Problem 68 - Chemistry The Science in Context

Question

The concentrations of six important elements in a sample of river water are \(0.050 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{Al}^{3+}, 0.040 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{Fe}^{3+}, 13.4 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{Ca}^{2+}, 5.2 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{Na}^{+}, 1.3 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{K}^{+}\), and \(3.4 \mathrm{mg} / \mathrm{kg}\) of \(\mathrm{Mg}^{2+} .\) Express each of these concentrations in molality units.

Step-by-Step Solution

Answer

Answer: The molality values for the different elements are: - Al³⁺: 1.85×10⁻⁶ mol/kg - Fe³⁺: 7.16×10⁻⁷ mol/kg - Ca²⁺: 3.35×10⁻⁴ mol/kg - Na⁺: 2.26×10⁻⁴ mol/kg - K⁺: 3.32×10⁻⁵ mol/kg - Mg²⁺: 1.40×10⁻⁴ mol/kg

Step 1: Convert mass from mg to grams

To convert the mass from mg to grams, divide the mass value by 1000. \(0.050 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 5.0\times10^{-5} \mathrm{g}\) of \(\mathrm{Al}^{3+}\) \(0.040 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 4.0\times10^{-5} \mathrm{g}\) of \(\mathrm{Fe}^{3+}\) \(13.4 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 0.0134 \mathrm{g}\) of \(\mathrm{Ca}^{2+}\) \(5.2 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 0.0052 \mathrm{g}\) of \(\mathrm{Na}^{+}\) \(1.3 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 0.0013 \mathrm{g}\) of \(\mathrm{K}^{+}\) \(3.4 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} = 0.0034 \mathrm{g}\) of \(\mathrm{Mg}^{2+}\)

Step 2: Determine molar masses

The molar masses of the elements are as follows: \(\mathrm{Al}^{3+}\): 26.98 g/mol \(\mathrm{Fe}^{3+}\): 55.85 g/mol \(\mathrm{Ca}^{2+}\): 40.08 g/mol \(\mathrm{Na}^{+}\): 22.99 g/mol \(\mathrm{K}^{+}\): 39.10 g/mol \(\mathrm{Mg}^{2+}\): 24.31 g/mol

Step 3: Calculate molality

To calculate molality, divide step 1 value (grams) by step 2 value (molar mass) then divide by the mass of solvent (1 kg) in mol/kg units. \(\mathrm{Al}^{3+}\): \(\frac{5.0\times10^{-5} \mathrm{g}}{26.98 \mathrm{g/mol}}= 1.85\times10^{-6} \mathrm{mol}\) and \(1.85\times10^{-6} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 1.85\times10^{-6} \mathrm{mol/kg}\) \(\mathrm{Fe}^{3+}\): \(\frac{4.0\times10^{-5} \mathrm{g}}{55.85 \mathrm{g/mol}}= 7.16\times10^{-7} \mathrm{mol}\) and \(7.16\times10^{-7} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 7.16\times10^{-7} \mathrm{mol/kg}\) \(\mathrm{Ca}^{2+}\): \(\frac{0.0134 \mathrm{g}}{40.08 \mathrm{g/mol}}= 3.35\times10^{-4} \mathrm{mol}\) and \(3.35\times10^{-4} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 3.35\times10^{-4} \mathrm{mol/kg}\) \(\mathrm{Na}^{+}\): \(\frac{0.0052 \mathrm{g}}{22.99 \mathrm{g/mol}}= 2.26\times10^{-4} \mathrm{mol}\) and \(2.26\times10^{-4} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 2.26\times10^{-4} \mathrm{mol/kg}\) \(\mathrm{K}^{+}\): \(\frac{0.0013 \mathrm{g}}{39.10 \mathrm{g/mol}}= 3.32\times10^{-5} \mathrm{mol}\) and \(3.32\times10^{-5} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 3.32\times10^{-5} \mathrm{mol/kg}\) \(\mathrm{Mg}^{2+}\): \(\frac{0.0034 \mathrm{g}}{24.31 \mathrm{g/mol}}= 1.40\times10^{-4} \mathrm{mol}\) and \(1.40\times10^{-4} \mathrm{mol}\times\frac{1}{1 \mathrm{kg}} = 1.40\times10^{-4} \mathrm{mol/kg}\)

Key Concepts

Conversion of Units

Understanding the conversion of units is a crucial step in many chemistry calculations. In this exercise, the concentration of elements in river water is initially given in milligrams per kilogram (mg/kg). To work with these values in a molality context, we need to convert milligrams into grams. This is because the unit of molality is moles per kilogram.
The conversion factor for milligrams to grams is straightforward: 1 gram is equal to 1000 milligrams. Thus, to convert milligrams to grams, you divide the value in milligrams by 1000. For example, for the aluminum ion

• 0.050 mg is converted to 0.000050 grams by dividing by 1000.

This simple division helps standardize units for further calculations.

Molar Mass

The concept of molar mass is central to converting the mass of a substance to the number of moles, which is necessary for calculating molality. The molar mass of a substance is the mass of one mole of its atoms, ions, or molecules and is expressed in grams per mole (g/mol).

Every element has a specific molar mass; for instance, aluminum (\(Al^{3+}\) has a molar mass of 26.98 g/mol, and iron (\(Fe^{3+}\) has a molar mass of 55.85 g/mol.

• To find the number of moles, divide the mass in grams by the molar mass.

For example, if you have 5.0×10^-5 grams of aluminum, dividing by its molar mass of 26.98 g/mol gives the number of moles. This value will then be used in calculating the molality.

Solution Concentration Calculations

Solution concentration calculations, specifically molality, involve determining the amount of a solute compared to the mass of the solvent. Molality is expressed in moles per kilogram of solvent (mol/kg).

The calculation begins with determining the moles of the solute using its mass in grams and its molar mass. Next, molality is calculated by taking these moles and dividing by the mass of the solvent. In the example of river water, it's assumed there is 1 kg of solvent. Thus,

• Molality for each ion: divide moles of the ion by 1 kg of solvent.

The ease of this calculation hinges on having already converted masses and determined the moles of each solute in the solution.