A student weighs out a 14.0 g sample of Fe(NO3)2, transfers it to a 500 mL volumetric flask, adds enough water to dissolve it and then adds water to the 500 mL tic mark.
What is the molarity of iron(II) nitrate in the resulting solution?
Calculate the molar mass of Fe(NO3)2 (F)
number of moles of Fe(NO3)2 = 14.0 g/F
If the number of moles were dissolved in 1 L the molarity would be
14.0/F mol L^-1
but this is in only 500 ml so the concentration is
(14.0 x 1000)/(F x 500)
i.e. twice the concentration if the final volume had been 1 L.
To find the molarity of iron(II) nitrate in the resulting solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the moles of Fe(NO3)2 in the 14.0 g sample:
Molar mass of Fe(NO3)2 = atomic mass of Fe + (atomic mass of N + 3 * atomic mass of O) * 2
Molar mass of Fe(NO3)2 = 55.85 g/mol + (14.01 g/mol + 3 * 16.00 g/mol) * 2
Molar mass of Fe(NO3)2 = 55.85 g/mol + (14.01 g/mol + 48.00 g/mol) * 2
Molar mass of Fe(NO3)2 = 55.85 g/mol + 124.02 g/mol
Molar mass of Fe(NO3)2 = 179.87 g/mol
Now, we can calculate the moles of Fe(NO3)2:
moles = mass / molar mass
moles = 14.0 g / 179.87 g/mol
moles ≈ 0.0778 mol
Next, let's calculate the volume of the solution in liters:
500 mL = 500 / 1000 = 0.5 L
Finally, we can calculate the molarity:
Molarity (M) = moles / volume
Molarity (M) = 0.0778 mol / 0.5 L
Molarity (M) ≈ 0.155 M
Therefore, the molarity of iron(II) nitrate in the resulting solution is approximately 0.155 M.