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261 EXAM 1, Summaries of Stoichiometry

University of Washington (UW) - SeattleStoichiometry

ple of PbBr2 (Ksp = 4.6 × 10−6) be the least soluble? 1. 100 mL of 0.01 M KOH. 2. 100 mL of 0.07 M HCl. 3. 100 mL of pure water. 4. 100 mL of 0.002 M NaBr.

Typology: Summaries

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signature: 261
version number
Juan Maldonado (jam8776)
EXAM 1
McCord Classes ·Spring 2015
(uniques: 50140 and 50145)
Water Data
Kf= 1.86 ◦C/m
Kb= 0.512 ◦C/m
Cice = 2.09 J/g ◦C
Cwater = 4.184 J/g ◦C
Csteam = 2.09 J/g ◦C
∆Hfusion = 334 J/g
∆Hvapor = 2260 J/g
Please refer to the back of the bubble sheet for more info.
NOTE: Please keep your Exam copy intact (all pages still stapled). You must turn in your
exam copy, plus your bubble sheet, and any scratch paper.
pf3
pf4
pf5
pf8

Partial preview of the text

Download 261 EXAM 1 and more Summaries Stoichiometry in PDF only on Docsity!

signature: 261

version number

Juan Maldonado (jam8776)

EXAM 1

McCord Classes · Spring 2015

(uniques: 50140 and 50145)

Water Data

Kf = 1. 86 ◦C/m

Kb = 0. 512 ◦C/m

Cice = 2.09 J/g ◦C

Cwater = 4.184 J/g ◦C

Csteam = 2.09 J/g ◦C

∆Hfusion = 334 J/g

∆Hvapor = 2260 J/g

Please refer to the back of the bubble sheet for more info.

NOTE: Please keep your Exam copy intact (all pages still stapled). You must turn in your exam copy, plus your bubble sheet, and any scratch paper.

This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

001 4.0 points In which of the following will a given sam- ple of PbBr 2

Ksp = 4. 6 × 10 −^6

be the least soluble?

  1. 100 mL of 0.01 M KOH
  2. 100 mL of 0.07 M HCl
  3. 100 mL of pure water
  4. 100 mL of 0.002 M NaBr
  5. 100 mL of 0.05 M HBr correct

Explanation: PbBr 2 will have a common ion effect with any solution with Pb2+^ or Br−^ in them. Two of the solutions have Br−^ and those will cause PbBr 2 to be less soluble. The 0.05 M HBr solution has a higher concentration of Br− than the NaBr solution, which means it will have the bigger effect on solubility, meaning LESS PbBr 2 will dissolve in that solution.

002 4.0 points The combustion of methane (CH 4 ) in an oxy- gen rich environment results in the production of 6.0 g of water vapor. What was the initial mass of methane?

  1. 4.2 g
  2. 2.7 g correct
  3. 0.33 g
  4. 3.0 g
  5. 5.3 g

Explanation: CH 4 + 2O 2 → CO 2 + 2H 2 O Ratio of methane to water is 1:2 in moles or 16:36 in grams.

6/36 = 1/6, 1/6 of 16 = 2.67 or 2.7 g of methane

003 4.0 points A compound is extremely soluble in water. You would expect ∆G for this dissolution process to be (negative/positive) and that this compound would have a (large/small) Ksp value.

  1. positive ; small
  2. negative ; small
  3. negative ; large correct
  4. positive ; large

Explanation: extremely soluble = spontaneous = nega- tive ∆G extremely soluble = much higher concen- trations when saturated which means a large value for Ksp

004 4.0 points Lead(II) chloride dissolves in water to form a saturated solution. The chloride ion con- centration in this saturated solution is 0. mol/L. What is the molar solubility of lead(II) chloride in water?

    1. 3 × 10 −^1 mol/L
    1. 4 × 10 −^5 mol/L
    1. 6 × 10 −^2 mol/L
    1. 0 × 10 −^3 mol/L correct
    1. 0 × 10 −^3 mol/L

Explanation: Let x be the amount of lead(II) chloride that dissolves in one liter of solution - aka: the molar solubility. Then the stoichiometry is: xPbCl 2 (s) → xPb2+^ + 2xCl− The 2x will be the concentration of chloride ion which is 0.008 M.

correct

  1. CaSO 4 < Cd 3 (AsO 4 ) 2 < BiI < AlPO 4

Explanation: Molar solubility can be approximated by taking the nth^ root of the Ksp where n is the number of ions in the salt. Doing so results in approximate molar solubilities of 10−^10 , 10−^7 , 10 −^11 and 10−^3 for bismuth iodide, cadmium arsenate, aluminum phosphate and calcium sulfate, respectively. Arranging these from least to greatest produces: AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4.

