Question

A scientist has 2 liters of a 30% saline solution. How many liters of pure water would he need to dilute the saline concentration to 8.5%? Round to the nearest hundredth. Define a variable and solve.

A) 0.84 liters
B) 1.16 liters
C) 1.32 liters
D) 1.52 liters

Answer 1

To dilute a 2-liter 30% saline solution to an 8.5% concentration, the equation 0.30 × 2 L = 0.085 × (2 L + x L) is used, resulting in approximately 5.06 liters of pure water needed, which does not match the provided options.

Step-by-step explanation:

To determine how many liters of pure water need to be added to dilute a 30% saline solution to an 8.5% concentration, we can set up an equation based on the concept of dilution. The amount of pure water to be added will be our variable, let's call it x liters. The total volume of the new solution will be the original 2 liters plus the added x liters of water.

The amount of salt in the original solution (which is 30% of 2 liters) remains the same; only the total volume of liquid changes. Our equation becomes:

0.30 × 2 L = 0.085 × (2 L + x L)

This simplifies to:

0.60 = 0.085 × (2 + x)

Solving for x, we have:

0.60 = 0.17 + 0.085x

0.60 - 0.17 = 0.085x

0.43 = 0.085x

x = 0.43 / 0.085

x = 5.06

Therefore, approximately 5.06 liters of pure water are needed to dilute the solution, which does not match with any of the answer choices. It is possible there was a typo or misunderstanding regarding the options provided.