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adamp1988

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The following question has got me completely confused. I know there was someone previous that had posted a question almost identical to mine however there are slight but significant changes.

Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADH to the quite toxic product of formaldehyde. The toxic effects of ingesting methanol (a component of many commercial solvents) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of methanol by displacing it from LADH. This provides sufficient time for the methanol to be harmlessly excreted by the kidneys. If an individual has ingested 30 mL of methanol (a lethal dose), how much 80 proof whiskey (40% ethanol by volume) must be imbibed to reduce the activity of his LADH towards methanol to 1% of its original value? The adult human body contains ~40L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g/cm3. Assume the KM values of LADH for ethanol and methanol to be 10-3 M and 10-2 M, respectively, and that Ki = KM for ethanol.

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%?

I'm completely lost.