Solution - Chem-is
advertisement
Marcus Hong, Steven Jiang Ch. 4,13 Solution Test 1) You are in a study group with Daisy who asks you to solve homework problems 13.19. πΆπ»3 ππ»g/mL in 0.786 g/mL and 0.793 g/mL respectively. A solution contains 20.0 mL πΆπ»3 ππ» and 100.0 mL πΆπ»3 πΆπ. [15 points] a) What is the mole fraction of methanol is the solution? 20 mL MeOH 0.793π ππππ» 1 πππ ππππ» 100 mL πΆπ»3 ππ» ii) X = ππΏ ππππ» 32π ππππ» 0.786π πππ = 1.917 mL πΆπ»5 πΆπ ππΏ πππππ» πππππ» + ππππ‘ππ = = 0.494 mole MeOH 41π 0.494 0.494 + 1.917 = 0.205 mol b) What is the molality of methanol is the solution? molality= πππππ» = 0.494 πππ ππΏ # ππ πΆπ»3 πΆπ 100 ππΏ πΆπ»3 πΆπ 0.286π 103 π ππ = 6.28 molality c) What is the molarity of the methanol in the solution? 3 πππππ» 0.494 πππ 10 ππΏ molarity = = = 4.12 M ππ πππ 120 ππΏ π πππ πΏ 2) Emily is to prepare… in her group’s Beer’s law lab; describe and justify how Emily can prepare such a solution [15 points] a) 100.0 mL of a 250.0 mM πΆπ’(ππ3 )2 aqueous solution using water and solid πΆπ’(ππ3 )2* 3 π»2 π. i) Cu 1*68.5=68.5 ππ3 2*62=124 π»2 π 3*18=54 68.5+124+54=241.5 ii) # mol = [ ] V = 25.0 iii) 0.025 mol 241.5π πππ ππππ πΏ 100 mL πΏ πππ 103 ππΏ 103 ππππ = 0.025 mol = 6.04g iv) mix 6.04g πΆπ’(ππ3 )2 3 π»2 π ππ 100 mL soln b) Using the preceding 250.0 mM solution of πΆπ’(ππ3 )2, prepare 75.0 mL of 200.0 mM solution. i) [dil] ππππ = [conc’d] πππππ′π 200 mM (75 mL) = 250 mM * πππππ′π πππππ′π = 200 ππ (75 ππΏ) 250 ππ = 60 mL ii) mix 60 mL 250mM πΆπ’(ππ3 )2 gs 25 mL (or 15 mL) 3) Nathan added excess sodium phosphate to 65.0 mL solution containing magnesium chloride, which formed 1.5g of a precipitate. [15 points] a) What was the concentration of magnesium chloride before the addition of sodium phosphate? i) 2ππ43− + 3ππ2+ → ππ3 (ππ4 )2 ii) 1.5g ππ(ππ4 )2 1 πππ ππ(ππ4 )2 3 πππ ππ2+ 262π ππ(ππ4 )2 πππ ππ(ππ4 )2 πππ2+ 0.0172 πππ 103 ππΏ iii) [πππΆπ2 ] = [ππ2+ ] or π = 1.5 ππΏ πΏ = 0.0172 mol ππ2+ = 0.26 M b) What is the effect on your determination of magnesium chloride if excess sodium phosphate was not added to the solution? Be specific and justify/rationalize your answer. 3− ↓ ππ4 → ↓ ππ2 (ππ4 )2 → ↓ πππ2+ → ↓ [πππΆπ2 ] 4) Haley’s group collected experimental datat to determine the empirical formula of a sample of hydrated tin chloride using the same protocol as in your copper chloride lab. What is the value of x, y, and x in πππ₯ πΆππ¦ ∗ π§π»2 π? [20 points] #g crucible #g crucible + sample; before heat #g crucible + sample, after 1st heat #g crucible + sample, after 2nd heat #g crucible + sample, after 3rd heat 