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 calculate the molality of a solution, you need to know the moles of solute (ethyl alcohol) and the mass of the solvent (water).
To determine the moles of ethyl alcohol in the solution, you use its molarity and the volume of the solution.
Given:
Molarity of ethyl alcohol solution (M) = 0.068 M
Volume of solution (V) = not specified (we'll assume 1 liter for simplicity)
First, convert the volume from liters to milliliters:
1 L = 1000 mL
Next, use the definition of molarity to calculate the moles of ethyl alcohol:
moles of ethyl alcohol (n) = Molarity (M) x Volume (V)
n = 0.068 mol/L x 1 L = 0.068 mol
Now, to determine the mass of the solvent (water), the density of the solution is used. Since the density of the solution is equal to that of pure water, you can assume that the mass of the solution is 1 kg (1000 g) since 1 L of water weighs approximately 1000 g.
Therefore, the molality (m) of the solution is calculated as:
molality (m) = moles of solute (n) / mass of solvent (in kg)
Since 1 kg of water has a mass of 1000 g:
m = 0.068 mol / 1000 g = 0.000068 mol/g
However, molality is typically expressed as moles of solute per kilogram of solvent, so we need to convert grams to kilograms:
1 kg = 1000 g
m = 0.000068 mol / 1000 g = 6.8 x 10^-8 mol/g
Therefore, the molality of the 0.068 M ethyl alcohol solution in water at 25 degrees Celsius is 6.8 x 10^-8 mol/g.