Molecular Weight Of Solutes

[Music] the topic of this film is the molecular weight of nonvolatile solutes and the method for determining s we found previously that we can determine the molecular weight of a gas by determining the weight of 22 and 410 L of that gas at standard conditions but this method is Impractical for substances which don't form Vapors at any reasonable temperatures or which should decompose upon heating for these materials however a way is out as provided by means of R's law which states that the vapor pressure of a solution is lowered and therefore its freezing point is correspondingly lowered by an amount which is directly proportional to the molar concentration of the nonsol solute in that solution or that is to the mity of the solution a mality of a solution is of course defined as the number of moles of solute per 1,000 G of solvent and the amount at the freezing point of the solution is below the freezing point of the pure solvent of that solution in a one mow solution is known as the molale freezing point depression our first job will be to calibrate the thermometer that we're going to use in carrying out this experiment this thermometer is graduated from -10° centigrade to+ 40 and each degree is divided into fifths so that we can read the temperature recorded by this thermometer rather accurately in this test tube I've placed in distilled water and in this Beaker we have a mixture of ice and salt and water we're going to determine the freezing point of pure distilled water on this thermometer we know this freezing point ought to be 0° but it's possible the thermometer calibrations might be slightly an error and we need to know what the actual freezing point is as measured by this thermometer to do this we'll lower the test tube containing the thermometer and the distilled water into the ice salt mixture and stir until the water in the test tube freezes we've now been stirring the tube in its contents for a few minutes considerable ice is formed in the tube we'll raise the tube from the beaker I think you can see the ice crystals inside the tube but some water Still Remains and we'll read the freezing ing point of water as recorded by the thermometer this turns out to be 1/10 of a de cenra therefore we'll record the figure 1/10 of a degree as a starting point in this experiment because we're going to be concerned with how much the freezing point of water is lowered by the presence of the solute we've placed a clean dry test tube on the left hand pan of the Ballance and have determined its weight the weight of the empty tube is found to be 39. 657 G we'll now uh place an approximately 4 G sample of Ura into the weighed test tube and then reway in order to obtain the exact weight of the sample of Ura taken the tube containing the sample of Ura has now been reweighed and found this time to weigh 43885 G this means that the sample of Ura taken weighs 4228 G to the sample of Ura in the test tube we will now add 25 mL of distilled water using this pipet we filled the pipet with water to the Mark is a 25 ml pipet by by removing the finger we can cause the water to flow into the test tube and onto the Ura we'll rinse down the side of the tube with the water in order to dissolve any fragments of Ura that might be on the side of the tube we will then place the thermometer in the tube containing the Ura and stir until Ura has all dissolved in the water and while the final traces of Ura are dissolving we'll lower the tube into the beaker of ice water and determine the freezing point of this solution of Ura and water after several minutes of stirring the solution in the tube has begun to freeze think you can plainly see the crystals of Frozen pure solvent in this case ice and the fact that there is plenty of solution left the temperature at this point will be the freezing point of this solution and reading it from the thermometer we find it to be -4.4 de we know then that the amount of Ura that we added to 25 mL of water depressed the freezing point so that it now reads -4.4 on this thermometer the data which we've collected in this experiment is now on the board The observed freezing point of our solution was -4.4 de The observed freezing point of water was. 1° and this means that the freezing point of our solution was lowered 4.5 de the weight of our tube containing the Ura 43.8 A5 G weight of the empty tube 39.67 G meaning that the weight of Ura present in this experiment was 4228 G and the weight of the water in which this Ura was dissolved was 25 G because we use 25 milliliters of water and the density of water is approximately one from this data we will now proceed to calculate the molecular weight of Ura we know that the molale freezing point depression for water is 1.86 de now the mality of this solution can be easily calculated then because our solution had a freezing point depression of 4.5 de and if a one mol solution depresses the freezing point 1.86 de by dividing 4.5 by 1.86 we see that our solution is 2.4 Mol now let's let x equal the weight of Ura dissolved in 1,000 G of water we know then that 4228 G were dissolved in 25 G of water so we can calculate how many grams will be dissolved in 1,000 G of water when this comes out to be 169 G so 169 G of Ura dissolved in 1,000 G of water then gives us the solution which is 2.4 M and our problem is to find out how much Ura it would take to form a solution which is 1 molal so we need to divide the 169 by the 2.4 and this gives us 70 which then is the approximate molecular weight of Ura now actually Ura has a molecular weight of 60 so we have an error here of 10 parts in 60 or about [Music] 177%