12.750 15.860 15.578 15.484 15.483 #g filter paper #g filter paper + precipitate 7.885 13.896 i) #g π»2 π= 15.860 - 15.481= 0.377g π»2 π ii) #g AgCl = 13.896 - 7.885 = 6.011g %Cl in AgCl = #π πΆπ ππ π΄ππΆπ #π π΄ππΆπ = 35.5 143.5 = #π πΆπ ππ π΄ππΆπ 6.011π #g Cl in cpd = #g Cl in AgCl = 1.487g Cl iii) #g cpd = 15.485 - 12.750 = 2.733g #g Sn = 2.733 - 1.487 = 1.246g Sn iv) 0.377g π»2 π 1.487g Cl ππΏ π»2 π 18π π»2 π πππ πΆπ = 0.04189 35.5π πΆπ πππ ππ 1.246g Sn = 0.02094 118.7 ππ = 0.0105 v) Sn : Cl : π»2 π 0.0105 : 0.04184 : 0.02094 1 : 4 : 2 πππΆπ4 ∗ 2π»2 π 5) Tony’s group has an unknown ionic sample, which is a colorless solution and forms a precipitate when mixed with either ππ2 ππ4or πΆπ(ππ3 )2. Identify the unknown sample; justify your answer, it has to be from the following table. [20 points] π΅ππΆπ2 πΆπ’πΆπ2 ππ2 πΆπ3 π΅π(ππ»)2 πΆπ’(ππ3 )2 πππΌ πΆπ’ππ4 ππππ» ππ2 ππ4 i) ππππππππ π → πππ‘ πΆπ’ πππ it’s Na or Ba cpd ii) πππ + ππ2 ππ4 → πππ‘ ⇒ πππ‘ ππ πππ It’s Ba cpd iii) πππ + πΆπ’(ππ3 )2 → πππ‘ ⇒ πππ‘ π΅ππΆπ2 it’s π΅π(ππ»)2 ********************************** 1) Ch.4 & 13 Solution(Partners: Ralph, Stephen & Gloria) 1. You are in a study group with Daisy, who asks you to solve homework problem 13.19. The density of acetonitrile, CH3CN, and methanol, CH3OH is .786 g/mL and .791 g/mL, respectively. A solution contains 20.0 mL CH3OH and 100.0 mL CH3CN a. What is the mole fraction of methanol in the solution? i) 20mL CH3OH(0.791CH3OH/1mL CH3OH) (1molCH3OH/32g CH3OH) = 0.494mol CH3OH 100mL CH3CN(0.786g/mL)(1mol/41g) = 1.917mol CH3CN ii) XCH3OH = nCH3OH/nCH3OH + nCH3CN = 0.494/(.494+1.917)=0.205 b. What is the molality of methanol in the solution? Molality = nCH3OH/#kg CH3CN =0.494 mol/100mL CH3CN(mL/.786g)(10^3g/kg) = 6.28 c. What is the molarity of methanol in the solution? Molarity= nCH3OH/Vsoln = .494mol/120mL soln(10^3mL/L) = 4.12 M 2. Emily is to prepare…in her group’s Beer’s Law lab; describe and justify how Emily can prepare such a solution. a. 100.0 mL of a 250.0 mM Cu(NO3)2 aqueous solution using water and solid Cu(NO3)2*3H2O i) Cu : 1x63.5 = 63.5 NO3: 2x62 = 124 H2O: 3x18 = 54 241.5 ii) # mol = [ ]V= 250mmol/L(100mL)(L/10^3mL)(1mol/10^3mmol) = .025 mol iii) .025mol(241.5g/mol)=6.04g iv) Mix 6.04g Cu(NO3)2*3H2O as 100 mL soln. b. Using the preceding 250.0 mM solution of Cu(NO3)2, prepare 75.0 mL of 200.0 mM solution. i) [dil]Vdil = [conc’d]Vconc’d 200mM(75mL)=250mM(Vconc’d) Vconc’d = 200mM(75mL)/(250mM)= 60mL ii) Mix 60 mL 250mM Cu(NO3)2 as 75 mL (or 15 mL H2O) 3. Nathan added excess sodium phosphate to 65.0 mL solution containing magnesium