What volume of 18.0 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid?
Group of answer choices
52.0 mL
1.87 mL
1.04 L
208 mL
104 mL
To solve this problem, we can use the equation:
(M1)(V1) = (M2)(V2)
Where:
M1 = concentration of the initial solution
V1 = volume of the initial solution
M2 = concentration of the final solution
V2 = volume of the final solution
In this case, M1 = 18.0 M, V1 is what we're trying to find, M2 = 0.156 M, and V2 = 12.0 L.
By plugging these values into the equation, we can solve for V1:
(18.0 M)(V1) = (0.156 M)(12.0 L)
Dividing both sides by 18.0 M, we get:
V1 = (0.156 M)(12.0 L) / 18.0 M
V1 = 0.104 L
So, the volume of 18.0 M H2SO4 required to prepare 12.0 L of 0.156 M sulfuric acid is 0.104 L, or 104 mL.
Therefore, the correct answer is 104 mL.