Chapter 11, Problem 11.67QP

Interpretation Introduction

Interpretation: The concentration of six important elements that are present in river water is given. The molality of these six different elements 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 elements that are present in the river water.

Answer to Problem 11.67QP

Solution

The molality of Ammonia is $\underset{\_}{0.07\times 10^{-3}mol/kg}$.

The molality of ${\text{NO}}_2{}^{-}$ is $\underset{\_}{0.009\times 10^{-3}mol/kg}$.

The molality of ${\text{NO}}_3{}^{-}$ is $\underset{\_}{21.98\times 10^{-3}mol/kg}$.

Explanation of Solution

Explanation

Given

The lethal concentration of Ammonia is $1.1\text{mg}/\text{L}$.

The lethal concentration of Nitrite ion is $0.40\text{mg}/\text{L}$.

The lethal concentration of Nitrate ion is $1361\text{mg}/\text{L}$.

Density of solution is $1\text{g}/\text{mL}$.

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

$1\text{mg}=0.000001\text{g}$

Hence, the conversion of $1.1\text{mg}$ into $\text{kg}$ is done as,

$\begin{array}{c}1.1\text{mg}=1.1\times 0.000001\text{kg}\\ =1.1\times {10}^{-6}\text{kg}\end{array}$

Similarly, the conversion of $0.40\text{mg}$ into $\text{kg}$ is done as,

$\begin{array}{c}0.40\text{mg}=0.40\times 0.000001\text{kg}\\ =0.4\times {10}^{-6}\text{kg}\end{array}$

Similarly, the conversion of $1361\text{mg}$ into $\text{kg}$ is done as,

$\begin{array}{c}1361\text{mg}=1361\times 0.000001\text{kg}\\ =1361\times {10}^{-6}\text{kg}\end{array}$

The mass of ${\text{NH}}_3$ in $1\text{L}$ of solution is $1.1\times {10}^{-6}\text{kg}$.

The mass of Nitrite ion in $1\text{L}$ of solution is $0.4\times {10}^{-6}\text{kg}$.

The mass of Nitrate ion in $1\text{L}$ of solution is $1361\times {10}^{-6}\text{kg}$.

Total mass of solute $\begin{array}{c}=\text{Mass of Ammonia}+\text{Mass of Nitrite ion}+\text{Mass of Nitrate ion}\\ =1.1\times {10}^{-6}\text{kg}+0.4\times {10}^{-6}\text{kg}+1361\times {10}^{-6}\text{kg}\\ =1362.5\times {10}^{-6}\text{kg}\\ \text{=1}\text{.363}\times {10}^{-3}\text{kg}\end{array}$

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

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

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

The mass of solution is for $0.001\text{L}$ of the solution $={10}^{-3}\text{kg}$.

The mass of solution is for $1\text{L}$ of the solution $\begin{array}{c}=\frac{{10}^{-3}}{0.001}\text{kg}\\ \text{=1}\text{kg}\end{array}$

The total mass of solution is given as,

$\text{Total mass of solution}=\text{Mass of solvent}+\text{Mass of solute}$

Substitute the mass of solution and solvent in the above equation,

$\begin{array}{c}\text{Total mass of solution}=\text{Mass of solvent}+\text{Mass of solute}\\ 1\text{kg}=\text{Mass of solvent}+\text{1}\text{.363}\times {10}^{-3}\text{kg}\\ \text{Mass of solvent}=0.999\text{kg}\end{array}$

The molar mss of ${\text{NH}}_3$ is calculated as,

$\begin{array}{c}{\text{Molar mass of NH}}_3=\left(1\times \text{Molar mass of N}+3\times \text{Molar mass of H}\right)\\ =\left(1\times \text{14}\text{.0}\text{g/mol}+3\times \text{1}\text{.008}\text{g/mol}\right)\\ =17.024\text{g}/\text{mol}\end{array}$

The molar mss of ${\text{NO}}_2{}^{-}$ is calculated as,

$\begin{array}{c}{\text{Molar mass of NO}}_2{}^{-}=\left(1\times \text{Molar mass of N}+2\times \text{Molar mass of O}\right)\\ =\left(1\times \text{14}\text{.0}\text{g/mol}+2\times \text{15}\text{.99}\text{g/mol}\right)\\ =45.98\text{g}/\text{mol}\end{array}$

The molar mss of ${\text{NO}}_3{}^{-}$ is calculated as,

$\begin{array}{c}{\text{Molar mass of NO}}_3{}^{-}=\left(1\times \text{Molar mass of N}+3\times \text{Molar mass of O}\right)\\ =\left(1\times \text{14}\text{.0}\text{g/mol}+3\times \text{15}\text{.99}\text{g/mol}\right)\\ =61.97\text{g}/\text{mol}\end{array}$

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 Ammonia in the above equation as,

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{1.1\times {10}^{-6}\text{kg}}{17.024\times {10}^{-3}\frac{\text{kg}}{\text{mol}}}\\ =0.065\times {10}^{-3}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{NO}}_2{}^{-}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.4\times {10}^{-6}\text{kg}}{45.98\times {10}^{-3}\frac{\text{kg}}{\text{mol}}}\\ =0.0087\times {10}^{-3}\text{mol}\end{array}$

Similarly, substitute the value of given mass and molar mass of ${\text{NO}}_3{}^{-}$ in the equation (1),

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{1361\times {10}^{-6}\text{kg}}{61.97\times {10}^{-3}\frac{\text{kg}}{\text{mol}}}\\ =21.96\times {10}^{-3}\text{mol}\end{array}$

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 Ammonia and mass of solvent in the above equation,

$\begin{array}{c}\text{Molality}=\frac{0.065\times {10}^{-3}\text{mol}}{0.999\text{kg}}\\ =\underset{\_}{0.07\times 10^{-3}mol/kg}\end{array}$

The molality of Ammonia is $\underset{\_}{0.07\times 10^{-3}mol/kg}$.

Similarly, substitute the number of moles of ${\text{NO}}_2{}^{-}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{0.0087\times {10}^{-3}\text{mol}}{0.999\text{kg}}\\ =\underset{\_}{0.009\times 10^{-3}mol/kg}\end{array}$

The molality of ${\text{NO}}_2{}^{-}$ is $\underset{\_}{0.009\times 10^{-3}mol/kg}$.

Similarly, substitute the number of moles of ${\text{NO}}_3{}^{-}$ and mass of solvent in the equation (2),

$\begin{array}{c}\text{Molality}=\frac{21.96\times {10}^{-3}\text{mol}}{0.999\text{kg}}\\ =\underset{\_}{21.98\times 10^{-3}mol/kg}\end{array}$

The molality of ${\text{NO}}_3{}^{-}$ is $\underset{\_}{21.98\times 10^{-3}mol/kg}$.

Conclusion

The molality of Ammonia is $\underset{\_}{0.07\times 10^{-3}mol/kg}$.

The molality of ${\text{NO}}_2{}^{-}$ is $\underset{\_}{0.009\times 10^{-3}mol/kg}$.

The molality of ${\text{NO}}_3{}^{-}$ is $\underset{\_}{21.98\times 10^{-3}mol/kg}$.