Answer: To find the molecular weight of the solute, we can use the formula for the boiling point elevation (∆Tb) caused by a nonionic solute:
∆Tb = Kb * m
Where:
∆Tb = Boiling point elevation
Kb = Boiling point elevation constant (a colligative property)
m = Molality of the solution (moles of solute per kilogram of solvent)
We can first calculate the molality (m) of the solution using the given data:
Mass of solute (m_solute) = 27.9 g
Mass of solvent (m_solvent) = 367 g
Boiling point elevation constant (Kb) = 2.53°C/m
First, convert the given boiling point elevation (83.74°C - 80.1°C) to ∆Tb in degrees Celsius:
∆Tb = 83.74°C - 80.1°C = 3.64°C
Now, calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
Step 1: Calculate moles of solute using its mass and molecular weight (M):
Moles of solute = mass of solute / molecular weight of solute
Step 2: Convert the mass of the solvent to kilograms:
Mass of solvent (m_solvent) = 367 g = 0.367 kg
Step 3: Calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
Now, let's calculate the molecular weight of the solute (M) using the derived molality and the given boiling point elevation constant (Kb):
3.64°C = 2.53°C/m * m
Now, rearrange the equation to solve for m:
m = 3.64°C / 2.53°C/m ≈ 1.44 m
Now, let's calculate the moles of solute using the molality and mass of solvent:
moles of solute = m * mass of solvent (in kg)
moles of solute = 1.44 * 0.367 kg ≈ 0.52848 mol
Finally, calculate the molecular weight (M) of the solute using the moles of solute and the mass of solute:
Molecular weight (M) = mass of solute / moles of solute
M = 27.9 g / 0.52848 mol ≈ 52.82 g/mol
The molecular weight of the solute is approximately 52.82 g/mol.