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{3.3\times {10}^{-4}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{3.3\times 10^{-4}mol/kg}\end{array}$

The concentration of ${\text{Ca}}^{2+}$ is $\underset{\_}{3.3\times 10^{-4}mol/kg}$.

Similarly, substitute the number of moles of ${\text{Na}}^{+}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{2.3\times {10}^{-4}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{2.3\times 10^{-4}mol/kg}\end{array}$

The concentration of ${\text{Na}}^{+}$ is $\underset{\_}{2.3\times 10^{-4}mol/kg}$.

Similarly, substitute the number of moles of ${\text{K}}^{+}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{3.3\times {10}^{-5}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{3.3\times 10^{-5}mol/kg}\end{array}$

The concentration of ${\text{K}}^{+}$ is $\underset{\_}{3.3\times 10^{-5}mol/kg}$.

Similarly, substitute the number of moles of ${\text{Mg}}^{2+}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{1.4\times {10}^{-4}\text{mol}}{\text{1}\text{kg}}\\ =\underset{\_}{1.4\times 10^{-4}mol/kg}\end{array}$

The concentration of ${\text{Mg}}^{2+}$ is $\underset{\_}{1.4\times 10^{-4}mol/kg}$.

Conclusion

The concentration of ${\text{Al}}^{3+}$ is $\underset{\_}{0.19\times 10^{-5}mol/kg}$.

The concentration of ${\text{Fe}}^{3+}$ is $\underset{\_}{0.07\times 10^{-5}mol/kg}$.

The concentration of ${\text{Ca}}^{2+}$ is $\underset{\_}{3.3\times 10^{-4}mol/kg}$.

The concentration of ${\text{Na}}^{+}$ is $\underset{\_}{2.3\times 10^{-4}mol/kg}$.

The concentration of ${\text{K}}^{+}$ is $\underset{\_}{3.3\times 10^{-5}mol/kg}$.

The concentration of ${\text{Mg}}^{2+}$ is $\underset{\_}{1.4\times 10^{-4}mol/kg}$.