A chemistry student was asked to make 100 L of alcohol solution by mbing
alcohol solution and alcohol solution. How much of each solution must the student
use?

Ask by Black Stanley. in Canada

Jan 23,2025

Question

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?

UpStudy AI · Accepted Answer

To determine how much of each alcohol solution the chemistry student needs to prepare 100 liters of a 48% alcohol solution by mixing 40% and 60% alcohol solutions, we can set up a system of equations.

**Let:**
- $$ x $$ = liters of 40% alcohol solution
- $$ y $$ = liters of 60% alcohol solution

**We have two conditions:**

1. **Total Volume:**
   $$
   x + y = 100 \quad \text{(since the total volume needed is 100 liters)}
   $$

2. **Total Alcohol Content:**
   $$
   0.40x + 0.60y = 0.48 \times 100 = 48 \quad \text{(since the final mixture needs to be 48% alcohol)}
   $$

**Solving the Equations:**

From the first equation:
$$
y = 100 - x
$$

Substitute $$ y $$ into the second equation:
$$
0.40x + 0.60(100 - x) = 48
$$
$$
0.40x + 60 - 0.60x = 48
$$
$$
-0.20x + 60 = 48
$$
$$
-0.20x = -12
$$
$$
x = \frac{-12}{-0.20} = 60
$$

Now, find $$ y $$:
$$
y = 100 - 60 = 40
$$

**Conclusion:**
- **60 liters** of the **40% alcohol solution**
- **40 liters** of the **60% alcohol solution**

are needed to prepare 100 liters of a 48% alcohol solution.

**Answer:**
Use 60 litres of the 40 % solution and 40 litres of the 60 % solution to make 100 litres.
To make 100 liters of a 48% alcohol solution, the student needs to use 60 liters of the 40% solution and 40 liters of the 60% solution.

A chemistry student was asked to make 100 L of alcohol solution by mbing
alcohol solution and alcohol solution. How much of each solution must the student
use?

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