chloride, which formed 1.5g of a precipitate. a. What was the concentration of magnesium chloride before the addition of sodium phosphate? i) 2PO43- + 3Mg+2 ο Mg3(PO4)2 ii) 1.5g Mg3(PO4)2 (1molMg3(PO4)2)/(262g Mg3(PO4)2)(3mol Mg+2/1mol Mg3(PO4)2 ) = .0172 mol iii) [MgCl2] = [Mg+2]= (nMg+2/V) = 0.0172mol/65mL(10^3mL/L) = .26M b. What is the effect on your determination of magnesium chloride if excess sodium phosphate was not added to the sample? Be specific and justify/rationalize your answer. Small PO43- ο small Mg3(PO4)2 ο small nMg+2 ο small [MgCl2] (i) (ii) (iii) Above Above Above 4. Haley’s group collected experimental data to determine the empirical formula of a sample of hydrated tin chloride using the same protocol as in your copper chloride lab. What is the value of x, y, and z in SnxCly * zH2O? #g crucible #g crucible + #g crucible + #g crucible + #g crucible + sample ; before sample; after sample; after sample; after st nd 3rd heat heat 1 heat 2 heat 12.750 15.860 15.578 15.484 15.483 #g filer paper #g filter paper + precipitate 7.885 13.896 i) #g H2O = 15.860 – 15.483 = 0.377g H2O ii) #g AgCl = 13.896 – 7.885 = 6.011g % Cl in AgCl = #g Cl in AgCl/#g AgCl = 35.5/143.5 = #g Cl in AgCl/ 6.011g #g Cl in cpd = #g Cl in AgCl = 1.487g Cl iii) #g cpd = 15.483 – 12.750 = 2.733g #g Sn = 2.733 – 1.487 = 1.246g Sn iv) 0.377g H2O(1mL H2O/18g H2O) = .02094 1.487g Cl(1mol Cl/35.5g Cl) = 0.04189 1.246g Sn(1mol Sn/118.7g Sn) = .0105 v) Sn: Cl: H2O .0105 : .04189: .02094 1 : 4 : 2 SnCl4*2H2O 5. Tony’s group has an unknown ionic sample, which is a colorless solution and forms a precipitate when mixed with either Na2SO4 or Co(NO3)2, but not Ba(NO3)2. Identify the unknown sample; justify your answer; it has to be from the following table. Na2CO3 BaCl2 CuCl2 Ba(OH)2 Cu(NO3)2 NaI CuSO4 NaOH Na2SO4 i) ii) iii) iv) Colorless ο not Cu cpd; it’s Na or Ba cpd Cpd + Na2SO4 ο ppt ο not Na cpd; it’s Ba cpd Cpd + Co(NO3)2 ο ppt ο not BaCl; it’s Ba(OH)2. Cpd + Ba(NO3) ο no ppt ο not needed Ba(OH)2 ************************************************************************* Gloria Lau Solution 1. You are in a study group with Daisy, who asks you to solve homework problem 13.19. The density of acetonitrile, CH3OH and 100.0mL CH3CN. [15 points] a. What is the mole fraction of methanol in the solution? 