Determine how to prepare 250.0 mL of 0.000362 M Fe(NH4)2(SO4)2 using solid Fe(NH4)2(SO4)2 *6H2O (392.13 g/mol)
To prepare 250.0 mL of a 0.000362 M Fe(NH4)2(SO4)2 solution using solid Fe(NH4)2(SO4)2 * 6H2O (with a molar mass of 392.13 g/mol), you will need to follow these steps:
Step 1: Calculate the moles of Fe(NH4)2(SO4)2 needed:
- Molarity is defined as moles of solute per liter of solution.
- Since we want to prepare 250.0 mL (or 0.2500 L) of the solution, the moles of Fe(NH4)2(SO4)2 can be calculated using the following equation:
Moles of Fe(NH4)2(SO4)2 = Molarity x Volume
Moles of Fe(NH4)2(SO4)2 = 0.000362 mol/L x 0.2500 L
Moles of Fe(NH4)2(SO4)2 = 0.0000905 mol
Step 2: Calculate the mass of Fe(NH4)2(SO4)2 * 6H2O needed:
- The molar mass of Fe(NH4)2(SO4)2 * 6H2O is given as 392.13 g/mol.
- The moles of Fe(NH4)2(SO4)2 calculated in step 1 can now be converted to grams using the molar mass:
Mass of Fe(NH4)2(SO4)2 = Moles x Molar mass
Mass of Fe(NH4)2(SO4)2 = 0.0000905 mol x 392.13 g/mol
Mass of Fe(NH4)2(SO4)2 = 0.0355 g
Step 3: Dissolve the calculated mass of Fe(NH4)2(SO4)2 * 6H2O in a sufficient volume of distilled water to reach a final volume of 250.0 mL.
- Weigh out 0.0355 g of Fe(NH4)2(SO4)2 * 6H2O using a balance and transfer it to a clean container.
- Add a small amount of distilled water to dissolve the solid completely.
- Transfer the solution to a 250.0 mL volumetric flask.
- Add distilled water to the flask up to the mark on the neck of the flask to reach a final volume of 250.0 mL.
- Mix the solution thoroughly by inverting the flask several times.
You now have a 250.0 mL solution of 0.000362 M Fe(NH4)2(SO4)2.