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 \(115 \mathrm{~g}\) 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?

Short answer

(a) The mass percentage of iodine \(\left(\mathrm{I}_{2}\right)\) in the solution is 7.17%. (b) The concentration of \(\mathrm{Sr}^{2+}\) in seawater is 7.9 ppm.

Step-by-step solution

Part (a) - Calculating the mass percentage of iodine \(\left(\mathrm{I}_{2}\right)\) in the solution.

To find the mass percentage of iodine in the solution, we'll first need to calculate the mass of iodine present in the solution and the mass of the entire solution, and then find the percentage. Step 1: Calculate the mass of iodine present in the solution. Given, the solution contains \(0.035\) moles of \(\mathrm{I}_{2}\). To find the mass, we'll multiply the number of moles with the molar mass of \(\mathrm{I}_{2}\). Molar mass of \(\mathrm{I}_{2}\) = 2 × (Molar mass of \(\mathrm{I}\)) = 2 × 126.90 g/mol = 253.80 g/mol Mass of \(\mathrm{I}_{2}\) = Number of moles × Molar mass = \(0.035 \mathrm{~ mol} \times 253.80 \mathrm{~ g/mol} \) = 8.883 g Step 2: Calculate the mass of the entire solution. The mass of CCl4 in the solution is given as 115 g. So, to find the mass of the entire solution, we'll add the mass of iodine to the mass of CCl4. Mass of the solution = Mass of \(\mathrm{I}_{2}\) + Mass of CCl4 = 8.883 g + 115 g = 123.883 g Step 3: Calculate the mass percentage of iodine in the solution. Mass percentage of \(\mathrm{I}_{2}\) = \(\frac{\text{Mass of }\mathrm{I}_{2}}{\text{Mass of solution}} \times 100\) Mass percentage of \(\mathrm{I}_{2}\) = \(\frac{8.883 \mathrm{~g}}{123.883 \mathrm{~g}} \times 100\) = 7.17% Thus, the mass percentage of iodine \(\left(\mathrm{I}_{2}\right)\) in the solution is 7.17%.

Part (b) - Calculating the concentration of \(\mathrm{Sr}^{2+}\) in seawater in ppm.

Given that seawater contains \(0.0079 \mathrm{~g} \mathrm{Sr}^{2+}\) per kilogram of water, we need to calculate the concentration of \(\mathrm{Sr}^{2+}\) in parts per million (ppm). Step 1: Convert kilograms to grams. 1 kg (kilogram) of water = 1000 g, so given concentration = 0.0079 g of Sr++ per 1000 g of water. Step 2: Convert the concentration to ppm (parts per million). To represent a concentration in ppm, we essentially need to express it as a ratio of the solute (in this case, Sr++) mass to the total mass of the solution (in grams) and then multiply by one million. Concentration of \(\mathrm{Sr}^{2+}\) in ppm = \(\frac{\text{Mass of Sr}^{2+}} {\text{Total mass of water}} \times 1,000,000\) Concentration of \(\mathrm{Sr}^{2+}\) in ppm = \(\frac{0.0079 \mathrm{~g}}{{1000 \mathrm{~g}}} \times 1,000,000\) = 7.9 ppm Thus, the concentration of \(\mathrm{Sr}^{2+}\) in seawater is 7.9 ppm.

Key concepts

Solution Concentration

When dealing with solutions, understanding concentration is key. Solution concentration tells us how much solute is present in a given amount of solvent or solution. There are various ways to express concentration, such as mass percentage, molarity, and parts per million (ppm).

Mass percentage, for example, is used when you want to know what fraction of a solution's weight is due to the solute. It's calculated by dividing the mass of the solute by the total mass of the solution, then multiplying by 100 to get a percentage. This method is handy when working with solid solutes like iodine in a heavy solvent like carbon tetrachloride ( CCl_4 ).

Molar Mass

Molar mass is a fundamental concept used to calculate the weight of molecules. It represents the mass of one mole of a substance. To find the molar mass, you sum up the atomic masses of all atoms in a molecule. Consider iodine ( I_2 ) in our problem:

• Each iodine atom has a molar mass of about 126.90 g/mol.
• Since iodine is diatomic ( I_2 ), the molar mass is calculated as 2 × 126.90 g/mol, equaling 253.80 g/mol.

This value helps us convert moles to grams, a crucial step in determining the mass percentage of iodine in a solution.

Parts Per Million (ppm)

Parts per million (ppm) is a concentration unit useful for very dilute solutions, like trace elements in water. It expresses the number of parts of solute per million parts of solution, making it excellent for environmental and chemical analysis.

To calculate ppm, use the equation:\[ \text{ppm} = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 1,000,000\]
In our example, ppm is used to express the concentration of Sr^{2+} ions in seawater. When we say 7.9 ppm, it means there are 7.9 grams of Sr^{2+} per million grams of seawater.

Chemical Calculations

Chemical calculations form the backbone of solution concentration and molar mass determinations. They help in accurately determining amounts needed for reactions, and concentrations in mixtures.

Key to these calculations is the relationship between moles, mass, and molar mass:

• Mass (g) = Moles × Molar Mass (g/mol)

This formula allows conversion between moles and grams, crucial for calculating things like mass percent.

For ppm, expressing small amounts in large totals involves using the \(1,000,000\) multiplier, reflecting the minute concentration of solutes like Sr^{2+} in water. This nuanced understanding aids both in everyday lab experiments and large-scale industrial applications.