010 4.0 points The vapor pressure of pure water at 27◦C is 27 torr. Enough fructose (a non-electrolyte) is dissolved into the water at this temperature such that the vapor pressure changes to 24 torr. What is the mole fraction of the fructose in this solution?

  1. 6/
  2. 2/
  3. 4/
  4. 1/9 correct
  5. 9/
  6. 8/
  7. 3/
  8. 7/
  9. 5/

Explanation: Raoult’s Law: Pwater = χwaterP (^) water◦ χwater = P Pwater ◦ water χwater = 24/27 = 8/ 9 χfructose = 1 − χwater = 1 − 8 /9 = 1/ 9

011 4.0 points 100 mL of 0.005 M NaOH is poured into 100 mL of 0.0025 M Mg(NO 3 ) 2 solution.

What happens next? (Ksp = 9. 0 × 10 −^12 for Mg(OH) 2 , Ksp = 1. 4 × 102 for NaNO 3 )

  1. a precipitate of Mg(OH) 2 forms after mix- ing. correct
  2. Both Mg(OH) 2 and NaNO 3 form precipi- tates after mixing.
  3. nothing, the two solutions mix completely and no precipitate forms.
  4. a precipitate of NaNO 3 forms after mix- ing.

Explanation: NaNO 3 is very very soluble and will not precipitate. All concentrations 1/2 when mixed due to the volume doubling. Q for Mg(OH) 2 = 0.00125(.0025)^2 = 7. 8 × 10 −^9 which is greater than Ksp and therefore pre- cipitates.

012 4.0 points What is Ksp for Ag 3 PO 4 , if its molar solubility is 2. 7 × 10 −^6 mol/L?

    1. 4 × 10 −^21 correct
    1. 0 × 10 −^17
    1. 3 × 10 −^12
    1. 7 × 10 −^14
    1. 3 × 10 −^23
    1. 3 × 10 −^16
    1. 8 × 10 −^22

Explanation: S = 2. 7 × 10 −^6 mol/L The solubility equilibrium is

Ag 3 PO 4 (s) ⇀↽ 3 Ag+(aq) + PO^34 − (aq)

[Ag+] = 3 S = 8. 1 × 10 −^6 mol/L [PO^34 − ] = S = 2. 7 × 10 −^6 mol/L

Ksp = [Ag+]^3 [PO^34 − ]

8. 1 × 10 −^6

(2. 7 × 10 −^6 )

= 1. 43489 × 10 −^21

013 4.0 points The normal freezing point of camphor is 176 ◦C. When 0.220 mol of a certain solute is dissolved in 10 kg of camphor (Kf =

  1. 8 ◦C/m), the freezing point is decreased to 173. 50 ◦C. What is the identity of the so- lute?

  2. Na 3 PO 4

  3. sugar

  4. Ba 3 (AsO 3 ) 2

  5. KBr

  6. CaBr 2 correct

Explanation: ∆Tf = i · Kf · m ∆Tf = 176 − 173 .5 = 2. 50 ◦C m =. 22 /10 =. 022 i =

∆Tf Kf ·m

i =

  1. 8 · 0. 022

So the solute has to have i = 3 which only matches CaBr 2 which dissociates into 3 ions (1 cation and 2 anions).

014 4.0 points Consider the following chemical equation. If 0.2453 g of LaCl 3 is dissolved in 10 g of wa- ter, what is the boiling point of the solution? Assume complete dissociation of the solid.

LaCl 3 −→ La3+^ + 3Cl−

  1. 100.205 ◦C correct
  2. 99.875 ◦C
  3. 101.524 ◦C

4. 102.358 ◦C

5. 98.641 ◦C

Explanation: none

015 4.0 points What is the molar solubility for Co (OH) 3 (Ksp= 2. 5 × 10 −^43 )?

