Freezing-point-depression-problems PDF

**Problem \#1:** How many grams of pyrazine (C~4~H~4~N~2~) would have to be dissolved in 1.50 kg of carbon tetrachloride to lower the freezing point by 4.4 °C? The freezing point constant for carbon tetrachloride is 30. °C/m. **Solution:** Δt = i K~f~ m 4.4 °C = (1) (30. °C kg mol^-1^) (x / 1.50...

**Problem \#1:** How many grams of pyrazine (C~4~H~4~N~2~) would have to be dissolved in 1.50 kg of carbon tetrachloride to lower the freezing point by 4.4 °C? The freezing point constant for carbon tetrachloride is 30. °C/m. **Solution:** Δt = i K~f~ m 4.4 °C = (1) (30. °C kg mol^-1^) (x / 1.50 kg) 4.4 °C = (1) (20. °C mol^-1^) (x) x = 0.22 mol 0.22 mol times 80.0896 g/mol = 17.6 g (I\'ll ignore sig figs and leave it at three. I\'m such a rebel!) **Problem \#2:** When 0.258 g of a molecular compound, benzoic acid, was dissolved in 40.0 g of benzene, the freezing point of the solution was lowered to 5.23 °C. What is the molecular weight of the benzoic acid? **Solution:** We look up the K~f~ for benzene, finding it to be 5.12 °C/m. The freezing point of benzene is found to be 5.5 °C. Δt = i K~f~ m 0.27 °C = (1) (5.12 °C kg mol^-1^) (x / 0.0400 kg) 0.27 °C = (128 °C mol^-1^) (x) x = 0.00211 mol 0.258 g / 0.00211 mol = 122 g/mol (to three sig figs) **Problem \#3:** When 92.0 g of a molecular compound was dissolved in 1000. g of water, the freezing point of the solution was lowered to -3.72 °C. What is the molecular weight of the compound? **Solution:** Δt = i K~f~ m 3.72 °C = (1) (1.86 °C kg mol^-1^) (x / 1.000 kg) 3.72 °C = (1.86 °C mol^-1^) (x) x = 2.00 mol 92.0 g /2.00 mol = 46.0 g/mol (to three sig figs) **Problem \#4:** What is the freezing point depression when 62.2 g of toluene (C~7~H~8~) is dissolved in 481 g of naphthalene? The freezing point constant for naphthalene is 7.00 °C/m. **Solution:** 62.2 g / 92.1402 g/mol = 0.675058 mol Δt = i K~f~ m x = (1) (7.00 °C kg mol^-1^) (0.675058 mol / 0.481 kg) x = 9.82 °C \ 2.4 g / 1.008 g/mol = 2.38 mol N \-\--\> 16.6 g / 14.007 g/mol = 1.185 mol O \-\--\> 38.1 g / 16.00 g/mol = 2.38 mol Divide through by the smallest: C \-\--\> 3.57 mol / 1.185 mol = 3 H \-\--\> 2.38 mol / 1.185 mol = 2 N \-\--\> 1.185 mol / 1.185 mol = 1 O \-\--\> 2.38 mol / 1.185 mol = 2 The empirical formula is C~3~H~2~NO~2~ 5\) Determine the molecular formula: C~3~H~2~NO~2~ weighs 84 168 / 82 = 2 The molecular formula is C~6~H~4~N~2~O~4~ **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 °C. The freezing point of pure benzene is 5.444 °C and the K~f~ for benzene is 5.12 °C/m. What is the molar mass of the unknown compound? **Solution:** 1\) Determine temperature change: 5.444 - 3.77 = 1.674 °C 2\) Determine how many moles of the compound dissolved: Δt = i K~f~ m 1.674 °C = (1) (5.12 °C kg mol^-1^) (x / 0.500 kg) 2.16 °C = (10.24 °C mol^-1^) (x) x = 0.2109375 mol 3\) Determine molecular weight: 20.0 g / 0.2109375 mol = 94.8 g/mol **Problem \#12:** Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 °C. What s the approximate molar mass of lauryl alcohol? **Solution:** Δt = i K~f~ m 1.344 °C = (1) (5.12 °C kg mol^-1^) (x / 0.100 kg) 1.344 °C = (51.2 °C mol^-1^) (x) x = 0.02625 mol 5.00 g / 0.02625 mol = 190. g/mol Note that the freezing point constant is not provided. Such values can be easily looked up in standard reference materials. **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 °C? K~f~ for water is 1.86 °C/m. **Solution:** Δt = i K~f~ m 0.50 °C = (1) (1.86 °C kg mol^-1^) (x / 0.350 kg) 0.50 °C = (5.3143 °C mol^-1^) (x) x = 0.094086 mol 35.0 g / 0.094086 mol = 372 g/mol **Problem \#14:** What is the freezing point of a solution of ethyl alcohol, that contains 20.0 g of the solute (C~2~H~5~OH), dissolved in 590.0 g of water? **Solution:** 20.0 g / 46.07 g/mol = 0.434122 mol Δt = i K~f~ m x = (1) (1.86 °C kg mol^-1^) (0.434122 mol / 0.5900 kg) x = 1.37 °C The solution freezes at -1.37 °C. **Problem \#15:** A 300. mg sample of caffeine was dissolved in 10.0 g of camphor (K~f~ = 39.7 °C/m), decreasing the freezing point of camphor by 3.07 °C. What is the molar mass of caffeine? **Solution:** Δt = i K~f~ m 3.07 °C = (1) (39.7 °C kg/mol) (x / 0.0100 kg) 0.0307 kg °C = (1) (39.7 °C kg/mol) (x) x = 0.00077329975 mol 0.300 g / 0.00077329975 mol = 388 g/mol From other sources, we know the molar mass of caffeine to be 194 g/mol. 