In: Chemistry
What is the approximate temperature of a 0.1 molar solution of benzoic acid in acetone that is heated in a boiling water bath?
Ans. Step 1: Calculate molality of solution
Given, solution = 0.1 M benzoic acid in acetone or 0.1 mol acetone in 1.0 L acetone.
It’s assumed that there is no change in volume after addition of benzoic acid. Also, the solution is initially at room temperature (25.00C).
Moles of benzoic acid in 1.0 L solution = 0.1 mol
Volume of of acetone (solvent) = 1.0 L
Mass of acetone (solvent) = Volume x density
= 1.0 L x (0.78576 kg/ L)
= 0.78576 kg
Now, molality of 0.1 M benzoic acid in acetone = Moles of benzoic acid / Mass of acetone
= 0.1 mol / 0.78576 kg
= 0.127 mol/ kg
= 0.127 m
Step 2: Elevation in boiling point of a solution is given by-
dTb = i Kb m - equation 1
where, i = Van’t Hoff factor = [i = 1 for non-electrolyte solute].
Kb = molal boiling point elevation constant of solvent, acetone = 1.710C/m
m = molality of the solution
dTb = Boiling point of solution - Boiling point of pure solvent
Boiling point of acetone = 56.050C
Putting the values in equation 1-
dTb = 1 x (1.710C/m) x 0.127 m = 0.2170C
Now,
Boiling point of solution = Boiling point of pure solvent + dTb
= 56.050C + 0.2170C
= 56.720C
Therefore, boiling point of resultant solution = 56.720C
Boiling point is the minimum temperature at which the liquid phase transits into vapor phase. It remains constant till all the liquid is converted into vapor. So, whatever the temperature of water bath, here temperature of solution remains constant at 56.270C until all of it is vaporized.