What is the boiling point of a 1.5 aqueous solution of fructose? The boiling point elevation constant of water is 0.512C/m. Assume the solute is a no electrolyte.
To find the boiling point of a solution, we can use the formula:
ΔTb = i * Kf * m
Where:
ΔTb = Boiling point elevation
i = Van't Hoff factor (1 for non-electrolytes)
Kf = Boiling point elevation constant of water (0.512C/m)
m = Molality of the solution
First, we need to calculate the molality of the solution:
Molar mass of fructose = 180.16 g/mol
1.5 moles of fructose in 1 L of solution = 1.5 * 180.16 = 270.24 g
Molality (m) = moles of solute / kg of solvent
m = 1.5 / (1000 mL * 1.0 g/mL) = 0.0015 mol/g
Now we can plug the values into the formula:
ΔTb = 1 * 0.512 * 0.0015 = 0.000768 C
To find the boiling point of the solution, we add the boiling point elevation to the boiling point of pure water:
Boiling point of pure water = 100 C
Boiling point of solution = 100 C + 0.000768 C = 100.000768 C
Therefore, the boiling point of a 1.5 aqueous solution of fructose is 100.000768 C.