Commercial vinegar is a $5.00\%$ by mass aqueous solution of acetic acid, $\ce{CH3CO2H}$ $(M=60.0)$. What is the molarity of acetic acid in vinegar? [density of vinegar = 1.00g/mL)

$ \textbf{(A)}\hspace{.05in}0.833 M\qquad\textbf{(B)}\hspace{.05in}1.00 M\qquad\textbf{(C)}\hspace{.05in}1.20 M\qquad\textbf{(D)}\hspace{.05in}3.00 M\qquad$
To find the molarity of acetic acid in vinegar, we need to follow these steps:

1. **Determine the mass of acetic acid in a given mass of solution:**
   Given that the solution is 5.00% by mass acetic acid, this means that in 100 grams of vinegar, there are 5 grams of acetic acid.

2. **Calculate the number of moles of acetic acid:**
   The molar mass of acetic acid ($\ce{CH3CO2H}$) is given as 60.0 g/mol. Therefore, the number of moles of acetic acid in 5 grams is:
   \[
   \text{Moles of acetic acid} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{60.0 \text{ g/mol}} = \frac{5}{60} = \frac{1}{12} \text{ mol}
   \]

3. **Determine the volume of the solution:**
   The density of vinegar is given as 1.00 g/mL. Therefore, 100 grams of vinegar has a volume of:
   \[
   \text{Volume of solution} = \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{1.00 \text{ g/mL}} = 100 \text{ mL} = 0.100 \text{ L}
   \]

4. **Calculate the molarity of the acetic acid solution:**
   Molarity (M) is defined as the number of moles of solute per liter of solution. Using the values calculated:
   \[
   \text{Molarity} = \frac{\text{moles of acetic acid}}{\text{volume of solution in liters}} = \frac{\frac{1}{12} \text{ mol}}{0.100 \text{ L}} = \frac{1}{12} \times \frac{1}{0.100} = \frac{1}{1.2} \approx 0.833 \text{ M}
   \]

Therefore, the molarity of acetic acid in vinegar is approximately \(0.833 \text{ M}\).

The final answer is \(\boxed{0.833 \text{ M}}\)