Problem 40 - Chemistry The Central Science

Question

(a) What is the mass percentage of iodine \(\left(\mathrm{I}_{2}\right)\) in a solution containing \(0.035 \mathrm{~mol} \mathrm{I}_{2}\) in \(125 \mathrm{~g}\) of \(\mathrm{CCl}_{4} ?\) (b) Seawater contains \(0.0079 \mathrm{~g} \mathrm{Sr}^{2+}\) per kilogram of water. What is the concentration of \(\mathrm{Sr}^{2+}\) measured in ppm?

Step-by-Step Solution

Answer

(a) Mass percentage of iodine in the solution: Step 1: Mass of I₂ = \(0.035 \text{ mol} \times 253.8 \frac{\text{g}}{\text{mol}} = 8.883 \text{ g}\) Step 2: Total mass = 8.883 g I₂ + 125 g CCl₄ = 133.883 g Step 3: Mass percentage = \(\frac{8.883 \text{ g}}{133.883 \text{ g}} \times 100\% = 6.64 \% \) (b) Concentration of Sr²⁺ in seawater: Step 1: Mass of water = 1 kg × 1000 \( \frac{\text{g}}{\text{kg}} = 1000 \text{ g}\) Step 2: Concentration in ppm = \(\frac{0.0079 \text{ g Sr}^{2+}}{1000 \text{ g water}} \times 1,000,000 \ \text{ppm} = 7.9 \ \text{ppm}\)

Step 1: Calculate the mass of I₂

To find the mass of I₂, we will multiply the moles of I₂ given (0.035 mol) by the molar mass of I₂ (253.8 g/mol): Mass of I₂ = moles of I₂ × molar mass of I₂ Mass of I₂ = \(0.035 \text{ mol} \times 253.8 \frac{\text{g}}{\text{mol}}\)

Step 2: Find the total mass of the solution

The total mass of the solution is the sum of the masses of I₂ and CCl₄ Total mass = mass of I₂ + mass of CCl₄

Step 3: Calculate the mass percentage of I₂

Now, we will use the formula to find the mass percentage of I₂: Mass percentage = \(\frac{\text{mass of I₂}}{\text{total mass of solution}} \times 100\% \) (b) Concentration of Sr²⁺ in seawater

Step 1: Convert mass of water to grams

First, we will convert the mass of water from kilograms to grams: Mass of water = 1 kg × 1000 \( \frac{\text{g}}{\text{kg}}\)

Step 2: Calculate concentration in ppm

Now, to find the concentration of Sr²⁺ in ppm, use the formula: Concentration in ppm = \(\frac{\text{mass of Sr}^{2+}}{\text{total mass of water}} \times 1,000,000 \ \text{ppm}\)

Key Concepts

Molar Mass

Understanding molar mass is key to solving problems in chemistry involving moles and mass. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule.

For diatomic iodine, \( ext{I}_2\), each iodine atom has an atomic mass of approximately 126.9 u (atomic mass units). Therefore, the molar mass of \( ext{I}_2\) is calculated as follows:

• Add the atomic masses of the two iodine atoms: \[ 126.9 \text{ u} + 126.9 \text{ u} = 253.8 \text{ u} \]
• Express the result in grams per mole: \(253.8 \text{ g/mol}\)

Molar mass acts as a conversion factor between moles and grams. This is useful in determining the amount of a substance in a chemical equation when given its mass, as shown in the calculation of the mass of \(0.035 \text{ moles}\) of \(\text{I}_2\).

By multiplying the number of moles by the molar mass, you can find the total mass, which is crucial for further calculations such as determining mass percentages.

Concentration

Concentration describes how much of a substance is present in a mixture. It tells us the quantity of solute (here, \(\text{Sr}^{2+}\)) dissolved in a solvent (like water). Concentration is commonly expressed in several forms including molarity, percent composition, and parts per million (ppm).

In our example, the mass of \(\text{Sr}^{2+}\) in seawater is given as 0.0079 grams per kilogram of water. To express this concentration in ppm:

• We convert the mass of water into grams, as concentration in ppm accounts for grams of solute per million grams of solution.
• 1 kilogram of water equals 1000 grams.
• Thus, understanding the concentration of \(\text{Sr}^{2+}\) as 7.9 ppm, meaning there are 7.9 grams of \(\text{Sr}^{2+}\) per one million grams of seawater.

Expressing concentrations in ppm is particularly useful for very dilute solutions, where the amount of solute is much smaller compared to the solvent.

ppm (parts per million)

Concentration can be measured in many ways, with parts per million (ppm) being an effective unit for expressing small concentrations. PPM stands for parts per million, indicating how many parts of a solute are present in one million parts of a solution.

To calculate ppm, especially useful in environmental and chemical sciences, you use the following formula:

• \[ \text{ppm} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 1,000,000 \]

This method is advantageous in detecting trace levels of contaminants in water, air, and soil. Take for example, understanding the concentration of \(\text{Sr}^{2+}\) in seawater as 7.9 ppm means that for every million parts of seawater, there are 7.9 parts of \(\text{Sr}^{2+}\).

PPM simplifies the expression of concentrations, especially when the value is so small it would be cumbersome to use standard percentage forms. This is why ppm is extensively used in measuring pollutants, impurities, and components of mixtures in various scientific analyses.