Freezing Point Deperession Essay Sample

Decision: My Lab unsuccessfully found the molar mass of the unknown given to us in the lab. We were tasked with happening the molar mass of an unknown compound utilizing the colligative belongings of Freezing Point Depression. The manner we did this was first happening the temperature at which ice crystals form for merely regular BHT. Fortunately thanks to our TA we were given temperature investigations so we could nail exact temperatures. We found that the temperature at which crystallisation of BHT was 68. 9 Celsius in our lab. After happening temperature of BHT we were tasked with adding the solute cetyl acid to our BHT mixture. We so heated and melted the two substances than found the freezing point of the two substances. We found the temperature at which crystals formed of the two to be 63. 8 grades Celsius. Now the chief ground we mixed these two substances was to happen the freezing point of BHT. Now to make this we need to take the gms of cetyl acid and happen moles of solute and split it by kgs of our dissolver.

This gave the molal concentration of our solution which turned out to be. 515 mol/kg. After happening molal concentration we could happen kfp or the freezing point. To make this you would utilize the expression ?Tfp = kfpm. so we take the alteration in temperature which is 5. 1 grades Celsius equal to kfp times. 515 mol/kg all which turns out to be 9. 90 oc/molal. After we find this important piece of information we can add the unknown to our BHT melt the two down and happen the ?T of it which comes out to be 3. 3. With all this information given we can eventually find the molar mass of our unknown. To make this we would utilize the equation MM ( unknown solute ) = kfp x g ( solute ) / kilogram ( dissolver ) x ?Tfp now when we plugged all our information into said expression and crunched the Numberss we found our molar mass to be 386 g/mol. We found out that the right molar mass was 284 g/mol and the terra incognita was stearic acid.