how many grams are present in 20ml of .500M of Fe(NO3)3?
To find the number of grams present in 20 mL of a solution with a given concentration, you need to use the molarity (M) and the molar mass of the solute. In this case, the solute is Fe(NO3)3.
First, let's determine the number of moles of Fe(NO3)3 in the solution:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Rearranging the equation, we get:
Moles of solute = Molarity × Volume of solution (in liters)
Since the volume is given in milliliters (mL), we need to convert it to liters by dividing by 1000:
Volume of solution (in liters) = 20 mL / 1000 mL/L = 0.02 L
Now we can calculate the moles of Fe(NO3)3:
Moles of Fe(NO3)3 = 0.500 M × 0.02 L = 0.01 moles
Next, we need to find the molar mass of Fe(NO3)3. The molar mass of Fe is 55.845 g/mol, and the molar mass of NO3 is 62.0049 g/mol. Since there are three nitrate ions (NO3-) in Fe(NO3)3, we multiply the molar mass of NO3 by 3:
Molar mass of Fe(NO3)3 = (55.845 g/mol) + (62.0049 g/mol × 3) = 241.8527 g/mol
Finally, we can calculate the grams of Fe(NO3)3:
Grams of Fe(NO3)3 = Moles of Fe(NO3)3 × Molar mass of Fe(NO3)3
Grams of Fe(NO3)3 = 0.01 moles × 241.8527 g/mol
Therefore, there are 2.4185 grams present in 20 mL of a 0.500M solution of Fe(NO3)3.