what volume of butyric acid would be needed to make 25.0mL of a 1.0M solution?
I doubt that buryric acid is soluble enough in water to make a 1.0 M solution; however, as a problem it is done this way.
How many moles do you need? That is moles = M x L = ?
Then moles of the acid = grams/molar mass
Solve for grams; you have moles and molar mass.
To calculate the volume of butyric acid needed to make a 1.0M solution, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial concentration (in this case, 1.0M)
V1 = initial volume (unknown)
M2 = final concentration (also 1.0M)
V2 = final volume (25.0mL)
Plugging in the values into the equation:
1.0M x V1 = 1.0M x 25.0mL
Simplifying the equation:
V1 = (1.0M x 25.0mL) / 1.0M
V1 = 25.0mL
Therefore, to make a 25.0mL solution of 1.0M butyric acid, you would need 25.0mL of butyric acid.
To calculate the volume of butyric acid needed to make a 1.0M solution with a volume of 25.0mL, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (unknown)
V1 = initial volume of the solution (unknown)
C2 = final concentration of the solution (1.0M)
V2 = final volume of the solution (25.0mL)
First, let's rearrange the formula to solve for V1:
V1 = (C2 × V2) / C1
Now we can plug in the values:
V1 = (1.0M × 25.0mL) / C1
The volume of the butyric acid needed (V1) will depend on the initial concentration of the solution (C1). Since the initial concentration is not provided in the question, we cannot determine the exact volume of butyric acid needed without this information.
If you have the initial concentration of the butyric acid solution, you can substitute it into the equation to calculate the volume (V1).