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. Prior to coming across this problem and adding it to my web site, the ChemTeam did not know that caffeine dimerizes. Good stuff! **Problem \#16:** A 7.85 g sample of a compound with the empirical formula C~5~H~4~ is dissolved in 301 g of benzene. The freezing point of the solution is 1.04 °C below that of pure benzene. Determine the molar mass & molecular formula of the compound. **Solution:** 1\) Determine moles of the compound: Δt = i K~f~ m 1.04 °C = (1) (5.12 °C kg mol^-1^) (x / 0.301 kg) 1.04 °C = (17.01 °C mol^-1^) (x) x = 0.06114 mol 2\) Determine the molar mass: 7.85 g / 0.06114 mol = 128.4 g/mol 3\) Determine molecular formula: The weight of C~5~H~4~ is 64.0866. 128.4 / 64.0866 = 2 The molecular formula is C~10~H~8~. **Problem \#17:** A 10.180 g sample of benzophenone is found to freeze at 46.8 °C. 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 °C. Calculate the molecular weight for the unknown. K~f~ for benzophenone is 9.80 °C/m. **Solution:** 1\) Determine moles of unknown compound: Δt = i K~f~ m 4.2 °C = (1) (9.80 °C/m) (x / 0.010180 kg) 4.2 °C = (962.672 °C mol¯^1^) (x) x = 0.004362857 mol 2\) Determine the molecular weight: 0.680 g / 0.004362857 mol = 156 g/mol (to three sig figs) **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 °C. Calculate the molar mass of the unknown compound. Express your answer using two significant figures. **Solution:** 1\) Determine moles of unknown compound: Δt = i K~f~ m 1.3 °C = (1) (1.86 °C/m) (x / 0.1600 kg) 1.3 °C = (11.625 °C mol¯^1^) (x) x = 0.111828 mol 2\) Determine the molecular weight: 34.3 g / 0.111828 mol = 306.72 g/mol To three sig figs, 307 g/mol **Problem \#19:** Calculate the freezing point of a solution of 5.00 g of diphenyl C~12~H~10~ and 7.50 g of naphthalene, C~10~H~8~ dissolved in 200.0 g of benzene (fp = 5.5 °C) **Solution** 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. 1\) The key to this problem is to calculate moles of each substance and add then together: (5.00 g / 154.2 g mol¯^1^) + (7.50 g / 128.2) = 0.0909 mol 2\) Calculate the molality: 0.0909 mol / 0.200 kg = 0.455 m 3\) Calculate the freezing point depression: ΔT = i K~f~ m x = (1) (5.12 °C m¯^1^) (0.455 m) = 2.33 °C The solution freezes at 5.5 - 2.33 = 3.17 °C **Problem \#20:** Vitamin K is involved in the blood clotting mechanism. When 0.500 g is dissolved in 10.0 g of camphor, the freezing point is lowered by 4.43 °C. Calculate the molecular weight of vitamin K. **Solution** 1\) To solve this problem, I\'d like to engage in an analysis of the units. We will start with the freezing point depression equation: ΔT = i K~f~ m Replacing the right side with units gives: ΔT = (°C kg mol¯^1^) times (mol kg¯^1^) Notice that i goes away since it is unitless. Next I will replace mol with g / g mol¯^1^: ΔT = (°C kg mol¯^1^) times (g / g mol¯^1^ kg¯^1^) 2\) Now, let\'s insert numbers in the proper place: 4.43 = (40.) times (0.500 / x 0.010) 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. This becomes: 4.43 = 2000 / x x = 8860 g / mol We know this is a reasonable answer since vitamins and proteins have molecular weights in the thousands or even tens of thousands. **Problem \#21:** A compound