How many grams of iron(III) nitrate (Fe(NO3)3, molar mass=241.88 g/mol) are required to make 198.0 mL of a 0.130 M solution? Enter
The correct answer and explanation is :
To determine the grams of iron(III) nitrate (Fe(NO₃)₃) required to make 198.0 mL of a 0.130 M solution, we need to follow these steps:
1. Use the formula for molarity:
Molarity (M) is defined as the number of moles of solute (in this case, Fe(NO₃)₃) per liter of solution. The formula is:
[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}}
]
Rearranging this formula to find moles of solute:
[
\text{moles of solute} = \text{Molarity (M)} \times \text{Volume (L)}
]
2. Convert volume to liters:
Given that the volume of the solution is 198.0 mL, we first convert it to liters:
[
198.0 \, \text{mL} = 198.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.198 \, \text{L}
]
3. Calculate moles of Fe(NO₃)₃:
Now, using the given molarity of 0.130 M and the volume in liters (0.198 L), we can calculate the moles of iron(III) nitrate:
[
\text{moles of Fe(NO₃)₃} = 0.130 \, \text{M} \times 0.198 \, \text{L} = 0.02574 \, \text{moles}
]
4. Convert moles to grams:
To find the mass of Fe(NO₃)₃ required, we use the molar mass of Fe(NO₃)₃, which is given as 241.88 g/mol. The mass can be calculated as:
[
\text{mass of Fe(NO₃)₃} = \text{moles of Fe(NO₃)₃} \times \text{molar mass}
]
[
\text{mass of Fe(NO₃)₃} = 0.02574 \, \text{moles} \times 241.88 \, \text{g/mol} = 6.23 \, \text{grams}
]
Final Answer:
You need 6.23 grams of iron(III) nitrate (Fe(NO₃)₃) to make 198.0 mL of a 0.130 M solution.
Explanation:
- Step 1: We use the molarity equation to relate the amount of solute (in moles) to the volume of solution.
- Step 2: We converted the volume from mL to L to align with the molarity unit of L.
- Step 3: The moles of Fe(NO₃)₃ were calculated by multiplying the molarity (0.130 M) by the volume in liters (0.198 L).
- Step 4: To convert moles to grams, we used the molar mass (241.88 g/mol) and multiplied it by the number of moles. This gives the required mass in grams.
Thus, the answer is clear and shows all the required steps to arrive at the final value of 6.23 grams of Fe(NO₃)₃ needed.