A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ________ (Integer answer) (Given : Molar mass in $\mathrm{g} \mathrm{~mol}^{-1}$ Ethyl alcohol : 46 water : 18)

<p>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.</p> <p>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.</p> <p>First, we calculate the mass of ethyl alcohol:</p> <p> <p>$\text{Mass of ethyl alcohol} = \text{Number of moles} \times \text{Molar mass}$</p> </p> <p> <p>$$\text{Mass of ethyl alcohol} = 1 \, \text{mol} \times 46 \, \text{g/mol} = 46 \, \text{g}$$</p> </p> <p>Next, we calculate the mass of water:</p> <p> <p>$\text{Mass of water} = \text{Number of moles} \times \text{Molar mass}$</p> </p> <p> <p>$$\text{Mass of water} = 9 \, \text{mol} \times 18 \, \text{g/mol} = 162 \, \text{g}$$</p> </p> <p>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:</p> <p> <p>$$\text{Total mass of solution} = 46 \, \text{g} + 162 \, \text{g} = 208 \, \text{g}$$</p> </p> <p>Then we calculate the mass percent of the solute (ethyl alcohol) using the formula:</p> <p> <p>$$\text{Mass percent of solute} = \left( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100\%$$</p> </p> <p> <p>$$\text{Mass percent of solute} = \left( \frac{46 \, \text{g}}{208 \, \text{g}} \right) \times 100\%$$</p> </p> <p> <p>$$\text{Mass percent of solute} = \left( \frac{46}{208} \right) \times 100\% \approx 22.12\%$$</p> </p> <p>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 <b>22%</b>.</p>