Answer:
The correct answer is 104.13ºC
Explanation:
When a solute is added to a solvent, the boiling point of the solvent (Tb) increases. That is a colligative property. The increment in Tb (ΔTb) is given by the following expression:
ΔTb = Tb - Tbº= Kb x m
Where Tb and Tbº are the boiling points of the solvent in solution and pure, respectively; Kb is a constant and m is the molality of the solution.
In this problem, the solvent is water (H₂O). It is well known that water has a boiling point of 100ºC (Tb). The value of Kb for water is 0.512ºC/m. So, we have to calculate the molality of the solution (m):
m = moles of solute/Kg solvent
The solute is C₂H₆O₂ and we have to calculate the number of moles of this component by dividing the mass into the molecular weight (Mw):
Mw(C₂H₆O₂)= (2 x 12 g/mol) + (6 x 1 g/mol) + (2 x 16 g/mol)= 62 g/mol
⇒ moles of C₂H₆O₂ = mass/Mw = 100 g/(62 g/mol) = 1.613 moles
Now, we need the mass of solvent (H₂O) in kilograms, so we divide the grams into 1000:
200 g x 1 kg/1000 g = 0.2 kg
Finally, we calculate the molality as follows:
m = 1.613 moles of C₂H₆O₂/0.2 kg = 8.06 moles/kg = 8.06 m
The increment in the boiling point will be:
ΔTb = Kb x m = 0.512ºC/m x 8.06 m = 4.13ºC
So, the boiling point of pure water (Tbº=100ºC) will increase in 4.13ºC:
Tb= 100ºC+4.12ºC= 104.13ºC
