Problem 40
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 ppm?
Step-by-Step Solution
Verified Answer
The mass percentage of iodine in the solution is 7.11%, and the concentration of Sr2+ in seawater is 7.9 ppm.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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:
4Step 1: 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.
Key Concepts
Mass PercentageMolar MassConcentration in ppm
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:
The formula for mass percentage can be written as:
- Mass percentage = (mass of solute / mass of solution) * 100
- Mass of iodine = 0.035 mol * 254 g/mol = 8.89 g
- Mass percentage = (8.89 g / 125 g) * 100 = 7.11%
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:
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.
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
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:
In the seawater example, the concentration of Sr 2+ was calculated as follows:
The formula for calculating concentration in ppm is:
- ppm = (mass of solute / mass of solution) * 106
In the seawater example, the concentration of Sr 2+ was calculated as follows:
- Ppm concentration = (0.0079 g / 1000 g) * 106 = 7.9 ppm
Other exercises in this chapter
Problem 37
The Henry's law constant for hydrogen gas \(\left(\mathrm{H}_{2}\right)\) in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{M} / \mathrm{kPa
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(a) Calculate the mass percentage of \(\mathrm{NaNO}_{3}\) in a solution containing \(13.6 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) in \(834 \mathrm{~g}\) of wate
View solution Problem 41
A solution is made containing \(50.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(1000 \mathrm{~g}\) of \(\mathrm{H}
View solution Problem 43
Calculate the molarity of the following aqueous solutions: (a) \(0.640 \mathrm{~g}\) of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(500.0 \mathrm{~mL}\
View solution