1 / 15

************************************************* BSAC Homework 3 Solutions and Solids

************************************************* BSAC Homework 3 Solutions and Solids Answer Key *************************************************. **************************************************************************************************

hagen
Download Presentation

************************************************* BSAC Homework 3 Solutions and Solids

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ************************************************* • BSAC Homework 3 • Solutions and Solids • Answer Key • *************************************************

  2. **************************************************************************************************************************************************************************************************** • BSAC General Chemistry I • Homework 3: Solutions and the Solid State • ************************************************************************************************** Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H2O] cube. Being careful with units and significant figures….. a) Calculate the solute molality and b) Calculate the mole fraction of the solute c) Calculate the volume of the solution (mL to 4 s.f.), given that it its molarity was measured as 1.0925 molL-1. Q4: Suppose 200 mL decane is used to dissolve 1.03 g of polyethylene. Given that the osmotic pressure against a membrane was observed to be 264 Pa at room temperature, calculate the molar mass of the polymer and hence the average number of repeating -C2H4- units in each hydrocarbon chain. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. a) Using Pythagoras where necessary, calculate the total volume of each unit cell in terms or r3 (r = radius) b) Calculate the volume occupied by atoms c) Calculate the percentage of volume occupied by atoms. d) Which crystal is most efficiently packed? Which has most empty space? e) Given that the density of polonium crystal is found to be 8.79 g/cm3 and the metal’s atomic radius is always 170 pm, determine which isotope of Po is present. Rb Po Ir Q2: By considering their colligative properties, decide which of these solutions... 0.5 M glucose 0.10 M Sr(NO3)20.15 M Al2(SO4)30.20 M HCl0.25 M dioxane a) ….. has the highest freezing point? b) …. has the highest boiling point? c) Which salt solution exerts the lower osmotic pressure? Q3: 3 g of 4-hydroxybenzoic acid was dissolved in 128 g benzene. constants for pure benzene: vapour pressure (25 ºC) is 99.5 mmHg molal boiling point elevation constant Kb = 5.12 Kkgmol-1 boiling point = 80.1 ºC a) What was the boiling point of the solution? b) Calculate the expected vapour pressure of the solution in Torr. Metal M f) Calculate the mass of each unit cells (in Dalton). Which unit cell is heaviest? Which is lightest? CaSe Q6: a) The ccp (face-centered cubic) unit cell of CaSe has a very similar mass as the unit cell of metal M shown. What is metal M? g) The unit cell edge length of Ir is known from X-ray crystal data to be 311.8 pm. Meanwhile, experiments with rubidium show it to have an atomic diameter of 0.50 nm. Calculate the density of each metal in g/cm3. Would the bulk metals float or sink in liquid mercury? b) CaSe is an ionic solid so has quite a high melting point. How do you think the melting point of CaS and SrSe compare with that of CaSe?

  3. Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H2O] cube. Being careful with units and significant figures….. a) Calculate the solute molality and b) Calculate the mole fraction of the solute c) Calculate the volume of the solution (mL to 4 s.f.), given that it its molarity was measured as 1.0925 molL-1. A: first work out number of moles of each: molar mass acetonitrile = 41.053 g/mol moles acetonitrile = 200 / 41.053 = 4.8717 mol molar mass water (H2O) = 18.016 g/mol moles water (solute) = 4 / 18.016 = 0.2220 mol a) Molality = no. of moles H2O mass acetonitrile = 0.2220 mol / 0.200 kg = 1.11 m or 1.11 mol/kg

  4. Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H2O] cube. Being careful with units and significant figures….. a) Calculate the solute molality and b) Calculate the mole fraction of the solute c) Calculate the volume of the solution (mL to 4 s.f.), given that it its molarity was measured as 1.0925 molL-1. A: b)Mole fraction = no. of moles H2O no. of moles acetonitrile + no. moles H2O = 0.2220 /(4.8717 + 0.2220 ) = 0.0436 (no units) just over 4% of molecules in the mixture are water molecules c) volume (solution) = no. of moles H2O molarity H2O = 0.2220 mol / (1.0925 mol/L) = 0.2032 L or 203.2 mL

