Problem 40 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) The mass percentage of iodine in the solution is approximately 6.64%. (b) The concentration of Sr2+ in seawater is 7.9 ppm.

Step 1: a: Finding the mass of I2 in the solution

First, let's find the mass of I2 in the solution. The molar mass of I2 is approximately 253.8 g/mol. mass of I2 = moles x molar mass mass of I2 = 0.035 mol x 253.8 g/mol mass of I2 = 8.883 g

Step 2: a: Calculating the mass percentage of I2 in the solution

Now, we can calculate the mass percentage of I2 in the solution using the given mass of carbon tetrachloride. mass percentage = (mass of solute / total mass of solution) x 100% First, let's find the total mass of the solution. total mass of solution = mass of I2 + mass of CCl4 total mass of solution = 8.883 g + 125 g total mass of solution = 133.883 g Now, we can calculate the mass percentage: mass percentage = (8.883 g / 133.883 g) x 100% mass percentage ≈ 6.64 %

Step 3: b: Finding the concentration of Sr2+ in seawater

Since the given mass of Sr2+ is 0.0079 g per kilogram of water, we need to convert the mass of solution to grams. mass of solution = 1 kg x 1000 g/kg mass of solution = 1000 g Now, we can calculate the concentration of Sr2+ ions in seawater: concentration (ppm) = (mass of solute / mass of solution) x 10^6 concentration (ppm) = (0.0079 g / 1000 g) x 10^6 concentration (ppm) = 7.9 ppm

Key Concepts

Mass Percentage

Understanding the mass percentage of a component in a solution is a fundamental aspect of chemistry. It represents the mass of the solute divided by the total mass of the solution, multiplied by 100. This calculation gives us a percentage which shows how much of the solution is made up of the solute in question.

Let's simplify this with an example. Suppose we dissolve a certain amount of substance (the solute) in a solvent. If the solute mass is 5 grams and the total solution mass, which includes the solute and the solvent, is 100 grams, then the mass percentage of the solute is given by \((5\, g / 100\, g) \times 100\% = 5\%\). This value is essential in preparing solutions in laboratories and various industries, ensuring precise concentrations for reactions or products.

Molar Mass

The molar mass of a substance is the weight of one mole of that substance. It is usually expressed in grams per mole (g/mol) and can be thought of as the bridge that connects mass and moles in chemical equations. To find the molar mass, one must sum up the atomic weights of all the atoms in a molecule.

For example, water (H₂O) has a molar mass of approximately 18.015 g/mol because it has two hydrogen atoms (each with an atomic weight of about 1.008 g/mol) and one oxygen atom (with an atomic weight of about 15.999 g/mol). This concept is critical in stoichiometry as it allows for conversions between moles of different substances in a chemical reaction.

Parts Per Million (ppm)

Parts per million (ppm) is a measurement of the concentration of a solution that denotes the number of parts of a solute per one million parts of the solution. It is often used for very dilute solutions where the solute concentration is too small to be expressed as a percentage.

To illustrate, if we have 1 gram of salt dissolved in a large tank containing 1 million grams of water, the concentration of salt would be 1 ppm. This unit is also commonly employed in measuring pollutant levels in air and water, where precision is vital for safety and regulatory purposes. The smaller the ppm value, the less the amount of solute present in the solution.

Stoichiometry

Stoichiometry is the area of chemistry that relates to the quantitative relationships and conversions based on the laws of conservation of mass and the stoichiometric coefficients from balanced chemical equations. It involves calculations that ensure the correct amounts of reactants and products are used or formed during a chemical reaction.

For instance, consider the reaction between hydrogen and oxygen to form water, where the balanced equation is \(\text{2H}_{2} + \text{O}_{2} \rightarrow \text{2H}_{2}\text{O}\). If we start with 2 moles of hydrogen, stoichiometry teaches us that we need 1 mole of oxygen to produce 2 moles of water, embracing the principle that matter is neither created nor destroyed in the reaction.