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

************************************************************************************************** BSAC General Chemistry I Homework 3: Solutions and the Solid State ************************************************************************************************** Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H 2 O] 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 mol L-1. Q2: By considering their colligative properties, decide which of these solutions M glucose 0.10 M Sr(NO 3 ) M Al 2 (SO 4 ) M HCl 0.25 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 K b = 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. 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 -C 2 H 4 - units in each hydrocarbon chain. Rb Po Ir g) The unit cell edge length of Ir is known from X-ray crystal data to be pm. Meanwhile, experiments with rubidium show it to have an atomic diameter of 0.50 nm. Calculate the density of each metal in g/cm 3. Would the bulk metals float or sink in liquid mercury? f) Calculate the mass of each unit cells (in Dalton). Which unit cell is heaviest? Which is lightest? 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 r 3 (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/cm 3 and the metal’s atomic radius is always 170 pm, determine which isotope of Po is present. Metal M CaSe 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? 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?

Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H 2 O] 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 molL -1. A: first work out number of moles of each: molar mass acetonitrile = g/mol moles acetonitrile = 200 / = mol molar mass water (H 2 O) = g/mol moles water (solute) = 4 / = mol a) Molality = no. of moles H 2 O mass acetonitrile = mol / kg = 1.11 m or 1.11 mol/kg

Q1: 200 g of the solvent acetonitrile dissolved a 4 g ice [H 2 O] 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 molL -1. A: b) Mole fraction = no. of moles H 2 O no. of moles acetonitrile + no. moles H 2 O = /( ) = (no units) just over 4% of molecules in the mixture are water molecules c) volume (solution) = no. of moles H 2 O molarity H 2 O = mol / ( mol/L) = L or mL

Q2: By considering their colligative properties, decide which of these solutions M glucose 0.10 M Sr(NO 3 ) M Al 2 (SO 4 ) M HCl 0.25 M dioxane a) ……….. has the highest freezing point? b) …….. has the highest boiling point? c) Which salt solution exerts the lower osmotic pressure? 0.10 M Sr(NO 3 ) 2 (particle concentration = 3 x 0.10M = 0.30 M  smaller osmotic pressure than Al 2 (SO 4 ) 3 solution 0.25 M dioxane (particle concentration = 0.25 M,  smallest depression of m.p.) 0.15 M Al 2 (SO 4 ) 3 (particle concentration = 5 x 0.15M = 0.75 M)  greatest raising of b.p.)

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 K b = 5.12 Kkgmol -1 boiling point = 80.1 ºC a) What was the boiling point of the solution?  T b = m solute K b molar mass (4-HB) = g/mol m solute = moles(4-HB) / mass benzene moles(4-HB) = mass 4-HB / molar mass 4-HB = mol m solute = / = molkg -1  T b = m solute K b = x 5.12 =  b.p is ºC

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 K b = 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: P soln =  solv.P 0 solv P soln = ? P 0 solv = 99.5 mmHg Mole fraction benzene (  solv ) = moles benzene / (moles benzene + moles 4-HB)  solv = (128 / 78.11) / [(128 / 78.11) ] = P soln =  solv.P 0 solv = x 99.5 Torr = 98.2 Torr

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 -C 2 H 4 - units in each hydrocarbon chain.  = MRT We need to find M M =  RT (use SI system of units) = 264 Pa / (8.314 J/Kmol x 298) = mol/m 3  moles in sample = mol/m 3 x 200 / 10 6 = 2.13 x g = 2.13 x mol  M r = g/mol Number of repeat -C 2 H 4 - units = g/mol / g/mol = 1723 mass of C 2 H 4 unit

RbPoIr Rb Po 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 r 3 (r = radius) face diagonal = 4r l 2 + l 2 = (4r) 2 l = r 8 v = l 3 = 8 3/2 r 3 = r 3 cell edge = 2r body diagonal = 4r v = l 3 = (2r) 3 = 8r 3 l 2 + l 2 + l 2 = (4r) 2 l = 16r 2 / 3 v = l 3 = (16r 2 / 3) 3/2 = r 3   Ir

RbPoIr 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. no. atoms = 4 volume of sphere = 4  r 3 / 3 volume occupied = 4 x 4  r 3 / 3 = r 3 no. atoms = 1 volume of sphere = 4  r 3 / 3 volume occupied = 1 x 4  r 3 / 3 = r 3 no. atoms = 2 volume of sphere = 4  r 3 / 3 volume occupied = 2 x 4  r 3 / 3 = r 3 % occupied = = 100 x r r 3 = 74.0% % occupied = = 100 x r 3 8 r 3 = 52.4% % occupied = = 100 x r r 3 = 68.0%

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/cm 3 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 cm v = l 3 = x cm 3  = m/v  m =  v = 8.79 x x cm 3 = x g per cell (and therefore g per atom)  molar mass = 6.02 x x x g = g/mol Rb is most efficiently packed (face center cubic) Po has most empty space (primitive cubic) isotope is 208 Po Avogadro number

Q5: Look at the unit cells of three different metals, Rb, Po and 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 Da = Da no. atoms = 1 mass = = 1 x = Da no. atoms = 2 mass = 2 x Da = Da heaviestlightest RbPoIr

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 pm. Meanwhile, experiments with rubidium show it to have an atomic diameter of 0.50 nm. Calculate the density of each metal in g/cm 3. Would the bulk metals float or sink in liquid mercury? RbIr r = 250 pm = 2.5 x cm from part a): v = 8 3/2 r 3 = r 3 = x cm 3 mass = 4 x g/cm 3 x x g = x g  = m/v = 1.61 g / cm 3 (floats on Hg) r = 250 pm = 2.5 x cm from part a): v = 8 3/2 r 3 = r 3 = x cm 3 mass = 4 x g/cm 3 x x g = x g  = m/v = 1.61 g / cm 3 (floats on Hg)

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 cations anions mass = 4 x ( ) = 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 = Da / 4 = Da M is tin (Sn, mass = 118.7) CaSe Metal M

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 S 2- and Se 2- anions are doubly charged. Ca 2+ and Sr 2+ 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.