  5. Q2: By considering their colligative properties, decide which of these solutions... 0.5 M glucose 0.10 M Sr(NO3)20.15 M Al2(SO4)30.20 M HCl0.25 M dioxane a) ……….. has the highest freezing point? b) …….. has the highest boiling point? c) Which salt solution exerts the lower osmotic pressure? 0.25 M dioxane (particle concentration = 0.25 M,  smallest depression of m.p.) 0.15 M Al2(SO4)3 (particle concentration = 5 x 0.15M = 0.75 M)  greatest raising of b.p.) 0.10 M Sr(NO3)2 (particle concentration = 3 x 0.10M = 0.30 M  smaller osmotic pressure than Al2(SO4)3 solution

  6. Q3: 3 g of 4-hydroxybenzoic acid was dissolved in 128 g benzene. constants for pure benzene: vapour pressure (25 ºC) is 99.5 mmHg molal boiling point elevation constant Kb = 5.12 Kkgmol-1 boiling point = 80.1 ºC a) What was the boiling point of the solution? DTb = msoluteKb molar mass (4-HB) = 138.12 g/mol msolute = moles(4-HB) / mass benzene moles(4-HB) = mass 4-HB / molar mass 4-HB = 0.02172 mol msolute = 0.02172 / 0.128 = 0.1697 molkg-1 DTb = msoluteKb = 0.1697 x 5.12 = 0.869  b.p is 80.97 ºC

  7. Q3: 3 g of 4-hydroxybenzoic acid was dissolved in 128 g benzene. constants for pure benzene: vapour pressure (25 ºC) is 99.5 mmHg molal boiling point elevation constant Kb = 5.12 Kkgmol-1 boiling point = 80.1 ºC b) Calculate the expected vapour pressure of the solution in Torr. Use Raoult’s Law for non-volatile solutes: Psoln = csolv.P0solv Psoln = ? P0solv= 99.5 mmHg Mole fraction benzene (csolv) = moles benzene / (moles benzene + moles 4-HB) csolv = (128 / 78.11) / [(128 / 78.11) + 0.02172] = 0.9869 Psoln = csolv.P0solv = 0.9869 x 99.5 Torr = 98.2 Torr

  8. Q4: 200 mL decane is used to dissolve 1.03 g of polyethylene. Given that the osmotic pressure against a membrane was observed to be 264 Pa at room temperature, calculate the molar mass of the polymer and hence the average number of repeating -C2H4- units in each hydrocarbon chain. P = MRT We need to find M M = P / RT (use SI system of units) = 264 Pa / (8.314 J/Kmol x 298) = 0.10655 mol/m3  moles in sample = 0.10655 mol/m3 x 200 / 106 = 2.13 x 10-5 1.03 g = 2.13 x 10-5 mol  Mr = 48331.4 g/mol Number of repeat -C2H4- units = 48331.4 g/mol / 28.052 g/mol = 1723 mass of C2H4 unit

  9. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. a) Using Pythagoras where necessary, calculate the total volume of each unit cell in terms or r3 (r = radius) Rb Po Ir Ir Rb Po body diagonal = 4r face diagonal = 4r l2 + l2 = (4r)2 l = r 8 v = l3 = 83/2r3 = 22.63 r3 cell edge = 2r l2 + l2 + l2 = (4r)2 l = 16r2 / 3 v = l3 = (16r2 / 3)3/2 = 12.32 r3 v = l3 = (2r)3 = 8r3 

