How many milliliters of pure acetic acid (density = 1.05 g/mL) would be required to prepare 1.00 L of the solution?
What solution?
Vinegar is a solution of acetic acid (C2H4O2) in water. Its concentration is 5.0%(w/w) and its density is 1.01 g/mL.
That was the information given before the question, then it asked for the molarity of vinegar and I got .84mol/L
To find the number of milliliters of pure acetic acid required to prepare 1.00 L of the solution, we need to determine the mass of acetic acid required.
The density of acetic acid is given as 1.05 g/mL. We can use this information to convert from volume (mL) to mass (g).
The equation for density is:
density = mass/volume
We rearrange the equation to solve for mass:
mass = density * volume
Given that the volume is 1.00 L, we can convert it to mL using the fact that 1 L = 1000 mL.
So, volume = 1.00 L = 1000 mL
Now we can calculate the mass:
mass = density * volume
= 1.05 g/mL * 1000 mL
= 1050 g
Therefore, 1050 milliliters of pure acetic acid would be required to prepare 1.00 L of the solution.