Answer: The changes are the mL of methanol ingestion (30mL) and the proof of the whiskey (40% ethanol; 80 proof), and the percentage of reduction (1%). I applied the formulas from the last post in reference to this question but am completely lost. The formulas used last time were: 2. Formulas Applied alpha=1 + ([etOH]/KetOH) (V[meOH]/V[etOH])= (Vmax*[meOH]/KmeOH+[meOH])/(Vmax*[meOH]/alpha*KmeOH+[meOH]) which reduces to (V[meOH]/V[etOH])=(alpha*KmeOH+[meOH])/(KmeOH+[meOH]) 3. My attempt Molarity of methanol: 30mL; which equates to 23.7g of methanol; in 40L that is equal to 0.5925 g/L Dividing the molecular weight by 32.04g/mol I get 0.0184925 which is approximately 0.02M; Km is 0.01M Since the molar mass of methanol and ethanol are two fold, I can multiply the g/l by 4. However, unlike the previous problem, I cannot multiply by 2 because I do not have 50% EtOH, so because 40 is less than 50 I assume to multiply by 2.5 yielding: (30mL)(4)(2.5)=300mL 300mL of EtOH to effectively reduce the Methanol to 1%.