How do you calculate molality?

How To Calculate Molality Given Pressure and Temperature?

• How To Calculate Molality Given Pressure and Temperature The vapor pressure of an aqueous glucose (C6H12O6) solution is 17.01 Torr at 20 degrees C, while that of pure water is 17.25 Torr at the same temperature. Calculate the molality of the solution. Possible Answers: a. 0.98 m b. 1.3 m c. 1.2 m d. 1.4 m e. 0.78 m This is the question and so far I'm so confused. First of all, I know that moles (n)=pressure*volume/R*temperature pressure: 17.01torr*(1/760atm)=.02238 atm R is a constant at .08205745 Temperature: must be in Kelvins so 20 + 273.15=293.15 K molar mass of C6H12O16=180.156 g But what about the volume? I'm not given a volume of the solution or a mass percent so that I can use the molar mass of both the solute and solvent H20 to figure it out and thus, it renders this equation completely useless. And I can't just convert 180.156 g to kg to liters and use that as a volume because that doesn't equal what the number of solutes in the solution and it doesn't account for the rest of the solution. And I tried plugging it in that way and I got some weird number after applying the same concept for H20 and going from moles to kg and then dividing it since molality = moles solute/mass solvent (kg). Can anyone help?

• Answer:

  The relation you want is that the vapor pressure of a component in a mixture (in this case, water and glucose) will decrease as the mole fraction of that component decreases (this is Raoult's Law). You know that the mole fraction of water in pure water is 1, and the vapor pressure for a mole fraction of 1 is 17.25 torr. With the glucose in solution the vapor pressure drops to 17.01 torr, meaning the addition of glucose decreased the mole fraction of water. So the new mole fraction of water in the solution is given by the ratio 17.01/17.25. You know the mole fractions of all components have to sum to 1, so the mole fraction of glucose is 1-17.01/17.25. Then, you have to convert mole fraction of the glucose to molality remembering that there are around 55.5 moles of water in a kilogram of water. I don't think you need to use the ideal gas law. This is just a Raoult's Law problem. I didn't solve this exactly, but one thing I do know is that you should not confuse the mole fraction of water in the mixture with the molality of glucose in the solution.