Question

If 5.28 g Fe(NO₃)₃ is dissolved in enough water to make exactly 298 mL of solution, what is the molar concentration of nitrate ion?

Answer 1

Answer 2

Calculation  :- 

1). Find Moles of Fe(NO₃)₃ (Iron Nitrate) :-

• Given: 5.28 g of Fe(NO₃)₃ dissolved.

• Molar mass of Fe(NO₃)₃:

• Iron (Fe) = 55.85 g/mol

• Nitrate (NO₃) = 62.00 g/mol (but there are 3 nitrates in the formula!)

• Total molar mass = 55.85 + (3 × 62.00) = 241.85 g/mol.

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• Moles of Fe(NO₃)₃:

  $$\text{Moles}=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{5.28\text{ g}}{241.85\text{ g/mol}}=0.0218\text{ moles}$$

2). Find Moles of Nitrate (NO₃⁻) Ions :

• Each Fe(NO₃)₃ molecule releases 3 NO₃⁻ ions when dissolved.

• Total moles of NO₃⁻:

  $$0.0218\text{ moles Fe(NO₃)₃}\times 3=0.0654\text{ moles NO₃⁻}$$

3). Calculate Molarity (Concentration) :

• Volume of solution: 298 mL = 0.298 L.

• Molarity (M):

  $$\text{Molarity}=\frac{\text{Moles of NO₃⁻}}{\text{Volume (L)}}=\frac{0.0654\text{ moles}}{0.298\text{ L}}=0.223\text{ M}$$

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Key Points:

• Why multiply by 3? Because Fe(NO₃)₃ is like a "nitrate delivery truck"—it carries 3 nitrates per molecule!

• Unit check: Always convert mL to liters (L) for molarity.

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 Answer: The nitrate ion concentration is 0.223 M.