Final answer:
The freezing point of a solution with 20 g of ethyl alcohol dissolved in 590 g of water is determined by calculating the molality and using this in the freezing point depression formula. The resulting freezing point of the solution is -1.37 °C.
Step-by-step explanation:
The question asks for the freezing point of a solution containing ethyl alcohol dissolved in water. To find this, we first calculate the molality of the solution, which is the number of moles of solute per kilogram of solvent. The formula for molality (m) is:
m = moles of solute / kilograms of solvent
Given that the molecular weight (MW) of ethyl alcohol (C2H5OH) is 46, we can calculate the moles of ethyl alcohol as follows:
moles of C2H5OH = mass (g) / MW
moles of C2H5OH = 20 g / 46 g/mol = 0.435 moles
Next, we calculate the molality:
molality (m) = 0.435 moles / 0.590 kg = 0.737 mol/kg
The freezing point depression (ΔTf) can be calculated using the equation:
ΔTf = i * Kf * m
Where 'i' is the van't Hoff factor which is 1 for non-electrolyte solutes like ethyl alcohol, 'Kf' is the freezing point depression constant for the solvent (water in this case), which is 1.86 °C/m, and 'm' is the molality calculated above. Thus:
ΔTf = 1 * 1.86 °C/m * 0.737 mol/kg = 1.37 °C
Since the normal freezing point of water is 0 °C, the new freezing point of the solution will be:
new freezing point = 0 °C - ΔTf
new freezing point = 0 °C - 1.37 °C
new freezing point = -1.37 °C
Therefore, the freezing point of the ethyl alcohol solution is -1.37 °C.