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.