    1. 1 × 10 −^38 M
    1. 3 × 10 −^22 M
    1. 0 × 102 M
    1. 2 × 1010 M
    1. 8 × 10 −^12 M correct

Explanation: Ksp = [Co2+][OH−]^3 = (x) (3x)^3 = 27x^4

x =

Ksp 27

= 9.8x10−^12

016 4.0 points 5 grams of ice at − 10 ◦C is added to 100 grams of water at 25◦C. What is the final tempera- ture and composition of the mixture once it has reached equilibrium?

  1. 1 g of ice and 104 g of water at 0◦C
  2. 2 g of ice and 103 g of water at 0◦C
  3. 105 g of water at 22.7◦C
  4. 105 g of water at 11.2◦C
  5. 105 g of water at 4.24◦C
  6. 105 g of water at 19.8◦C correct

Explanation: Convert all the ice to water at 25◦C and get the total amount of heat added. Then remove that amount of heat from the 105 g of 25◦C water. total heat from ice to water (3 steps)

has a density of 1.5091 g/mL. Calculate the molality of sulfuric acid in this solution.

  1. 15.95 m correct
  2. 6.81 m
  3. 12.84 m
  4. 19.62 m

Explanation: none

022 0.0 points This question starts out at zero points but could very well increase after the grading. Now, if more points are awarded (the curve) on this assignment, would you like them added to your score?

  1. NO, leave my score alone, I prefer the lower score
  2. YES, I would like the points and the higher score. correct

Explanation: This should be a no-brainer. Most students want higher scores. If you picked yes, you got credit for the question and you got the extra points you asked for (if they were granted by your instructor). If you answered NO, you also got what you wanted... no points awarded.

023 4.0 points Which of the following shows the right equa- tion for the ion product of iron(III) sulfate?

  1. Ksp = [2 · Fe3+]^2 [3 · SO^24 − ]^3
  2. Ksp = [Fe2+][SO^24 − ]
  3. Ksp = [Fe3+]^2 [SO^24 − ]^3 correct
  4. Ksp = [Fe3+]^3 [SO^24 − ]^2
  5. Ksp = 4 · [Fe3+]^2 27 · [SO^24 − ]^3

Explanation:

none

024 4.0 points Identify the FALSE statement about Ksp.

  1. Ksp is called the solubility product for a saturated solution.
  2. Ksp is dependent on temperature.
  3. Ksp is a function of the concentrations of ions coming from a given salt.
  4. Ksp is a constant that is specific to each compound.
  5. Ksp is used to describe salt solutions where the minimum amount of salt has been dissolved into the solvent. correct

Explanation: Ksp is used to describe a solution at satura- tion which is a condition with the maximum concentration of ions is present (not mini- mum).

025 4.0 points When 1.28 g of polypeptide are placed in 14.0 L of water at room temperature, the so- lution exerts an osmotic pressure of 4.23 torr. What is the molar mass of the polypeptide?

  1. more than 550. g/mol
  2. less than 25.0 g/mol
    1. g/mol
    1. g/mol
  5. 38.8 g/mol
  6. 40.1 g/mol
    1. g/mol correct

Explanation: Π = iM RT polypeptide is non electrolyte and i = 1 4.23/760 = (1)M (0.08206)(298.15)

M = 0.0002275 mol/L x 14L = 0.003185 mol 1.28g/0.003186 mol = 402. g/mol

026 4.0 points The vapor pressure of water at 37◦C is 47. torr and its enthalpy of vaporization is 44. kJ·mol−^1. Estimate the vapor pressure of wa- ter at 87◦C. Assume the enthalpy of vaporiza- tion of water is independent of temperature.

  1. 503 torr correct
  2. 713 torr
  3. 256 torr
  4. 52 torr
  5. 112 torr

Explanation: T 1 = 37◦C + 273.15 = 310.15 K T 2 = 87◦C + 273.15 = 360.15 K ∆H vap◦ = 44.0 kJ · mol−^1 P 1 = 47.1 torr Using the Clausius-Clapeyron equation,

ln

P 2

P 1

∆H vap◦ R

T 1

T 2

44 kJ · mol−^1 8 .314 J/mol−^1 · K−^1

1000 J

1 kJ

×

310 .15 K

360 .15 K

P 2

P 1

= e^2.^36896

P 2 = P 1 e^2.^36896 = (47.1 torr) e^2.^36896 = 503.323 torr