Final answer:
The percent by mass of CH3CH2OH in an aqueous solution with a molarity of 6.99 M and a density of 0.947 g/mL is approximately 99.1%. This calculation assumes that the volume remains constant when the solute (ethanol) is added.
Step-by-step explanation:
To calculate the percent by mass of CH3CH2OH (ethanol) in the solution, we need to find the mass of ethanol in a certain volume of solution and then compare this to the total mass of the solution.
Given the molarity of the ethanol solution is 6.99 M and the density of the solution is 0.947 g/mL, we first calculate the mass of ethanol in 1 liter (which is 1000 mL) of the solution:
- Mass of ethanol = volume of solution × density of solution × molarity of ethanol
- Mass of ethanol = 1000 mL × 0.947 g/mL × 6.99 moles/L
- Mass of ethanol = 6613.53 g
Next, we calculate the total mass of the solution, which includes the mass of the water and the ethanol. Assuming we are referring to 1 liter of solution, the mass of water is 60.0 g. So the total mass of the solution would be:
- Total mass = mass of ethanol + mass of water
- Total mass = 6613.53 g + 60.0 g
- Total mass = 6673.53 g
Lastly, we calculate the percent by mass of ethanol:
- Percent by mass = (mass of ethanol / total mass of solution) × 100
- Percent by mass = (6613.53 g / 6673.53 g) × 100
- Percent by mass = 99.1%
Note: This calculation assumes that the added mass of ethanol does not significantly alter the volume of the solution, which might not be the case in the real world due to volume contraction upon mixing.