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. g/mL in 0.786 g/mL and 0.793 g/mL respectively. A solution contains 20.0 mL and 100.0 mL . [15 points]

a) What is the mole fraction of methanol is the solution?

20 mL MeOH= 0.494 mole MeOH

100 mL = 1.917 mL

ii) X = = = 0.205 mol

b) What is the molality of methanol is the solution?

molality= =

= 6.28 molality

c) What is the molarity of the methanol in the solution?

molarity = = = 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) 100.0 mL of a 250.0 mM aqueous solution using water and solid * 3 .

i) Cu 1*68.5=68.5

2*62=124

3*18=54 68.5+124+54=241.5

ii) # mol = [ ] V = 25.0 100 mL = 0.025 mol

iii) 0.025 mol= 6.04g

iv) mix 6.04g 3 100 mL soln

b) Using the preceding 250.0 mM solution of , prepare 75.0 mL of 200.0 mM solution.

i) [dil] = [conc’d]

200 mM (75 mL) = 250 mM *

= = 60 mL

ii) mix 60 mL 250mM 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)

ii) 1.5g = 0.0172 mol

iii) [ = [ or = = 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.

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 ? [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 = 15.860 - 15.481= 0.377g

ii) #g AgCl = 13.896 - 7.885 = 6.011g

%Cl in AgCl = = =

#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 = 0.02094

1.487g Cl = 0.04189

1.246g Sn = 0.0105

v) Sn : Cl :

0.0105 : 0.04184 : 0.02094

1 : 4 : 2

5) Tony’s group has an unknown ionic sample, which is a colorless solution and forms a precipitate when mixed with either or . Identify the unknown sample; justify your answer, it has to be from the following table. [20 points]

i) it’s Na or Ba cpd

ii) It’s Ba cpd

iii) it’s

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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
  1. 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

  1. 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

  1. What is the molarity of methanol in the solution?

Molarity= nCH3OH/Vsoln = .494mol/120mL soln(10^3mL/L) = 4.12 M

  1. Emily is to prepare…in her group’s Beer’s Law lab; describe and justify how Emily can prepare such a solution.
  1. 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.

  1. 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)

  1. Nathan added excess sodium phosphate to 65.0 mL solution containing magnesium chloride, which formed 1.5g of a precipitate.
  1. What was the concentration of magnesium chloride before the addition of sodium phosphate?

i)2PO43- + 3Mg+2  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

  1. 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-  small Mg3(PO4)2  small nMg+2  small [MgCl2]

(i) (ii) (iii)

Above Above Above

  1. 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 + 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 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

  1. 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.

BaCl2 / CuCl2 / Na2CO3
Ba(OH)2 / Cu(NO3)2 / NaI
CuSO4 / NaOH
Na2SO4

i)Colorless  not Cu cpd; it’s Na or Ba cpd

ii)Cpd + Na2SO4 ppt not Na cpd; it’s Ba cpd

iii)Cpd + Co(NO3)2 ppt not BaCl; it’s Ba(OH)2.

iv)Cpd + Ba(NO3)  no ppt not needed

Ba(OH)2

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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]
  2. What is the mole fraction of methanol in the solution?
  3. 20mL MeOH = 0.494 mol MeOH
  4. 100mL CH3CN = 0.917 mol CH3CN
  5. X(MeOH) = = = 0.20g
  6. What is the molality of methanol in the solution?

Molality = = = 6.28 molal

  1. What is the molarity of methanol in the solution?

M = = = 4.12 M

  1. Emily is to prepare… in her group’s Beer’s Law lab. Describe and justify how Emily can prepare such a solution. [15 points]
  2. 100.0mL of a 250.0mM Cu(NO3)2 aqueous solution using water and solid Cu(NO3) 3 H2O.
  3. Cu: 63.5
    NO3: 124Cu + NO3 + H2O = 241,5
    H2O: 54
  4. #mol = [ ]V = 100mL = 2.5E-4 mol
  5. 2.5E-4 mol = 6.04 g
  6. mis 6.04 g Cu(No3)2 3H2O into 100mL soln
  7. Nathan added excess sodium phosphate to 65.0mL solution containing magnesium chloride, which formed 1.5 g of precipitate? [15 points]
  8. What was the concentration of magnesium chloride before the addition of sodium phosphate?
  9. 2PO43- + 3Mg2+ Mg3(PO4)2
  10. 1.5 g Mg(PO4)2 = 0.0172 mol Mg+
  11. [MgCl2] = [Mg+] = = = 0.26 M
  12. 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)2n(Mg2+) [MgCl2]

  1. 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 SnXClyzH2O? [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
  1. #g H2O = 15.860 - 15.483 = 0.377g H2O
  2. #g AgCl = 13.896 - 7.885 = 6.011g

%Cl in AgCl = = =

#g Cl in AgCl = 1.487g Cl

  1. #g cpd = 15.48g - 12.750 = 2.733g
    #g Sn = 2.733 - 1.487 = 1.246g Sn
  2. 0.377g H2O = 0.02094
    1.487g Cl = 0.04189
    1.246g Sn = 0.0105
  3. Sn : Cl : H2O
    0.0105 : 0.04184 : 0.02094
    SnCl4 2H2O
  1. 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
  1. colorless not Cu cpd (either Na or Ba)
  2. cpd + Na2So4ppt not Na (so Ba)
  3. cpd + Cu(NO3)2ppt not BaCl2 (so Ba(OH)2)