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Chapter 13: Problem 40

(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 ppm?

Short Answer

Expert verified
The mass percentage of iodine in the solution is 7.11%, and the concentration of Sr2+ in seawater is 7.9 ppm.

Step by step solution

01

Calculate the molar mass of iodine (#I_2#)

Using the periodic table, we find that the atomic mass of iodine (I) is approximately 127 g/mol. Since there are two iodine atoms in a molecule of iodine, the molar mass of iodine (I2) is 2 * 127 = 254 g/mol.
02

Convert moles of iodine into mass

We know that the solution contains 0.035 mol of I2. To find the mass, we can use the formula: mass of iodine = moles * molar mass. So, mass of iodine = 0.035 mol * 254 g/mol = 8.89 g.
03

Calculate the mass percentage of iodine

The mass of the solution is 125 g, and the mass of iodine in the solution is 8.89 g. To find the mass percentage, we can use the formula: mass percentage = (mass of iodine / mass of solution) * 100 So, mass percentage = (8.89 g / 125 g) * 100 = 7.11% b) Finding the concentration of Sr2+ in seawater in ppm:
04

Convert the mass of Sr2+ per mass of seawater into the mass of Sr2+ per million mass of seawater

We know that seawater contains 0.0079 g of Sr2+ per kilogram of water. To find the concentration of Sr2+ in ppm, we can use the formula: ppm concentration = (mass of solute / mass of solution) * 10^6 So, ppm concentration = (0.0079 g / 1000 g) * 10^6 = 7.9 ppm The mass percentage of iodine in the solution is 7.11%, and the concentration of Sr2+ in seawater is 7.9 ppm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percentage
The concept of mass percentage is essential for understanding the composition of solutions in chemistry. It gives us a way to express the concentration of a solute in a solution by comparing the mass of the solute to the total mass of the solution, then multiplying by 100.

The formula for mass percentage can be written as:
  • Mass percentage = (mass of solute / mass of solution) * 100
In our example, we calculated the mass percentage of iodine in a solution by first finding the mass of iodine using its molar mass. Iodine ( I 2 ) has a molar mass of 254 g/mol. For 0.035 moles of iodine, the mass is calculated as:
  • Mass of iodine = 0.035 mol * 254 g/mol = 8.89 g
With the mass of the total solution being 125 g, the mass percentage is determined as follows:
  • Mass percentage = (8.89 g / 125 g) * 100 = 7.11%
This tells us that 7.11% of the solution's mass is made up of iodine.
Molar Mass
Molar mass is a critical concept in chemistry that refers to the mass of one mole of a given substance. It is typically expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms present in a molecule.

To find molar mass, use the periodic table, where each element's atomic mass is listed. For instance, iodine's atomic mass is about 127 g/mol. Since iodine typically exists as a diatomic molecule ( I 2 ), its molar mass is:
  • I 2 : 2 * 127 g/mol = 254 g/mol
This allows us to convert moles of a substance into grams, which is especially useful in calculating concentrations or mass percentages.

Understanding molar mass also helps in determining other concentrations, such as molarity, and is a fundamental step in stoichiometric calculations - ensuring all calculations in chemical equations are balanced.
Concentration in ppm
Parts per million, or ppm, is a unit of concentration that is particularly useful for very dilute solutions. It measures the mass of a solute per million mass units of solution, essentially showing how many "parts" of solute are in a million "parts" of total solution.

The formula for calculating concentration in ppm is:
  • ppm = (mass of solute / mass of solution) * 106
In practical terms, this means for every gram of solute, there are a million grams of the solution.

In the seawater example, the concentration of Sr 2+ was calculated as follows:
  • Ppm concentration = (0.0079 g / 1000 g) * 106 = 7.9 ppm
This indicates that in each kilogram (or 1000 grams) of seawater, there are 7.9 grams of Sr 2+ ions, an extremely small concentration fitting for most environmental and chemical assessments in natural and lab settings.

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Most popular questions from this chapter

The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is \(0.786 \mathrm{~g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(25.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(100 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{CN}\) (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?

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