What is the molality of a 0.065M solution of ethyl alcohol in water at 25 degrees Celcius? (Assume the density of the solution at such low conentration is equal to that of water at 25 degrees Celcius, 0.997 g./mL)
Note the correct spelling of celsius.
0.065 M = 0.065moles/L solution.
Density of the solution = 0.997 g/mL; therefore, 1 L of solution has a mass of m = volume x density = 1000 x 0.997 = 997 grams.
How much of that is EtOH. 0.065moles x 46 g/mol = 2.99 grams EtOH. The total mass is 997; therefore, the water present must be 997-2.99 = ??g
molality = moles/kg solvent
To find the molality (m), you need two pieces of information: the moles of solute (ethyl alcohol) and the mass of the solvent (water). Here's how you can calculate the molality:
1. Convert the concentration of the solution from molarity (M) to moles of solute.
The concentration is given as 0.065 M, which means there are 0.065 moles of ethyl alcohol in every liter of the solution. Since moles = concentration x volume, we can calculate the moles of ethyl alcohol as follows:
Moles of ethyl alcohol = 0.065 M x 1 L = 0.065 moles
2. Calculate the mass of the solvent (water).
The density of the solution is given as 0.997 g/mL, which means 1 mL of the solution has a mass of 0.997 grams. Since the volume of the solution is not provided, we can assume a convenient value, such as 1000 mL (1 liter) since the density is similar to that of water. Therefore, the mass of the solvent (water) would be:
Mass of the solvent = 0.997 g/mL x 1000 mL = 997 grams
3. Calculate the molality.
Molality (m) is defined as moles of solute per kilogram of solvent.
First, convert the mass of the solvent to kilograms:
Mass of the solvent (in kg) = 997 g / 1000 = 0.997 kg
Now, divide the moles of ethyl alcohol by the mass of the solvent (in kg):
Molality (m) = Moles of solute / Mass of solvent (in kg)
Molality (m) = 0.065 moles / 0.997 kg
Therefore, the molality of the 0.065 M ethyl alcohol solution in water at 25 degrees Celsius is approximately 0.065 mol/kg.