How would you prepare 10.0 Liters of a solution that was 20 parts per million sodium perchlorate in water? What would be the molarity of the solution (assume density is 1.0 g/mL and 1ppm=1 mL/L)?
20 ppm = 20 mg NaClO4/L
10 L would be 200 mg NaClO4/10 L.
I would weigh 200 mg NaClO4 and dump it into 10L water. (That isn't quite right since it is mg/L soln and not mg/L H2O so if it is done right, you would weigh 200 mg NaClO4, dissolve it in some water, then make to a final volume of 10.0L).
M = ?
20 mg/L is 0.02 g/L
0.02 g = ? mols.
0.02/molar mass NaClO4
To prepare a solution with a concentration of 20 parts per million (ppm) of sodium perchlorate in water, you need to follow these steps:
Step 1: Determine the mass of sodium perchlorate required:
To calculate the mass, you need to know the concentration (ppm) and desired volume of the solution.
Mass (in grams) = Concentration (in ppm) * Volume (in liters) * Grams per liter (g/L)
Since 1 ppm is equivalent to 1 mg/L, and the density of water is 1.0 g/mL (or 1.0 g/L), the equation becomes:
Mass (in grams) = 20 ppm * 10.0 L * 1.0 g/L
Step 2: Convert the mass of sodium perchlorate to moles:
To convert grams of sodium perchlorate to moles, you need to know the molar mass of sodium perchlorate, which you can calculate by adding up the atomic masses of its constituent elements from the periodic table.
Molar mass of NaClO4 = Molar mass of Na + 4 * (Molar mass of Cl) + 16 * (Molar mass of O)
Step 3: Calculate the molarity of the solution:
Molarity (M) is calculated by dividing the moles of solute by the volume of the solution in liters.
Molarity (M) = Moles of sodium perchlorate / Volume of solution (in liters)
Now, let's do the calculations:
Step 1:
Mass (in grams) = 20 ppm * 10.0 L * 1.0 g/L
= 200 g
Step 2:
Molar mass of NaClO4 = 22.99 g/mol + 4 * (35.45 g/mol) + 16 * (16.00 g/mol)
= 122.44 g/mol
Step 3:
Moles of sodium perchlorate = Mass (in grams) / Molar mass (in g/mol)
= 200 g / 122.44 g/mol
Molarity (M) = Moles of sodium perchlorate / Volume of solution (in liters)
= (200 g / 122.44 g/mol) / 10.0 L
Therefore, the molarity of the solution would be:
Molarity (M) = 1.63 M (rounded to two decimal places)