In the laboratory you are asked to make a 0.659 m iron(II) sulfate solution using 315 grams of water.
How many grams of iron(II) sulfate should you add?___grams.
Show some work, or explain what you dont understand about this. We are not going to do your work, nor provide an answer key.
To find the mass of iron(II) sulfate needed, we need to use the given information about the desired concentration and the solvent.
The concentration of the iron(II) sulfate solution is given as 0.659 m, which means there is 0.659 moles of iron(II) sulfate per liter of solution.
First, we need to calculate the volume of water used in liters. Since the density of water is 1 gram per milliliter, we can convert 315 grams to 315 milliliters and then divide by 1000 to get the volume in liters:
Volume of water = 315 grams / 1000 = 0.315 liters.
Next, we can calculate the moles of iron(II) sulfate required:
Moles of iron(II) sulfate = Concentration × Volume of water
Moles of iron(II) sulfate = 0.659 mol/L × 0.315 L = 0.207435 moles.
Finally, we can calculate the mass of iron(II) sulfate using its molar mass. The molar mass of iron(II) sulfate (FeSO4) is 55.845 grams/mol.
Mass of iron(II) sulfate = Moles of iron(II) sulfate × Molar mass of iron(II) sulfate
Mass of iron(II) sulfate = 0.207435 mol × 55.845 g/mol ≈ 11.6 grams.
Therefore, you should add approximately 11.6 grams of iron(II) sulfate to make a 0.659 m solution using 315 grams of water.