  10. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. b) Calculate the volume occupied by atoms c) Calculate the percentage of volume occupied by atoms. Rb Po Ir no. atoms = 4 volume of sphere = 4pr3 / 3 volume occupied = 4 x 4pr3 / 3 = 16.755 r3 no. atoms = 1 volume of sphere = 4pr3 / 3 volume occupied = 1 x 4pr3 / 3 = 4.1888 r3 no. atoms = 2 volume of sphere = 4pr3 / 3 volume occupied = 2 x 4pr3 / 3 = 8.3776 r3 % occupied = = 100 x 16.755 r3 22.63 r3 = 74.0% % occupied = = 100 x 4.1888 r3 8 r3 = 52.4% % occupied = = 100 x 8.3776 r3 12.32 r3 = 68.0%

  11. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. d) Which crystal is most efficiently packed? Which has most empty space? e) Given that the density of polonium crystal is found to be 8.79 g/cm3 and the metal’s atomic radius is always 170 pm, determine which isotope of Po is present. Need to calculate mass of unit cell -first calculate cell length (l) cell edge l = 2r = 340 pm = 3.4 x 10-8 cm v = l3 = 3.9304 x 10-23 cm3 r = m/v  m = rv = 8.79 x 3.9304 x 10-23 cm3 = 3.4548 x 10-22 g per cell (and therefore g per atom)  molar mass = 6.02 x 1023 x 3.4548 x 10-22 g = 208.06 g/mol Rb is most efficiently packed (face center cubic) Po has most empty space (primitive cubic) isotope is 208Po Avogadro number

  12. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. Rb Po Ir f) Calculate the mass of each unit cells (in Dalton). Which unit cell is heaviest? Which is lightest? no. atoms = 4 mass = 4 x 85.47 Da = 341.88 Da no. atoms = 1 mass = = 1 x 208.1 = 208.1 Da no. atoms = 2 mass = 2 x 192.2 Da = 384.4 Da lightest heaviest

  13. Q5: Look at the unit cells of three different metals, Rb, Po and Ir. g) The unit cell edge length of Ir is known from X-ray crystal data to be 311.8 pm. Meanwhile, experiments with rubidium show it to have an atomic diameter of 0.50 nm. Calculate the density of each metal in g/cm3. Would the bulk metals float or sink in liquid mercury? Rb Ir r = 250 pm = 2.5 x 10-8 cm from part a): v = 83/2r3 = 22.63 r3 = 3.536 x 10-22 cm3 mass = 4 x 85.47 g/cm3 x 1.6605 x 10-24 g = 5.677 x 10-22 g r = m/v = 1.61 g / cm3 (floats on Hg) r = 250 pm = 2.5 x 10-8 cm from part a): v = 83/2r3 = 22.63 r3 = 3.536 x 10-22 cm3 mass = 4 x 85.47 g/cm3 x 1.6605 x 10-24 g = 5.677 x 10-22 g r = m/v = 1.61 g / cm3 (floats on Hg)

  14. Q6: a) The ccp (face-centered cubic) unit cell of CaSe has a very similar mass as the unit cell of metal M shown. What is metal M? A: First determine number of CaSe formula units per unit cell (cations are pink, anions are green) CaSe – formula units per unit cell = 1/2[1 + (12 x 1/4) + (8 x 1/8) + (6 x 1/2)] = 4 cationsanions mass = 4 x (40.08 + 78.96) = 476.16 Daltons/unit cell for metal M: (also fcc) count atoms per unit cell: = (8 x 1/8) + (6 x 1/2) = 4 mass of M atom = 677.44 Da / 4 = 119.04 Da M is tin (Sn, mass = 118.7) CaSe Metal M

  15. Q6: b) CaSe is an ionic solid so has quite a high melting point. How do you think the melting point of CaS and SrSe compare with that of CaSe? Both S2- and Se2- anions are doubly charged. Ca2+ and Sr2+ are both doubly charged also. Smaller ions have higher charge per surface area  greater electrostatic attraction  more energy required to disrupt salt lattice  higher melting point  SrSe has the lowest m.p. (largest ions) CaS has the highest m.p.

More Related