Calculate to four decimal places the mass of pure dry CaCO3 required to make 1 L of a 1000
ppm Ca stock solution.
1000 ppm, means 1000g
1000 ppm means .001g of Ca+2 ion, or .001/ 40.078 moles Ca, so you want then the same number of moles of C and O3
element, moles, mass
Ca, .001/40.078, .001g
C, ibid times 12.0107, 0.000299683118
3O, ibid times 3*15.9994, 0.00119761964
Now adding up the masses,
0.00119761964+.001+.000299683118
I get 0.00249730276 grams of CaCO3. Now, is the mass of a liter of solution to four decimal places really 1000g? No, depends on temperature, it varies some. So you have to have temperature given. http://water.usgs.gov/edu/density.html
check my work
Well, let's break it down. The units "ppm" stand for parts per million, which tells us how many parts of a substance (in this case, CaCO3) are in a million parts of a solution.
To calculate the mass of pure dry CaCO3 required, we first need to convert 1000 ppm to a concentration in grams per liter (g/L).
Since 1 ppm is equal to 1 mg/L (milligram per liter), we can convert 1000 ppm to 1000 mg/L.
Now, we can convert mg/L to g/L by dividing by 1000.
So, the concentration of CaCO3 in the stock solution is 1 g/L.
Since we want to make 1 L of this solution, the mass of pure dry CaCO3 required is simply 1 gram.
So, you would need 1 gram of pure dry CaCO3 to make 1 L of a 1000 ppm Ca stock solution. Just remember, this answer is not just a lot of CaCO3, it's a-moosing!
To calculate the mass of pure dry CaCO3 required to make a 1000 ppm Ca stock solution, we'll follow these steps:
Step 1: Determine the molar mass of CaCO3
The molar mass of CaCO3 can be calculated by summing the atomic masses of each element in the compound:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (there are 3 oxygen atoms in CaCO3)
Molar mass of CaCO3 = (1 * 40.08) + (1 * 12.01) + (3 * 16.00) = 100.09 g/mol
Step 2: Calculate the mass of CaCO3 required to make 1 L of 1000 ppm solution.
To convert the concentration in parts per million (ppm) to grams per liter (g/L), we need to consider that 1 ppm is equal to 1 mg/L.
Concentration of CaCO3 in ppm = 1000 ppm
Concentration of CaCO3 in g/L = 1000 mg/L = 1 g/L
So, 1 L of the Ca stock solution would contain 1 g of CaCO3.
Therefore, the mass of pure dry CaCO3 required to make 1 L of a 1000 ppm Ca stock solution is 1 g, with a molar mass of 100.09 g/mol.
To calculate the mass of pure dry CaCO3 required to make a 1000 ppm Ca stock solution, we need to use the formula:
PPM = (mass of solute / volume of solution) x 10^6
First, let's rearrange the formula to find the mass of solute:
mass of solute = (PPM * volume of solution) / 10^6
Given that we want to make a 1000 ppm Ca stock solution with a volume of 1 L, we can substitute the values into the formula:
mass of solute = (1000 ppm * 1 L) / 10^6
To solve this, we need to recognize that 10^6 is equivalent to 1 million:
mass of solute = (1000 ppm * 1 L) / 1,000,000
mass of solute = 1 ppm * 1 L / 1,000
Next, let's convert the ppm unit to a decimal:
1 ppm = 1/1,000,000
mass of solute = (1/1,000,000) * 1 L / 1,000
mass of solute = 0.000001 * 1 L / 1,000
mass of solute = 0.000001 kg
Finally, since 1 kg is equal to 1000 g, we can convert the mass of solute to grams:
mass of solute = 0.000001 kg * 1000 g/kg
mass of solute = 0.001 g
Therefore, the mass of pure dry CaCO3 required to make 1 L of a 1000 ppm Ca stock solution is 0.001 grams.