What volume of 18.0 M sulfuric acid must be used to prepare 15.5 L of 0.195 M H2SO4?
[A] 0.336 L
[B] 92.3 mL
[C] 168 mL
[D] 226 mL
[E] none of these
The Correct Answer and Explanation is :
To determine the volume of 18.0 M sulfuric acid (H₂SO₄) required to prepare 15.5 L of 0.195 M H₂SO₄, we use the dilution equation:
[
M_1 V_1 = M_2 V_2
]
Where:
- ( M_1 ) = initial concentration = 18.0 M
- ( V_1 ) = volume of the concentrated acid (what we’re solving for)
- ( M_2 ) = final concentration = 0.195 M
- ( V_2 ) = final volume = 15.5 L
Step 1: Rearrange the equation for ( V_1 ):
[
V_1 = \frac{M_2 V_2}{M_1}
]
Step 2: Plug in the values:
[
V_1 = \frac{(0.195)(15.5)}{18.0}
]
Step 3: Perform the calculations:
- Multiply ( M_2 ) and ( V_2 ):
[
0.195 \times 15.5 = 3.0225
] - Divide by ( M_1 ):
[
\frac{3.0225}{18.0} \approx 0.168 \, \text{L}
]
Step 4: Convert to mL (optional):
Since ( 1 \, \text{L} = 1000 \, \text{mL} ), we convert:
[
0.168 \, \text{L} \times 1000 = 168 \, \text{mL}
]
Final Answer:
The volume of 18.0 M H₂SO₄ needed is 168 mL, which corresponds to option [C].
Explanation:
This problem involves diluting a concentrated acid to achieve a lower concentration. The dilution equation ensures the number of moles of solute remains constant before and after dilution. The concentrated sulfuric acid is very strong, so only a small volume (168 mL) is required to prepare the 15.5 L solution. Always remember to add acid to water for safety.