The diagram to the right shows plots of vapor pressure versus temperature for a solvent and a solution.
c. What is the approximate concentration of the solution in mol/kg, if 1 mol of solute particles raises the boiling point of 1 kg of solvent by 3.63 °C?

Question

The diagram to the right shows plots of vapor pressure versus temperature for a solvent and a solution.

c. What is the approximate concentration of the solution in mol/kg, if 1 mol of solute particles raises the boiling point of 1 kg of solvent by 3.63 °C?

Accepted Answer

Welcome back, everyone. Plots of vapor pressure versus temperature for a solvent and a solution are shown in the graph below. If one mole of solute dissolved in one kg of solvent raises its boiling point by 2.55 Celsius, determine the approximate concentration of the solution in moles per kilogram. So the first thing we should take note of is that the vapor pressure of a liquid at its boiling point is going to equal the atmospheric pressure. And so recall that that would be at 760 millimeters of mercury. And since vapor pressure is reported in millimeters of mercury on the y axis, we observe the intersection on the two curves given for the pure solvent and the solution where the blue curve has a vapor pressure of 7 60 millimeters of mercury at approximately degrees Celsius. Whereas the orange curve has a vapor pressure of 760 millimeters of mercury at a higher temperature being 70 degrees Celsius. And we're going to recall that in determining which is the solvent versus solution, we would recall that the curve that has a lower boiling point would be the pure solvent So because the boiling point temperature is at 60 degrees Celsius for the blue curve, then the blue curve would be the purest solvent since it has the lower boiling point temperature. And so that means that therefore the orange curve is going to represent the solution with a higher boiling point temperature at 70 Celsius. So with that understood and actually just to make a more clear interpretation for the boiling point of the blue curve, our pure solvent, we actually should note a boiling point temperature at 58 degrees Celsius to be more accurate since it's just before which is marked at that point. And so that means that for the orange curve, the boiling point to be a bit more accurate as far as it's reading, we would say that it is at approximately two units before 70. So we'll say 68 degrees Celsius. So now we're going to calculate the boiling point elevation delta TB where we've got the difference between Celsius for the boiling point of our solution minus the lower boiling 680.0.58 Celsius as the boiling point of our pure solvent. And that would mean that our boiling point elevation is approximately 10 degrees Celsius based on that difference between those two boiling points. Now let's also recall that we can calculate the boiling point elevation by taking the boiling point constant K B and that is the boiling point elevation constant to be specific specific K B which is then multiplied by the modality of the solution. So that is another way of calculating boiling point elevation. And because we need to determine the concentration of our solution in moles per kilogram, we ultimately need to determine the modality of our solution. So we need to isolate this formula to solve for modality. Meaning we would divide both sides by the boiling point elevation constant K B. So that it cancels out on the right. And that would give us the modality of the solution equal to the boiling point elevation, which is then divided by the boiling point elevation constant. So plugging in our nodes, what we would have is then in our num 10 degrees Celsius for the boiling point elevation, which is divided by the boiling point elevation constant, which we will determine by referring to the prompt where it states that the one mole of solute dissolved in one kg of solvent raises the boiling point by 2.55 Celsius. And that would correlate to our boiling point elevation constant, 2.55 Celsius. Now, we should recall that the boiling point elevation constant is in terms of moles of our solute per kilogram of solvent. And so we would have 2.55 degrees Celsius where our Celsius temperature is multiplied by kilograms of our solvent per mole of our solute. I'm sorry, that's backwards. So our boiling point constant is sorry degrees Celsius multiplied by moles of our solute per kilogram of solvent. So now with that understood, we will then cancel out our units of Celsius. And we are left with moles per kilogram as our final units. And this will simplify so that we find that we have a modality of the solution equal to a value of 3.92 moles per kilogram and 3.92 moles per kilogram would be our final answer to complete this example as the morality of the solution or concentration of the solution. I hope that this made sense and let us know if you have any questions.

The diagram to the right shows plots of vapor pressure versus temperature for a solvent and a solution.
c. What is the approximate concentration of the solution in mol/kg, if 1 mol of solute particles raises the boiling point of 1 kg of solvent by 3.63 °C?
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Key Concepts

Vapor Pressure

Boiling Point Elevation

Molality