0.791π ππππ» 1 πππ ππππ» i. 20mL MeOH ii. 100mL CH3CN iii. X(MeOH) = 1 ππΏ ππππ» 32 π ππππ» 0.786 π 1 πππ 1 ππΏ 41 π π(ππππ») π(ππππ») + π(πΆπ»3πΆπ) = 0.494 mol MeOH = 0.917 mol CH3CN = 0.494 0.494 + 1.917 = 0.20g b. What is the molality of methanol in the solution? Molality = π # ππ = 0.494 πππ ππΏ 103 π 100 ππΏ πΆβ3πΆπ 0.786 π ππ = 6.28 molal c. What is the molarity of methanol in the solution? M= π π = 0.494 πππ 103 ππΏ 120 ππΏ π πππ πΏ = 4.12 M 2. Emily is to prepare… in her group’s Beer’s Law lab. Describe and justify how Emily can prepare such a solution. [15 points] a. 100.0mL of a 250.0mM Cu(NO3)2 aqueous solution using water and solid Cu(NO3) × 3 H2O. i. Cu: 63.5 NO3: 124 Cu + NO3 + H2O = 241,5 H2O: 54 250 ππ πΏ #mol = [ ]V = iii. 2.5E-4 mol iv. mis 6.04 g Cu(No3)2 × 3H2O into 100mL soln πΏ 241.5 π πππ 100mL πππ ii. 103 ππΏ 103 ππ = 2.5E-4 mol = 6.04 g 3. Nathan added excess sodium phosphate to 65.0mL solution containing magnesium chloride, which formed 1.5 g of precipitate? [15 points] a. What was the concentration of magnesium chloride before the addition of sodium phosphate? i. 2PO43- + 3Mg2+ → Mg3(PO4)2 ii. 1.5 g Mg(PO4)2 iii. [MgCl2] = [Mg+] = πππ ππ(ππ4)2 3 πππ ππ2+ 262 π ππ(ππ4)2 πππ ππ(ππ4)2 π(ππ2+) π = = 0.0172 mol Mg+ 0.0172 πππ 103 ππΏ 65 ππΏ πΏ = 0.26 M b. What is the effect on your determination of magnesium chloride if excess sodium phosphate was not added to the solution? Be specific and justify/ rationalize your answer. ↓Po43- →↓Mg3 (PO4)2 →↓n(Mg2+) →↓[MgCl2] 4. Hailey’s group collected experimental data to determine the empirical formula of a sample of hydrated tin chloride using the same protocol as in your copper chloride lab. What is the value of x, y, and z in SnXCly ×zH2O? [20 points] #g crucible #g crucible + sample before heat #g crucible + sample after 1st heat #g crucible + sample after 2nd heat #g crucible + sample after 3rd heat 12.750 15.860 15.578 15.484 15.483 #g filter paper #g filter paper + precipitate 7.885 13.896 i. #g H2O = 15.860 - 15.483 = 0.377g H2O ii. #g AgCl = 13.896 - 7.885 = 6.011g %Cl in AgCl = #π πΆπ ππ π΄ππΆπ #π π΄ππΆπ = 35.5 143.5 = #ππΆπ ππ π΄ππΆπ 6.011 π #g Cl in AgCl = 1.487g Cl iii. #g cpd = 15.48g - 12.750 = 2.733g #g Sn = 2.733 - 1.487 = 1.246g Sn iv. 0.377g H2O 1.487g Cl 1.246g Sn v. ππΏ π»2π 18π π»2π πππ πΆπ 35.5π πΆπ = 0.02094 = 0.04189 πππ ππ 118.7π ππ = 0.0105 Sn : Cl : H2O 0.0105 : 0.04184 : 0.02094 SnCl4 × 2H2O 5. Tony’s group has an unknown ionic sample, which is a colorless solution and a precipitate when mixed with either Na2So4 or Co (NO3)2, but not Ba(NO3)2. Identify the unkown sample; justify your answer; it has to be from the following table. [20 points] BaCl2 CuCl2 Na2CO3 Ba(OH)2 Cu(NO3)2 NaI CuSO4 NaOH Na2SO4 i. colorless → not Cu cpd (either Na or Ba) ii. cpd + Na2So4 →ppt →not Na (so Ba) iii. cpd + Cu(NO3)2 →ppt →not BaCl2 (so Ba(OH)2)
