A chemistry student was asked to make 100 L of \( 48 \% \) alcohol solution by mbing \( 40 \% \) alcohol solution and \( 60 \% \) alcohol solution. How much of each solution must the student use?

To make a \( 48\% \) alcohol solution, we can use the concept of mixing solutions. Let \( x \) be the volume of the \( 40\% \) alcohol solution and \( y \) be the volume of the \( 60\% \) alcohol solution. We know that: 1. \( x + y = 100 \) 2. The alcohol content can be expressed as \( 0.40x + 0.60y = 0.48(100) \) Solving these equations gives us: 1. From \( x + y = 100 \), we can express \( y = 100 - x \). 2. Substitute \( y \) in the second equation: \[ 0.40x + 0.60(100 - x) = 48 \] Solving this results in \( x = 40 \) and \( y = 60 \). Thus, you need \( 40 \) liters of the \( 40\% \) alcohol solution and \( 60 \) liters of the \( 60\% \) alcohol solution to create \( 100 \) liters of an \( 48\% \) alcohol solution. Mixing solutions is a bit like being a cocktail connoisseur—it's all about hitting the right balance! Just like in bartending, you've got to pay attention to how each ingredient contributes to the final flavor. In this case, we're focusing on alcohol concentrations instead of fruity garnishes! Next time you're measuring into a graduated cylinder or beaker, remember the importance of precision! It's crucial not to confuse \( \% \) with the absolute amounts. A common mistake is to think that just pouring a bit of one solution into another can simply yield the desired \( \% \). It's always good to double-check your math to avoid any dilution disasters!