What volume of 90% alcohol by weight `(d = 0.8 g mL^(-1))` must be used to prepared `80 mL` of 10% alcohol by weight `(d = 0.9 g mL^(-1))`

Explanation

To solve the problem of determining what volume of 90% alcohol by weight (with a density of 0.8 g/mL) is needed to prepare 80 mL of 10% alcohol by weight (with a density of 0.9 g/mL), we can follow these steps:

### Step 1: Understand the Given Information
- We have two solutions:
1. **Solution A**: 90% alcohol by weight, density = 0.8 g/mL
2. **Solution B**: 10% alcohol by weight, density = 0.9 g/mL, volume = 80 mL

### Step 2: Calculate the Mass of Alcohol in Solution B
To find the mass of alcohol in the 10% solution:
- The percentage by weight means that in 100 g of solution, there are 10 g of alcohol.
- For 80 mL of solution B, we first need to find its mass using its density.

\[ \text{Mass of Solution B} = \text{Volume} \times \text{Density} = 80 \, \text{mL} \times 0.9 \, \text{g/mL} = 72 \, \text{g} \]

Now, calculate the mass of alcohol in this solution:
\[ \text{Mass of Alcohol in B} = 10\% \, \text{of} \, 72 \, \text{g} = \frac{10}{100} \times 72 = 7.2 \, \text{g} \]

### Step 3: Set Up the Equation for Solution A
Let \( V \) be the volume of the 90% alcohol solution needed. The mass of alcohol in this solution can be calculated as follows:
\[ \text{Mass of Solution A} = V \times 0.8 \, \text{g/mL} \]
The mass of alcohol in solution A is:
\[ \text{Mass of Alcohol in A} = 90\% \, \text{of} \, (\text{Mass of Solution A}) = 0.90 \times (V \times 0.8) = 0.72V \, \text{g} \]

### Step 4: Equate the Masses of Alcohol
Since the mass of alcohol in both solutions must be equal:
\[ 0.72V = 7.2 \]

### Step 5: Solve for \( V \)
Now, we can solve for \( V \):
\[ V = \frac{7.2}{0.72} = 10 \, \text{mL} \]

### Conclusion
The volume of 90% alcohol by weight needed to prepare 80 mL of 10% alcohol by weight is **10 mL**.