containing only boron, nitrogen, and hydrogen was found to be 40.3% B, 52.2% N, and 7.5% H by mass. When 3.301 g of this compound is dissolved in 50.00 g of benzene, the solution produced freezes at 1.30 °C. The freezing point of pure benzene is 5.48 °C; K~b~ for benzene is 5.12 °C m^-1^. What is the molecular weight of this compound? a\. Determine the molecular weight of the solid. b\. Determine the molecular formula of the solid c\. Determine the mole fraction of the solid in the solution d\. If the density of this solution is 0.8989 g/mL , calculate the molarity of the solution **Solution:** 1\) Determine moles of solute using freezing point depression data: Δt = K~f~ m 4.18 °C = 5.12 °C-kg/mol (x / 0.0500 kg) 4.18 ° = (102.4 °C/mol) (x) x = 0.04082 mol 2\) Use moles and 3.301 g to determine molecular weight (answer to part a): 3.301 g / 0.04082 mol = 80.867 g/mol 80.9 g/mol (to 3 sig figs) 3\) Use an [[on-line empirical formula calculator]](http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml) to determine the empirical formula: BNH~2~ 4\) Determine the molecular formula (answer to part b): the \"empirical formula weight\" of BNH~2~ is 26.8338 g the molecular weight divided by the \"EFW\" yields 3 therefore, the molecular formula is: B~3~N~3~H~6~ Comment: the particular substance in this question is [[Borazine]](http://en.wikipedia.org/wiki/Borazine), often referred to as the \"inorganic benzene.\" Its formula is written (BH)~3~(NH)~3~. 5\) Determine the mole fraction of the solute (answer to part c): moles of solvent = 50.00 g / 78.1134 g/mol = 0.6401 mol χ~solute~ = 0.04082 mol / (0.04082 mol + 0.6401 mol) χ~solute~ = 0.060 6\) Determine molarity of solution (answer to part d): i\) our solution weighs 53.301 g; calculate its volume: 0.8989 g/mL = 53.301 g / x x = 59.30 mL ii\) calculate the molarity: 0.04082 mol / 0.05930 L = 0.6884 M (to 4 sig figs) **Problem \#22:** A 1.60-g sample of a mixture of naphthalene (C~10~H~8~) and anthracene (C~14~H~10~) is dissolved in 20.0 g benzene (C~6~H~6~). The freezing point of the solution is found to be 2.81 °C. What is the composition as mass percent of the sample mixture? (The freezing point of benzene is 5.51 °C and K~f~ is 5.12 °C-kg/mol.) **Solution path \#1:** 1\) Determine the fp depression if all 1.6 g were naphthalene: x = (5.12) \[(1.6/128.174)/0.020)\] x = 3.1956559 °C 2\) Determine the fp depression if all 1.6 g were anthracene: x = (5.12) \[(1.6/178.234)/0.020)\] x = 2.29810 °C 3\) What combination of naphthalene and anthracene provides a Δt = 2.70 °C? 2.70 = (3.1956559) (x) + (2.29810) (1.6 - x) Some algebra reults in: x = 0.716 g You may calculate the anthracene on your own. Don\'t forget to include the weight of the benzene if you calculate the mass percents. **Solution path \#2:** Δt = 2.70 °C A and B in the following are placeholders for the molalities. 2.70 = (5.12)(A) + (5.12)(B) You could also write 2.70 = (5.12)(A + B) A is as follows: numerator is x over molar mass of naphthalene and the denominator is 0.020 kg B is as follows: numerator is 1.6 - x over molar mass of anthracene and the denominator is 0.020 kg Keep in mind that both A and B have a fraction in the numerator. Be careful with the algebra, if you decide to work it out. Don\'t forget the benzene, if you do the mass percents. **Problem \#23:** A 0.265m MgSO~4~ solution has a freezing point of -0.61 °C. What is the van \'t Hoff factor for this solution? K~f~ = 1.86 °C/m **Solution:** Δt = i K~f~ m 0.61 °C = (x) (1.86 °C/m) (0.265m) i = 1.24 The theoretical van \'t Hoff factor for MgSO~4~ is 2. Because of our value of 1.24, we must conclude that there is a significant amount of ion-pairing in aqueous solutions of MgSO~4~. Is this truly the case or is this just a fake problem? [[Look at this