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Chapter 13: Problem 40

(a) What is the mass percentage of iodine in a solution containing 0.035 \(\mathrm{mol}_{2}\) in 125 \(\mathrm{g}\) of \(\mathrm{CCl}_{4} ?\) (b) Seawater contains 0.0079 \(\mathrm{g}\) of 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 6.64%, and the concentration of Sr虏鈦� ions in seawater is 7.9 ppm.

Step by step solution

01

(Part a: Calculate the mass of iodine)

First, we need to calculate the mass of iodine in grams given the number of moles (0.035 moles). The molecular weight of iodine is approximately 254 g/mol. So, to find the mass of iodine in grams, we use the formula: mass (g) = number of moles x molecular weight mass (g) = 0.035 mol x 254 g/mol mass (g) = 8.89 g So, the mass of iodine in the solution is 8.89 g.
02

(Part a: Calculate the mass percentage of iodine)

Now that we know the mass of iodine, we can find the mass percentage of iodine in the solution using the formula: mass percentage = (mass of solute / mass of solution) x 100 The solution is made up of iodine (8.89 g) and CCl4 (125 g). Therefore, the mass of the solution is: mass of solution = 8.89 g (iodine) + 125 g (CCl4) = 133.89 g Now we can calculate the mass percentage: mass percentage = (8.89 g / 133.89 g) x 100 = 6.64% So, the mass percentage of iodine in the solution is 6.64%.
03

(Part b: Convert grams per kilogram to ppm)

In this part, we have 0.0079 g of Sr虏鈦� ions per kilogram of seawater. To convert it to ppm (parts per million), we can use the following conversion: 1 g/kg = 1,000 ppm So, we can multiply the concentration in g/kg by 1,000 to find the concentration in ppm: concentration of Sr虏鈦� (ppm) = 0.0079 g/kg x 1,000 = 7.9 ppm Thus, the concentration of Sr虏鈦� ions 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.

Molecular Weight
Understanding molecular weight is crucial in chemistry, especially when dealing with processes like calculating the mass of a substance. Molecular weight refers to the mass of a given molecule based on the sum of the atomic masses of all the atoms it contains. This is typically expressed in grams per mole (g/mol).
For example, iodine (\(\mathrm{I_2}\)) has a molecular weight of approximately 254 g/mol, meaning one mole of iodine molecules weighs 254 grams.

To find the mass of a substance using its molecular weight:
  • Determine the number of moles of the substance you have.
  • Multiply the number of moles by the molecular weight of the substance.
This calculation lets you convert from the number of moles to the mass in grams, a crucial step in determining mass percentages in solutions.
Conversion to ppm
Parts per million (ppm) is a concentration metric, expressing the number of parts of a substance per million parts of a whole. It is especially used for very dilute solutions, such as impurities in water.
Think of ppm as a way to scale down the concentration, similar to percentages but a thousand times smaller since 1 percent is equal to 10,000 ppm.

To convert a concentration from grams per kilogram (g/kg) to ppm:
  • Understand that 1 g/kg is equivalent to 1,000 ppm.
  • Multiply the concentration in g/kg by 1,000 to find the value in ppm.
For instance, a concentration of 0.0079 g of \(\mathrm{Sr^{2+}}\) per kilogram of seawater is equivalent to 7.9 ppm.
This conversion helps in quantifying elements or compounds within a solution, widely used in environmental sciences.
Mole Calculations
Mole calculations are a fundamental part of stoichiometry in chemistry, relating masses, numbers of particles, and volumes of gases. The mole is defined as the amount of substance that contains as many entities (like atoms or molecules) as there are atoms in 12 grams of carbon-12.
When you need to calculate the mass of a substance in a chemical solution:
  • First, determine the number of moles of the substance using its given quantity or concentration.
  • Use the substance's molecular weight to convert the moles to grams, multiplying them together.
These steps are foundational for further calculations such as computing mass percentages and concentrations, essential in academic and practical laboratory settings.

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Most popular questions from this chapter

(a) Calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in a solution containing 10.6 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 483 \(\mathrm{g}\) of water. (b) An ore contains 2.86 \(\mathrm{g}\) of silver per ton of ore. What is the concentration of silver in ppm?

Indicate whether each statement is true or false: (a) If you compare the solubility of a gas in water at two different temperatures, you find the gas is more soluble at the lower temperature. (b) The solubility of most ionic solids in water decreases as the temperature of the solution increases. (c) The solubility of most gases in water decreases as the temperature increases because water is breaking its hydrogen bonding to the gas molecules as the temperature is raised. (d) Some ionic solids become less soluble in water as the temperature is raised.

Most fish need at least 4 ppm dissolved \(\mathrm{O}_{2}\) in water for survival. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain 4 \(\mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C} ?\) (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(1.71 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) -atm.)

Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.

Consider water and glycerol, CH \(_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\) . (a) Would you expect them to be miscible in all proportions? (b) List the intermolecular attractions that occur between a water molecule and a glycerol molecule.
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