A pharmacist needs to strengthen a 15% alcohol solution to one of 32% alcohol. How much pure alcohol should be added to 400 mL of the 15% solution?

Options

A. 1000 ml
B. 68 ml
C. 100ml
D. 128 ml

Answer

To solve the problem of how much pure alcohol should be added to 400 mL of a 15% alcohol solution to strengthen it to a 32% alcohol solution, we can follow these steps:

### Step 1: Calculate the amount of alcohol in the initial solution
The initial solution is 15% alcohol. We can calculate the amount of alcohol in 400 mL of this solution.

\[ \text{Amount of alcohol} = \text{Volume of solution} \times \text{Concentration} \]
\[ \text{Amount of alcohol} = 400 \, \text{mL} \times \frac{15}{100} = 60 \, \text{mL} \]

### Step 2: Set up the equation for the final concentration
Let \( y \) be the amount of pure alcohol to be added. After adding \( y \) mL of pure alcohol, the total volume of the solution will be \( 400 + y \) mL, and the total amount of alcohol will be \( 60 + y \) mL.

The final concentration of the solution should be 32%. We can express this as:

\[ \frac{60 + y}{400 + y} = \frac{32}{100} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ 100(60 + y) = 32(400 + y) \]

### Step 4: Expand both sides
Expanding both sides of the equation:

\[ 6000 + 100y = 12800 + 32y \]

### Step 5: Rearrange the equation
Now, we will rearrange the equation to isolate \( y \):

\[ 100y - 32y = 12800 - 6000 \]
\[ 68y = 6800 \]

### Step 6: Solve for \( y \)
Now, divide both sides by 68 to find \( y \):

\[ y = \frac{6800}{68} = 100 \, \text{mL} \]

### Conclusion
The amount of pure alcohol that should be added to the 400 mL of 15% alcohol solution is **100 mL**.

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