1.50 g of ethyl benzoate (150.17 g/mol) is treated with a 3.00 mol/L solution of methylmagnesium…

Step 1. $n = \dfrac{1.50}{150.17} = 9.99\ \text{mmol}$.

Step 2. Volume. Reagent: $2 \times 1.1 \times 9.99 = 21.98\ \text{mmol}$. Volume: $\dfrac{21.98\ \text{mmol}}{3.00\ \text{mol/L}} = 7.33\ \text{mL}$.

Step 3. Alcohol. $9.99\ \text{mmol} \times 136.19\ \text{g/mol} \times 0.85 = 1.16\ \text{g}$.

Step 4. Why two methyl groups. The first equivalent adds to the ester and ethoxide leaves, which gives the ketone, acetophenone. The ketone is more reactive than the ester, so a second equivalent adds to it at once. Both added groups are methyl.