Question

A student made a mistake by calculating molarities of Fe(NO3)3 assuming that anhydrous salt was used instead of Fe(NO3)3 ×9 H2O. What is the average value of K that he will obtain? Hint: Replace Mw of Fe(NO3)3 get this value.

Answer 1

Let the molarity calculated by student be M

Molar mass of anhydrous Fe(NO3)3 = Molar mass of Fe + 3 * (Molar mass of N + 3 * Molar mass of O) = 242 gm/mol

Molar mass of hydrous salt Fe(NO3)3.9H2O = 242 + 9 * Molar mass of water = 242 + 9 * 18 = 404 gm/mol

Molarity = number of moles of solute/Volume of solution in L

Since the student used the weight of Fe(NO3)3, hence the number of moles would be higher

Hence actual molarity = Old molarity * mass of Fe(NO3)3/mass of Fe(NO3)3.9H20

=> M * 242/404

=> 0.599M