You have a 250ppm solution of K with KCl, you want to prepare a 0.00100 mole of Cl, how many mililiter should be diluted to prepare 1 liter of solution
250 ppm is 250 mg KCl/L or 0.250 g KCl/L or 0.250/molar mass KCl/L = M
Then M = mols x L.
YOu know M and mols needed, solve for L and convert to mL.
thank u
To calculate the amount of KCl needed to prepare a 0.00100 mole of Cl in a 1 liter solution, we can use the molar mass of KCl and the molarity of the KCl solution (250 ppm).
First, let's calculate the molar mass of KCl:
K (potassium) has a molar mass of approximately 39.10 grams/mol
Cl (chlorine) has a molar mass of approximately 35.45 grams/mol
So, the molar mass of KCl is 39.10 + 35.45 = 74.55 grams/mol.
Next, we need to determine the amount of KCl (in grams) needed to prepare 0.00100 mole of Cl. We can use the molar ratio between KCl and Cl to convert moles of Cl to moles of KCl. Since there is 1 mole of Cl for every 1 mole of KCl, we can simply use the same value of 0.00100 moles.
Now, let's use the molar mass of KCl to calculate the grams of KCl needed:
0.00100 moles * 74.55 grams/mol = 0.07455 grams of KCl.
To find the volume of the 250 ppm (parts per million) KCl solution needed to provide 0.07455 grams of KCl, we need to use the definition of ppm.
ppm = (mass of solute / mass of solution) * 1,000,000
Since we know the ppm (250 ppm), the mass of solute (0.07455 grams), and we want to find the mass of the solution, we can rearrange the formula:
mass of solution = (mass of solute / ppm) * 1,000,000
mass of solution = (0.07455 grams / 250 ppm) * 1,000,000 = 298.2 grams
Finally, to convert grams of solution to milliliters, we need to know the density of the KCl solution. Assuming the density is 1 g/mL (which is close to the density of water), we can directly convert grams to milliliters.
Therefore, you would need to dilute approximately 298.2 milliliters of the 250 ppm KCl solution to prepare a 1 liter (1000 milliliters) solution with 0.00100 mole of Cl.
35.4grams/mole*.001mole/2=.0177 grams Cl
If you added to 4ml of the original solutions, you would get
4*250e-6*1g/ml=1e-3 grams of Cl ions, or .001grams
so you want to add not 4 ml, but 4ml(.0177/.001)=70.8ml to make up a volume of 1 liter of solution.