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 is present in 330 mL of beer with 5.0% (v/v) alcohol content.
Explanation:
To determine the volume of alcohol in a solution that has a certain percent by volume, the formula used is:
Volume of alcohol = (Alcohol % by volume / 100) × Total volume of solution
In this case:
Alcohol % by volume = 5.0
Total volume of beer = 330 mL
Substituting the values into the formula:
Volume of alcohol = (5.0 / 100) × 330
Volume of alcohol = 0.05 × 330
Volume of alcohol = 16.5 mL
This means that out of the total 330 mL of beer, 16.5 mL is pure alcohol, specifically ethanol.
Percent by volume (% v/v) indicates how many milliliters of a solute (in this case, ethanol) are present in 100 milliliters of the total solution. A 5.0% v/v alcohol content implies that every 100 mL of the beverage contains 5 mL of alcohol. By simple proportional reasoning, increasing the total volume to 330 mL means the alcohol volume also increases in the same proportion.
This method of expressing alcohol content is standard in beverages like beer, wine, and spirits. It allows consumers to estimate their alcohol intake. The knowledge of the actual alcohol volume helps in understanding the potential effects of consumption, making it relevant in both health and legal contexts such as driving regulations.
Understanding how to calculate alcohol volume from percentage content is essential in chemistry, nutrition, and responsible consumption practices. It is a straightforward multiplication problem based on the concept of ratios, specifically a volume-to-volume ratio in a homogeneous liquid mixture.