Express the concentration of a 0.0620M aqueous solution of fluoride, F-, in mass percentage and in parts per million. assume the density of the solution is 1.00g/mL.
0.0620M is 0.0620 mols/L solution.
g = mols x molar mass = 0.0620 x 19 = 1.18 g/L
mass% = (g solute/g solution)*100 =
(1.18/1000)*100 = ? = about 0.12 mass %.
A ppm is 1 mg/L so if we have 1.18g/L that is 1180 mg/L and that is 1180 ppm.
What the number of significant figures in the above, I didn't check them. Also, note I've estimated.
Oh, I love a good chemistry question! So you have a 0.0620M solution of fluoride, F-. Well, to express its concentration in terms of mass percentage, we need to calculate the mass of fluoride present in a given volume of the solution.
Since the density of the solution is 1.00g/mL, we can easily convert the volume to mass. Let's assume we have 1 mL of the solution. Since the molarity (M) is given, we can conclude that there is 0.0620 moles of fluoride present in 1 liter of the solution.
Now, to find the mass of fluoride in those 0.0620 moles, we need to multiply the molar mass of fluoride by the number of moles. The molar mass of fluoride (F-) is approximately 19.00 g/mol. So, by multiplying 0.0620 moles by 19.00 g/mol, we find that there are 1.178 grams of fluoride in 1 liter of this solution.
To express this concentration in mass percentage, we take the mass of fluoride (1.178 g) and divide it by the total mass of the solution (which in this case is 1 g, since the density is 1.00 g/mL). Multiplying this by 100 gives us a mass percentage of 117.8%.
But wait a minute, a mass percentage of over 100%? That's impossible! Looks like we have a bit of a problem here. The density assumption of 1 g/mL might not be accurate, or there might be an issue with the molarity provided. A solution cannot have more than 100% mass percentage.
Moving on to parts per million (ppm), this unit is often used in trace quantities. To calculate ppm, we take the mass of solute (in this case, fluoride) and divide it by the total mass of the solution. Then, we multiply the result by 1,000,000.
Using the same mass of fluoride (1.178 g) and the assumed total mass of the solution (1 g), we find that there are 1,178,000 parts of fluoride per million parts of the solution. Therefore, the concentration can be expressed as 1,178,000 ppm.
So there you have it, in terms of mass percentage, we encounter a problem with a concentration of over 100%, but in parts per million (ppm), we have a concentration of 1,178,000 ppm. I hope that adds a little chemistry fun to your day!
To express the concentration of a 0.0620M aqueous solution of fluoride, F-, in mass percentage and parts per million (ppm), you can follow the steps below:
Step 1: Determine the molar mass of F- (fluoride ion).
The molar mass of F- is 18.998 g/mol.
Step 2: Convert the molarity to grams per liter (g/L).
For a 0.0620M solution, there are 0.0620 moles of F- in 1 liter of solution.
Therefore, the mass of F- in 1 liter of solution is: 0.0620 moles x 18.998 g/mol = 1.139 g.
Step 3: Determine the mass percentage.
Mass percentage is calculated by dividing the mass of the solute (F-) by the mass of the solution and multiplying by 100.
In this case, the mass of the solution is 1.00g/mL (given the density is 1.00g/mL).
Mass percentage = (1.139 g/1.00 g) x 100% = 113.9%
Step 4: Convert mass percentage to parts per million (ppm).
Parts per million (ppm) represents the number of parts of the solute per 1 million parts of the solution.
To convert mass percentage to ppm, multiply the mass percentage by 10,000.
PPM = 113.9% x 10,000 = 1,139,000 ppm
Therefore, the concentration of the 0.0620M aqueous solution of fluoride, F-, is approximately 113.9% by mass and 1,139,000 ppm.
To express the concentration of a solution in mass percentage and parts per million (ppm), we need to understand the concepts and formulas related to these units.
1. Mass Percentage:
Mass percentage represents the mass of the solute (in this case, F-) as a percentage of the total mass of the solution. The formula for mass percentage is:
Mass Percentage = (Mass of solute / Mass of solution) × 100
To find the mass of the solute (F-), we can multiply the volume of the solution by its density, as density = mass / volume. Since the density of the solution is given as 1.00 g/mL, the mass of the solution will be equal to its volume.
Given that the concentration of the solution is 0.0620 M, this means there are 0.0620 moles of fluoride (F-) ions in one liter (1000 mL) of the solution. We need to calculate the mass of these 0.0620 moles of fluoride.
To do this, we need the molar mass of F-. Fluorine has an atomic weight of 19 g/mol, and since F- has an extra electron, we add one more atomic mass unit (amu) to the atomic weight. So the molar mass of F- is 19 + 1 = 20 g/mol.
Now we can calculate the mass of the solute (F-):
Mass of F- (in grams) = moles of F- × molar mass (20 g/mol)
2. Parts per Million (ppm):
Parts per million is another unit of concentration, representing the number of parts of the solute per million parts of the solution. The formula for ppm is:
ppm = (Mass of solute / Mass of solution) × 10^6
To calculate the ppm, we need the mass of the solute (F-) and the mass of the solution (which is the same as the mass of the solute since the density is given as 1.00 g/mL).
Now let's put all the calculations together:
Step 1: Mass Percentage:
Mass of F- = moles of F- × molar mass of F- (20 g/mol)
Mass percentage = (Mass of F- / Mass of solution) × 100
Step 2: Parts per Million (ppm):
ppm = (Mass of F- / Mass of solution) × 10^6
By following these steps, you can easily find the concentration of the 0.0620 M aqueous solution of fluoride (F-) in mass percentage and ppm.