Question Detail
Question
The mass (in grams) of FeSO₄·7H₂O required for preparation of 125 mL of 0.90 M solution is:
Correct Answer
Option D is the correct answer.
Detailed Explanation
Molar mass FeSO₄·7H₂O = 55.85 + 32.06 + (4×16) + 7×18 = 55.85+32.06+64+126 = 277.91 g/mol. Moles needed = 0.125 L × 0.90 M = 0.1125 mol. Mass = 0.1125 × 277.91 = 31.27 g ≈ 31 g.
Hint
Moles = M × V(L). Then mass = moles × molar mass.