The freezing point for tert-butanol in literature is 78 F, or approximately 25 C, which is quite close to what was observed, as stated in the Freezing Points Table. (NOAA, 2016). The molarity of the additive substance could be calculated by taking the moles used and diving by the volume of the solvent. So for benzoic acid, there would be two different values. For the first, the moles could be calculated by dividing the grams used by the molar mass of benzoic acid, 122.123. (NIH, 2016). .3/122.123= .0025 mol, which would then be divided by the volume of the solution, .065L. .0025/.0065=.378 M. This process is then repeated for each additive to get, .768 M of Benzoic acid and .384 M of Camphor. To find the molality of a substance, you divide the the moles of solute over the kg of the solvent. So, the moles of the additive are divided over the kg of the original substance, which in this case is 0.005kg of tert-butanol. For the first benzoic acid, it would be 0.0025 mol/0.005 kg, to get 0.491 m. For the second it would be 0.005 mol/ 0.005kg, to get 1 m. For camphor, it would be …show more content…
To calculate freezing point depression, the freezing point of the solution is subtracted by the freezing point of the pure solvent. The actual freezing point depressions would be, 6.3 for .3 g of benzoic acid, 7.1 for .61g of benzoic acid, and 4 for .38g of Camphor. This information develops an understanding of freezing points because it shows actual examples of what is happening. More can be drawn by knowing the original freezing points, and it can be seen how with different molecules, there are different bonds that are harder to bring together, and therefore are harder to freeze. With one solute being added twice, just in different amounts, the effects of the amount of solute added can be seen