what is the molality of a 0.068M solution of ethyl alcohol in water at 25 degrees celsius. (assume the density of the solution at such a low concentration is equal to that of water at 25degrees celsius, 0.997g/ml)
To find the molality of a solution, you need to know the amount of solute (in moles) and the mass of the solvent (in kilograms). In this case, the solute is ethyl alcohol and the solvent is water.
Given information:
- Concentration of the ethyl alcohol solution: 0.068 M
- Density of the solution: 0.997 g/ml
- Temperature of the solution: 25 degrees Celsius
Step 1: Convert the density of the solution to kg/L.
The given density is in grams per milliliter (g/ml). To convert it to kilograms per liter (kg/L), divide by 1000:
0.997 g/ml / 1000 = 0.997 kg/L
Step 2: Calculate the mass of the solvent.
The mass of the solvent can be found by multiplying the density by the volume of the solution:
Mass of solvent = Density x Volume
Since the concentration of the ethyl alcohol solution is given as 0.068 M, it means there are 0.068 moles of ethyl alcohol in 1 liter of the solution. Therefore, the volume of the solution is 1 L.
Mass of solvent = 0.997 kg/L x 1 L = 0.997 kg
Step 3: Calculate the molality.
Molality is defined as the moles of solute divided by the mass of the solvent in kg.
Molality = Moles of Solute / Mass of Solvent (in kg)
Since the concentration of the ethyl alcohol solution is given as 0.068 M, it means there are 0.068 moles of ethyl alcohol in 1 L of the solution.
Molality = 0.068 moles / 0.997 kg ≈ 0.068 mol/kg
So, the molality of the 0.068 M solution of ethyl alcohol in water at 25 degrees Celsius is approximately 0.068 mol/kg.