

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 ☌. Example: 0.00 has three significant figures, 0.000 has four significant figures ans so on.ChemTeam: Freezing Point Depression Problems #11-25 Freezing Point Depression

Therefore, any zeros after the decimal point are also significant.
