how many millimeters of a 5.15 M CH3OH stock solution are needed to prepare 375 mL of a solution having 7.50 mg of methanol per mL of solution

17 mL

I ran through it and obtained 17.06 mL which I would round to 17.1 mL.

To calculate the number of millimeters of a 5.15 M CH3OH stock solution needed to prepare 375 mL of a solution with 7.50 mg of methanol per mL of solution, we need to consider the molar mass of CH3OH and use the equation:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's calculate the moles of methanol (CH3OH) needed to achieve a concentration of 7.50 mg/mL in 375 mL of solution:

Mass of methanol = concentration × volume of solution
Mass of methanol = 7.50 mg/mL × 375 mL = 2812.5 mg

Next, we need to convert the mass of methanol to moles by dividing it by the molar mass of CH3OH.

The molar mass of CH3OH is:
(1 × atomic mass of C) + (4 × atomic mass of H) + (1 × atomic mass of O)
= (1 × 12.01 g/mol) + (4 × 1.01 g/mol) + (1 × 16.00 g/mol)
= 32.05 g/mol

Now, let's convert the mass of methanol to moles:
Moles of methanol = mass of methanol / molar mass of CH3OH
Moles of methanol = 2812.5 mg / 32.05 g/mol = 87.75 mmol

Finally, we can calculate the volume (in milliliters) of the 5.15 M CH3OH stock solution needed to obtain 87.75 mmol of methanol:

Volume of stock solution (mL) = moles of solute / molarity of stock solution
Volume of stock solution = 87.75 mmol / 5.15 M = 17 mL (rounded to the nearest whole milliliter)

Therefore, you would need approximately 17 milliliters of the 5.15 M CH3OH stock solution to prepare 375 mL of a solution containing 7.50 mg of methanol per mL.