If a student weighs out 0.744 g Fe (NO 3) 3 ⋅ 9 H 2 O, what is the final concentration of the ∼0.2 M Fe (NO 3) 3 solution that the student makes?

Molar concentration of Fe (NO3) 3. 9H2O = 0.12M

Explanation:

Fe (NO3).9H2O - -> Fe (NO3) 3 + 9H2O

By stoichiometry,

1 mole of Fe (NO3) 3 will be absorb water to form 1 mole of Fe (NO3) 3. 9H2O

Therefore, calculating the mass concentration of Fe (NO3) 3;

Molar mass of Fe (NO3) 3 = 56 + 3 * (14 + (16*3))

= 242 g/mol

Mass concentration of Fe (NO3) 3 = molar mass * molar concentration

= 242 * 0.2

= 48.4 g/L

Molar mass of Fe (NO3) 3. 9H2O = 56 + 3 * (14 + (16*3)) + 9 * ((1*2) + 16)

= 242 + 162 g/mol

= 404g/mol

Concentration of Fe (NO3) 3. 9H2O = mass concentration/molar mass

= 48.4 / 404

= 0.12 mol/l

Molar concentration of Fe (NO3) 3. 9H2O = 0.12M