We know this is a reasonable answer since vitamins and proteins have molecular weights in the thousands or even tens of thousands. X is the molecular weight of Vitamin K, 0.500 and 4.43 are from the problem and 0.010 is 10 g of camphor done as kilograms. ΔT = (☌ kg mol¯ 1) times (g / g mol¯ 1 kg¯ 1)Ģ) Now, let's insert numbers in the proper place: Next I will replace mol with g / g mol¯ 1: Notice that i goes away since it is unitless (and equal to 1, in any case). Replacing the right side with units gives: We will start with the freezing point depression equation: Calculate the molecular weight of vitamin K.ġ) To solve this problem, I'd like to engage in an analysis of the units. When 0.500 g is dissolved in 10.0 g of camphor, the freezing point is lowered by 4.43 ☌. Problem #20: Vitamin K is involved in the blood clotting mechanism. There is a tiny curve in this problem, but keep in mind that colligative properties are all about how many particles in solution and nothing else.ġ) The key to this problem is to calculate moles of each substance and add then together: Problem #19: Calculate the freezing point of a solution of 5.00 g of diphenyl C 12H 10 and 7.50 g of naphthalene, C 10H 8 dissolved in 200.0 g of benzene (fp = 5.5 ☌) Express your answer using two significant figures. Calculate the molar mass of the unknown compound. Problem #18: An aqueous solution containing 34.3 g of an unknown molecular (nonelectrolyte) compound in 160.0 g of water was found to have a freezing point of −1.3 ☌. Calculate the molecular weight for the unknown. When 0.680 g of an unknown are added to the 10.180 g of benzophenone the resulting solution is found to freeze at 42.6 ☌. Problem #17: A 10.180 g sample of benzophenone is found to freeze at 46.8 ☌. Determine the molar mass & molecular formula of the compound.ġ.04 ☌ = (1) (5.12 ☌ kg mol¯ 1) (x / 0.301 kg) The freezing point of the solution is 1.04 ☌ below that of pure benzene. Problem #16: A 7.85 g sample of a compound with the empirical formula C 5H 4 is dissolved in 301 g of benzene. Prior to coming across this problem and adding it to my web site, the ChemTeam did not know that caffeine dimerizes. The fact that we got exactly double that value shows that caffeine dimerizes in solution and that the van 't Hoff factor should be 0.5 for caffeine. What is the molar mass of caffeine?ģ.07 ☌ = (1) (39.7 ☌ kg/mol) (x / 0.0100 kg)įrom other sources, we know the molar mass of caffeine to be 194 g/mol. mg sample of caffeine was dissolved in 10.0 g of camphor (K f = 39.7 ☌/m), decreasing the freezing point of camphor by 3.07 ☌. Problem #14: What is the freezing point of a solution of ethyl alcohol, that contains 20.0 g of the solute (C 2H 5OH), dissolved in 590.0 g of water? Problem #13: What is the molar mass of 35.0 g of an unknown substance that depresses the freezing point of 0.350 kg of water 0.50 ☌? K f for water is 1.86 ☌/m.Ġ.50 ☌ = (1) (1.86 ☌ kg mol¯ 1) (x / 0.350 kg) Such values can be easily looked up in standard reference materials. Note that the freezing point constant is not provided. What s the approximate molar mass of lauryl alcohol?ġ.344 ☌ = (1) (5.12 ☌ kg mol¯ 1) (x / 0.100 kg) A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 ☌. Problem #12: Lauryl alcohol is obtained from coconut oil and is used to make detergents. What is the molar mass of the unknown compound?Ģ) Determine how many moles of the compound dissolved:ġ.674 ☌ = (1) (5.12 ☌ kg mol¯ 1) (x / 0.500 kg) The freezing point of pure benzene is 5.444 ☌ and the K f for benzene is 5.12 ☌/m.
Problem #11: When 20.0 grams of an unknown nonelectrolyte compound are dissolved in 500.0 grams of benzene, the freezing point of the resulting solution is 3.77 ☌. ChemTeam: Freezing Point Depression Problems #11-25 Freezing Point Depression
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