Answer:
Boiling point of the solution = 103.11 °C.
Step-by-step explanation:
To determine the boiling point of a solution, we need to use the equation:
ΔTb = Kb · m
where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant for water (0.512 °C/m), and m is the molal concentration of the solute (in mol/kg).
First, we need to calculate the molality of the solution, which is the number of moles of solute per kilogram of solvent. We can use the formula:
molality (m) = moles of solute / mass of solvent in kg
The mass of water is 425 g, which is 0.425 kg. To find the moles of ethanol, we can use the formula:
moles of solute = mass of solute / molar mass of solute
The molar mass of ethanol is 46.07 g/mol, so the moles of ethanol are:
moles of ethanol = 119 g / 46.07 g/mol = 2.58 mol
Therefore, the molality of the solution is:
m = 2.58 mol / 0.425 kg = 6.07 mol/kg
Now we can use the boiling point elevation equation to find ΔTb:
ΔTb = Kb · m
ΔTb = 0.512 °C/m · 6.07 mol/kg
ΔTb = 3.11 °C
The boiling point elevation is 3.11 °C, which means that the boiling point of the solution is higher than the boiling point of pure water. To find the boiling point of the solution, we need to add the boiling point elevation to the boiling point of water at standard pressure (1 atm), which is 100.00 °C. Therefore:
boiling point of solution = 100.00 °C + 3.11 °C = 103.11 °C
So the boiling point of the solution prepared by dissolving 119 g ethanol in 425 g water is 103.11 °C.