What mass of iron (III) nitrate, Fe (NO₃)₃, is needed to prepare a 125 mL solution of 0.250 M?

Question

Quizplus · Accepted Answer

To prepare a 125 mL solution of 0.250 M Fe(NO3)3, we need to calculate the number of moles of Fe(NO3)3 required:Moles of Fe(NO3)3 = Concentration × Volume= 0.250 M × 0.125 L= 0.03125 molesNow, we can calculate the mass of Fe(NO3)3 required using its molar mass:Molar mass of Fe(NO3)3 = 241.86 g/molMass of Fe(NO3)3 = Moles of Fe(NO3)3 × Molar mass of Fe(NO3)3= 0.03125 moles × 241.86 g/mol= 7.56 gramsTherefore, the correct answer is A) 7.56 grams.

What Mass of Iron (III) Nitrate, Fe (NO3)3, Is Needed

What Mass of Iron (III) Nitrate, Fe (NO₃)₃, Is Needed