My Question is:

Commercial vinegar usually has a concentration of 5%, per volume. Calculate the [H3O+], if the density of acetic acid is 1.051 g/mL.

What I did was:

mass = V * density
mass = 0.05 * 1.051 = 0.05 g

moles = grams/molar mass = 0.05/60.05 = 8.3 * 10^-4

M = moles/V = 8.3 * 10^-4/0.1 = 8.3 * 10^-3 M

I can do the rest on my own but is this much correct?

Thanks

I don't think so. I don't know why you converted anything to mass although it can be done that way but not by your calculation. (For example, the volume is not 0.05). You want the molarity and that is mols/L. You already have it as 5% w/v which means 5 g/100 mL or 50 g/L.

How many mols is in 50 g? That gives you the molarity (and note we never used the density). Then calculate (H3O^+) using Ka for acetic acid.

You CAN use the density this way.
mass = volume x density
mass = 100 mL x 1.051 = 105.1 grams for mass % of 5/105.1 = 0.04757% w/w.
Then 1.051 x 1000 mL x 0.04757 x (1/60.05) = ? M but this is far more complicated and more work than the above.

Yes, your calculation is correct up to the point where you found the molarity of the acetic acid. To find the molarity (M) of the acetic acid, you correctly calculated the moles of acetic acid (0.00083 moles) and divided it by the volume of the solution (0.1 L) to get 0.0083 M (or 8.3 * 10^-3 M) as the concentration of the acetic acid.

Now, to find the concentration of [H3O+], you need to consider the dissociation of acetic acid in water. Acetic acid, CH3COOH, is a weak acid that partially ionizes in water to form the hydronium ion, H3O+, and the acetate ion, CH3COO-.

The balanced equation for the dissociation of acetic acid is:
CH3COOH + H2O ⇌ H3O+ + CH3COO-

Since acetic acid is a weak acid, the concentration of [H3O+] will be equal to the concentration of acetic acid and can be considered as equal to 0.0083 M (or 8.3 * 10^-3 M).

Therefore, the concentration of [H3O+] in the commercial vinegar is approximately 0.0083 M.