page.]](http://www3.interscience.wiley.com:8100/legacy/college/brady/0471215171/int_dialogue/data/cycle_percent_ionization.html) It presents a different set of data (i = 1.12 at 0.1m), but it does show that this issue of ion-pairing in MgSO~4~ is not a made-up one. [[Take a look at this.]](http://pubs.acs.org/doi/abs/10.1021/jp034870p) The concept of ion-pairing in MgSO~4~ is very real and is of importance in chemical research, both for geochemical and medical reasons. **Problem \#24:** A 0.2436 g sample of an unknown substance was dissovled in 20.0 mL of cyclohexane. The density of cyclohexane is 0.779 g/mL. The freezing point depression was 2.50 °C. Calculate molar mass of the unknown substance. **Solution:** 1\) Determine grams of cyclohexane: (0.779 g/mL) (20.0 mL) = 15.58 g this is 0.01558 kg 2\) Determine moles of unknown substance present: Δt = i K~f~ m 2.50 °C = (1) (20.2) (x / 0.01558) x = 0.00192822 mol Note assumption of i = 1 for the solute. Also, the cryoscopic constant for cyclohexane needed to be [[looked up.]](http://en.wikipedia.org/wiki/Freezing-point_depression#Cryoscopic_constants) 3\) Determine molecular weight: 0.2436 g / 0.00192822 mol = 126.3 g/mol **Problem \#25:** A 0.124 m trichloracetic acid, CCl~3~COOH (aq), solution has a freezing point of -0.386 °C. What is the percentage ionization of the acid? **Solution:** 1\) Calculate van \'t Hoff factor: Δt = i K~b~ m 0.386 = i (1.86) (0.124) i = 1.6736 2\) Calculate value for \[H^+^\]: CCl~3~COOH \ H^+^ + CCl~3~COO¯ total concentration of all ions in solution equals: (1.6736) (0.124) = 0.20753 m This is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm^3^, which makes the molarity equal to the molality. 0.20753 = (0.124 - x) + x + x x = 0.08353 M 3\) Calculate the percent dissociation: 0.08353 / 0.124 = 67.4% Comment: I mention above that we effectively treat the molality as a molarity. If you were to do this problem: Calculate the percent dissociation of a 0.124 M (molarity, not molality) solution of trichloroacetic acid (K~a~ = 0.170), you would find the answer to be 67%. If you choose to try the problem, you will need to solve a quadratic equation to solve for the hydrogen ion concentration, since \'ignore the minus x\' doesn\'t work. Use a [[quadratic solver]](http://id.mind.net/~zona/mmts/miscellaneousMath/quadraticRealSolver/quadraticRealSolver.html%22) to get the roots, of which the positive root will be the correct (remember, you can\'t have a negative concentration) H^+^ concentration. **Bonus Problem \#1:** Cyclohexanol, C~6~H~11~OH, is sometimes used as the solvent in molar mass determination. If 0.235 g of benzoic acid C~7~H~6~O~2~, dissolved in 12.45 g of cyclohexanol, lowered the freezing point of pure cyclohexanol by 6.55 °C, what is the molal freezing point constant of the solvent? **Solution:** 1\) Determine moles of benzoic acid: 0.235 g / 122.1224 g/mol = 0.0019243 mol 2\) Substitute into the following equation: Δt = i K~b~ m 6.55 = (1) (x) (0.0019243 mol / 0.01245 kg) x = 42.4 °C/m Note: benzoic acid does not dissociate in cyclohexanol, therefore a van\'t Hoff factor of 1 is used. **Bonus Problem \#4:** An average value for the salinity of sea water is around 35 ppt. The Dead Sea, however, has an average value salinity around 287 ppt. Assuming that the only salt present is magnesium chloride (MgCl~2~), calculate the temperature required to ice skate on the Dead Sea. **Solution:** 1\) Convert 287 ppt to molality: 287 parts per thousand means 287 g solute per 1000 g of solution. 287 g / 95.211 g/mol = 3.01436 mol of MgCl~2~ 1000 g - 287 g = 713 g of water The molality is 3.01436 mol / 0.713 kg = 4.2277 m (I\'ll keep some guard digits.) 2\) Determine freezing point: Δt = i K~f~ m x = (3) (1.86 °C/m) (4.2277 m) x = 23.6 °C The Dead Sea freezes at -23.6 °C.