Chapter 16, Problem 94A

Interpretation Introduction

Interpretation: The mass of each ion present in the given table is to be calculated from its molarity.

Concept introduction: The molarity of a solution is defined as the mass of solute per liter of solution. The formula that can be used for calculating the molarity involves the ratio of the number of moles of the given solute and the volume of solution in liters.

That is,

The formula to calculate the molarity of the solution is given as:

  $\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}$   ...... (1)

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (2)

Answer to Problem 94A

Mass of chloride is $54\text{g}$ .

Mass of sodium is $0.470$ .

Mass of magnesium is $6.3\text{g}$ .

Mass of calcium is $2.06\text{g}$ .

Mass of potassium is $1.98\text{g}$ .

Explanation of Solution

Given:

Molarity of chloride is $0.546$ .

Molarity of sodium is $0.470$ .

Molarity of magnesium is $0.053$ .

Molarity of calcium is $0.0103$ .

Molarity of potassium is $0.0102$ .

Volume of solution is $5\text{L}$ .

The molar mass of chloride is 35.

The molecular mass of sodium is 23.

Substitute $0.546$ for molarity of chloride moles and $5\text{L}$ for the volume of solution in equation (1) to calculate moles of chloride.

  $\begin{array}{c}\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}\\ 0.546=\frac{\text{Moles solute}}{\text{5}\text{L}}\\ \text{Moles solute}=5\times 0.546\\ =2.73\text{moles}\end{array}$

Substitute, $35$ for molar mass of chloride and $2.73\text{moles}$ for moles of chloride in equation (2) to calculate the mass of chloride.

  $\begin{array}{c}\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}\\ 2.73\text{moles}=\frac{\text{ mass of chloride}}{35}\\ \text{mass of chloride}=2.73\times 35\\ =95.5g\end{array}$

Substitute $0.470$ for the molarity of sodium moles and $5\text{L}$ for the volume of solution in equation (1) to calculate moles of sodium.

  $\begin{array}{c}\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}\\ 0.470=\frac{\text{Moles solute}}{\text{5}\text{L}}\\ \text{Moles solute}=5\times 0.470\\ =2.35\text{moles}\end{array}$

Substitute, $23$ for molar mass of sodium and $2.35\text{moles}$ for moles of chloride in equation (2) to calculate the mass of sodium.

  $\begin{array}{c}\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}\\ 2.35\text{moles}=\frac{\text{ mass of sodium}}{23}\\ \text{mass of sodium}=2.35\times 23\\ =54g\end{array}$

Substitute $0.053$ for molarity of magnesium and $5\text{L}$ for the volume of solution in equation (1) to calculate moles of sodium.

  $\begin{array}{c}\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}\\ 0.053=\frac{\text{Moles solute}}{\text{5}\text{L}}\\ \text{Moles solute}=5\times 0.053\\ =0.265\text{moles}\end{array}$

Substitute, $24$ for molar mass of magnesium and $0.265\text{moles}$ for moles of chloride in equation (2) to calculate the mass of magnesium.

  $\begin{array}{c}\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}\\ 0.265\text{moles}=\frac{\text{ mass of magnesium}}{24}\\ \text{mass of magnesium}=0.265\times 24\\ =6.3g\end{array}$

Substitute $0.0103$ for the molarity of calcium and $5\text{L}$ for the volume of solution in equation (1) to calculate moles of calcium.

  $\begin{array}{c}\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}\\ 0.0103=\frac{\text{Moles solute}}{\text{5}\text{L}}\\ \text{Moles solute}=5\times 0.0103\\ =0.0515\text{moles}\end{array}$

Substitute, $40$ for molar mass of calcium and $0.0515\text{moles}$ for moles of calcium in equation (2) to calculate the mass of calcium.

  $\begin{array}{c}\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}\\ 0.0515\text{moles}=\frac{\text{ mass of calcium}}{40}\\ \text{mass of calcium}=0.0515\times 40\\ =2.06g\end{array}$

Substitute $0.0102$ for molarity of potassium and $5\text{L}$ for the volume of solution in equation (1) to calculate moles of potassium.

  $\begin{array}{c}\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume Litre solution}}\\ 0.0102=\frac{\text{Moles solute}}{\text{5}\text{L}}\\ \text{Moles solute}=5\times 0.0102\\ =0.051\text{moles}\end{array}$

Substitute, $39$ for molar mass of potassium and $0.051\text{moles}$ for moles of potassium in equation (2) to calculate the mass of potassium.

  $\begin{array}{c}\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}\\ 0.051\text{moles}=\frac{\text{ mass of potassium}}{39}\\ \text{mass of potassium}=0.051\times 39\\ =1.98g\end{array}$

Conclusion

Mass of chloride is $54\text{g}$ .

Mass of sodium is $0.470$ .

Mass of magnesium is $6.3\text{g}$ .

Mass of calcium is $2.06\text{g}$ .

Mass of potassium is $1.98\text{g}$ .