How many mL of alcohol is there in 330 ml of beer with 5.0% (v/v) alcohol content?
The correct answer and explanation is:
Correct Answer: 16.5 mL of alcohol
To determine how many milliliters of alcohol are present in 330 mL of beer with a 5.0% (v/v) alcohol content, use the definition of volume/volume percentage. Volume/volume percent expresses the volume of solute (in this case, alcohol) per 100 mL of solution (the beer).
The formula is:
Volume of alcohol = (Alcohol % v/v) × (Total volume of solution) / 100
Substituting in the given values:
Volume of alcohol = (5.0 × 330) / 100
= 1650 / 100
= 16.5 mL
This means that in 330 mL of beer, 16.5 mL is pure alcohol.
This calculation is based purely on the proportion defined by the percentage. Volume/volume percent is commonly used for liquid mixtures, especially beverages like beer, wine, and spirits. When the label on a beer bottle indicates 5.0% alcohol by volume, it tells you that 5.0 mL of pure ethanol is present in every 100 mL of beer. Therefore, scaling this ratio up to a 330 mL serving simply involves multiplying 330 mL by 0.05.
This calculation assumes ideal mixing, meaning the alcohol is uniformly distributed throughout the beer. It does not account for temperature changes or density differences between alcohol and water. However, for practical purposes, such approximations are sufficient in most everyday contexts.
Understanding this helps consumers be aware of their alcohol intake. It also helps in calculating limits for safe driving, knowing the approximate alcohol content ingested. Always remember that alcohol percentage can vary between different brands and types of alcoholic beverages.