What is the quantity, in moles, of CH3OH present in 150.0 mL of a 0.210 M CH3OH solution?

• A. 0.00315 mol
• B. 0.210 mol
• C. 0.0315 mol
• D. 0.0720 mol

Answer: (C)

Verified step by step guidance

1. Identify the given information: the volume of the solution is 150.0 mL and the molarity (M) of CH3OH is 0.210 M. Molarity is defined as moles of solute per liter of solution.
2. Convert the volume from milliliters to liters because molarity uses liters: use the conversion \$1\, \text{L} = 1000\, \text{mL}\(, so \)V(\text{L}) = \frac{150.0\, \text{mL}}{1000}$.
3. Use the molarity formula to find moles of CH3OH: \(\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\) which rearranges to \(\text{moles of solute} = \text{Molarity} \times \text{liters of solution}\).
4. Substitute the values into the equation: \(\text{moles of CH3OH} = 0.210\, \text{mol/L} \times V(\text{L})\) where \(V(\text{L})\) is the volume in liters from step 2.
5. Calculate the product to find the number of moles of CH3OH present in the 150.0 mL solution.