For example, if I dissolve 12 g of KCl in 400 mL of water, will the boiling point differ from a solution with doubled the amount of KCl (24 g) in 400 mL of water? Please explain, my teacher was not very clear on this subject.
Thanks in advance.
Thanks in advance.
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Your question has to do with the subject of boiling point elevation and dissociation in general. The basic concept is this: a nonvolatile solute (meaning one that generally stays in solution) will cause the solution to remain a liquid over a longer temperature range than that of the normal solvent. Water is the universal solvent, so its constants are on easily accessible charts. The Holt, Rinehart and Winston Chemistry book says that the Kb (boiling point constant) for water is .51 C/m (sorry I don't know how to do a degree sign on this system).
The equation you need is delta Tb = Kb * m (translation: the change in the boiling point temperature equals the boiling point constant multiplied by the solution's molality.)
In the example you gave, you would have to first find the molality for both KCl solutions.
Solution A (12 g KCl in 400. mL H2O); Solution B (24 g KCl in 400. mL H2O)
molality = moles solute / kg solvent
Step 1: conversions (g to mol solute and ml to kg solvent)
12 g KCl x 1 mol KCl/ 74.551 g KCl = .161 mol KCl
24 g KCl x 1 mol KCl/ 74.551 g KCl = .322 mol KCl
400. mL H2O x 1 g H2O/1 mL H2O x 1 kg x 1000 g = .400 kg H2O (same for both solns)
Step 2: molality calculations
m = .161 mol KCl / .400 kg = .4025 m
m = .322 mol KCl / .400 kg = .805 m
Step 3: plug into delta Tb = Kb * m
delta Tb = (.51 C/m)* .4025 m = .205275 degrees C
This means that the boiling point of the solution will be .205275 degrees celcius higher than that of normal water. (b.p. of water = 100. degrees C)
delta Tb = (.51C/m) * .805m = .41055 degrees C
This means that the boiling point of the solution will be .41055 degrees celcius higher than that of normal water.
Conclusion: doubling solute in a given amount of solvent will raise the boiling point elevation by roughly double.
The equation you need is delta Tb = Kb * m (translation: the change in the boiling point temperature equals the boiling point constant multiplied by the solution's molality.)
In the example you gave, you would have to first find the molality for both KCl solutions.
Solution A (12 g KCl in 400. mL H2O); Solution B (24 g KCl in 400. mL H2O)
molality = moles solute / kg solvent
Step 1: conversions (g to mol solute and ml to kg solvent)
12 g KCl x 1 mol KCl/ 74.551 g KCl = .161 mol KCl
24 g KCl x 1 mol KCl/ 74.551 g KCl = .322 mol KCl
400. mL H2O x 1 g H2O/1 mL H2O x 1 kg x 1000 g = .400 kg H2O (same for both solns)
Step 2: molality calculations
m = .161 mol KCl / .400 kg = .4025 m
m = .322 mol KCl / .400 kg = .805 m
Step 3: plug into delta Tb = Kb * m
delta Tb = (.51 C/m)* .4025 m = .205275 degrees C
This means that the boiling point of the solution will be .205275 degrees celcius higher than that of normal water. (b.p. of water = 100. degrees C)
delta Tb = (.51C/m) * .805m = .41055 degrees C
This means that the boiling point of the solution will be .41055 degrees celcius higher than that of normal water.
Conclusion: doubling solute in a given amount of solvent will raise the boiling point elevation by roughly double.