To determine the mass percent of the solute (ethyl alcohol) in the solution, we start by calculating the masses of ethyl alcohol and water using their respective molar masses.

The molar mass of ethyl alcohol (C₂H₅OH) is given as 46 g/mol and the molar mass of water (H₂O) is given as 18 g/mol.

First, we calculate the mass of ethyl alcohol:

$$\text{Mass of ethyl alcohol} = \text{Number of moles} \times \text{Molar mass}$$

$$\text{Mass of ethyl alcohol} = 1 \, \text{mol} \times 46 \, \text{g/mol} = 46 \, \text{g}$$

Next, we calculate the mass of water:

$$\text{Mass of water} = \text{Number of moles} \times \text{Molar mass}$$

$$\text{Mass of water} = 9 \, \text{mol} \times 18 \, \text{g/mol} = 162 \, \text{g}$$

Now, we have the masses of both components of the solution. The total mass of the solution is the sum of the masses of ethyl alcohol and water:

$$\text{Total mass of solution} = 46 \, \text{g} + 162 \, \text{g} = 208 \, \text{g}$$

Then we calculate the mass percent of the solute (ethyl alcohol) using the formula:

$$\text{Mass percent of solute} = \left( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100\%$$

$$\text{Mass percent of solute} = \left( \frac{46 \, \text{g}}{208 \, \text{g}} \right) \times 100\%$$

$$\text{Mass percent of solute} = \left( \frac{46}{208} \right) \times 100\% \approx 22.12\%$$

Since the integer answer is requested, we round 22.12% to the nearest whole number. Therefore, the mass percent of the solute (ethyl alcohol) in the solution is 22%.