You wish to prepare 100.0 mL of a 800.0 ppm w/v sodium (MW = 22.9897 g/mol) solution. How many grams of sodium dichromate dihydrate (Na2Cr2O7·2H2O, MW = 297.998 g/mol) are needed to prepare this solution?
You want 800 ppm Na w/v which is 800 mg Na/L or 80 mg Na/100 mL.
Convert 80 mg Na to ?mg Na2Cr2O7.2H2O That will be
80 mg x (molar mass Na2Cr2O7.2H2O/2Na) = ? mg Na2Cr2O7.2H2O to weigh out.
To calculate the number of grams of sodium dichromate dihydrate (Na2Cr2O7·2H2O) needed to prepare the solution, we need to use the formula:
ppm = (mass of solute / volume of solution) * 1,000,000
Let's break down the steps:
1. First, calculate the mass of sodium in the solution:
ppm = (mass of sodium / volume of solution) * 1,000,000
Rearranging the formula to solve for the mass of sodium:
mass of sodium = (ppm * volume of solution) / 1,000,000
Substituting the given values:
mass of sodium = (800.0 ppm * 100.0 mL) / 1,000,000
Note: ppm stands for parts per million, and since we want the mass in grams, the units will cancel out.
2. Calculate the number of moles of sodium:
Number of moles = mass of sodium / molar mass of sodium
The molar mass of sodium (Na) is 22.9897 g/mol.
Substituting the values:
Number of moles = mass of sodium / 22.9897 g/mol
3. Calculate the number of moles of sodium dichromate dihydrate (Na2Cr2O7·2H2O):
Since the molar ratio between sodium (Na) and sodium dichromate dihydrate (Na2Cr2O7·2H2O) is 2:1, the number of moles of Na2Cr2O7·2H2O is half the number of moles of sodium (Na).
Number of moles of Na2Cr2O7·2H2O = (Number of moles of sodium) / 2
4. Calculate the mass of sodium dichromate dihydrate (Na2Cr2O7·2H2O):
Mass of Na2Cr2O7·2H2O = Number of moles of Na2Cr2O7·2H2O * molar mass of Na2Cr2O7·2H2O
The molar mass of Na2Cr2O7·2H2O is 297.998 g/mol.
Substituting the values:
Mass of Na2Cr2O7·2H2O = Number of moles of Na2Cr2O7·2H2O * 297.998 g/mol
Now you can calculate the mass of sodium dichromate dihydrate (Na2Cr2O7·2H2O) needed to prepare the solution by following these steps.