PAC113F Test 1 Memo 2011 PDF

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PAC113F TEST 1 1. For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different? [4]

Solution For an acid and base, 1 normal = 1 molar when, HA (aq) → H+ (aq) + A – (aq) BOH (aq) → B+ (aq) + OH – (aq) For an acid and base, 1 normal ≠ 1 molar when, HnA (g, s) → n H+ (aq) + A n– (aq) B(OH)n (g, s) → B n+ (aq) + n OH – (aq) 2. A 1.37 M solution of citric acid (H 3C6H5O7) in water has a density of 1.10 g/mL. Calculate the mass percent, molality, mole fraction and normality of the citric acid. Citric acid has three acidic protons. [8]

Data Extraction Given: C(H3C6H5O7) = 1.37 mol/L, d(solution) = 1.10 g/mL R. T. C: C(H3C6H5O7) = ? mol/kg,( H3C6H5O7) = ? , and C(H3C6H5O7) = ? eq. mol/L (N)

Solution m ( H 3 C 6 H 5 O 7) n ( H 3 C 6 H 5 O 7 ) × Mm ( H 3 C 6 H 5 O 7 ) %m( H 3 C 6 H 5 O 7 )= × 100 % × 100 %= m ( solution ) V ( solution )×d ( solution ) %m( H 3 C 6 H 5 O 7 )=

C ( H 3 C 6 H 5 O7) × Mm( H 3 C 6 H 5 O7 ) × 100 % d ( solution ) 1.37

%m( H 3 C 6 H 5 O 7 )=

C ( H 3 C6 H 5 O 7) =

C ( H 3 C 6 H 5 O 7) =

g mol × 192.1 L mol ×100 %= 24.0 % 3 g 1.10 × 10 L

m ( H 3 C 6 H 5 O7) Mm ( H 3 C 6 H 5 O 7) × m ( H 2 O )

=

%m( H 3 C 6 H 5 O 7 ) Mm( H 3 C 6 H 5 O7) × %m( H 2 O )

24.0 % =1.64 ×10−3 mol /g=1.64 mol /kg =1.64 m g 192.1 × 76 % mol

1

χ ( H 3 C 6 H 5 O 7) =

n ( H 3 C6 H 5 O7 ) n ( H 3 C 6 H 5 O7 ) +n ( H 2 O )

χ ( H 3 C 6 H 5 O 7) =

=

1 1 = n ( H 2 O) m ( H 2 O ) × Mm ( H 3 C 6 H 5 O 7) 1+ 1+ n ( H 3 C 6 H 5 O 7) m ( H 3 C 6 H 5 O7 ) × Mm ( H 2 O )

1 1 =0.0290 = %m ( H 2 O ) × Mm ( H 3 C 6 H 5 O 7) 1+ 76 % × 192.1 g /mol 1+ 24 % ×18.02 g /mol %m( H 3 C 6 H 5 O7 ) × Mm ( H 2 O ) Since this is a 1:3 electrolyte H3C6H5O7 (aq) → 3H + (aq) + C6H5O7 3– (aq)

+¿ ¿ H ¿ 3 eq . n ¿ C ( H 3 C6 H 5 O 7) =¿ 3. Which solvent, water or carbon tetrachloride, would you choose to dissolve KrF2

[3]

Solution

Lewis structure

F

Kr

F

2 B. P. 3 L. P. 5 E. P.

Linear

Linear, non-polar, carbon tetrachloride 4. The lattice energy of NaI is – 686 kJ/mol, and the enthalpy of hydration is – 694 kJ/mol. Calculate the enthalpy of solution per mole of solid NaI. Describe the process for which this enthalpy applies. [4]

Data Extraction Given: Na + (g) + I – (g) → NaI (s) Na + (g) + I – (g)

Hlattice

H 2 O (l) Na + (aq) + I – (aq) →

R. T. C: Hsoln = ?, what does this process represent

Answer Hsoln = ( – Hlattice ) + Hhyd

∆ H soln =[ (−1 ×−686 )+ ( −694 ) ] kJ /mol=−8 kJ /mol 2

Hhyd

This process represents the reaction:

H 2 O(l) Na + (aq) + I – (aq)

NaI (s)

→

5. The weak electrolyte NH3 (g) does not obey Henry’s Law. Why? O 2 obeys Henry’s Law in water but not in blood (an aqueous solution) why? [4]

Solution In water NH3 (g) undergoes the transformation: NH3 (g) + H2O (l) → NH4 + (aq) + OH – (aq) and is expected not to obey Henry’s Law, whereas O 2 retains gaseous O2 properties when dissolved in water. In blood (a solution) it is possible that O 2 forms complex—ions with the heavier atoms (Fe) found in blood and these complex—ions are soluble in water. 6. What is the composition of a methanol (CH3OH)—propanol (CH3CH2CH2OH) solution that has a vapor pressure of 174 torr at 40 o C. At this temperature, the vapor pressures of pure methanol and pure propanol are 303 and 44.6 torr respectively. Assume the solution is ideal. [4]

Ptotal= χ CH

3

OH

P

Solution + (1− χ CH

o CH 3 OH

) PCH CH CH o

3

OH

3

2

2

OH

P o

o CH 3CH 2 CH 2 OH

)+ PoCH (¿ ¿ CH 3 OH −P P total= χ CH OH ¿

3

CH 2 CH 2 OH

3

P o o ¿ CH OH −P (¿ CH 3 χ CH

3

OH

=

χ CH OH = 3

χ CH CH 3

2

)

3 CH 2 CH 2 OH o CH 3 CH 2 CH 2 OH

Ptotal−P

¿

(174 − 44.6 )torr =0.501 (303−44.6)torr

CH 2 OH

=1− χ CH OH =1 −0.501=0.500 3

7. An aqueous solution of 10.0 g catalase, an enzyme found in the liver, has a volume of 1.00 L at 27 o C. The solutions osmotic pressure at 27 o C is 0.74 torr. Calculate the molar mass of catalase.

[3] Solution nRT Π =CRT = V

3

Π=

Mm (catalase )=

m × RT Mm× V

mRT = 10.0 g× 0.08206 L∙ atm /mol ∙ K ×300 K =2.53 ×105 g/mol Π ×V 1.00 L ×(0.74 / 760 )atm

8. Consider the following

1.0 M NaCl

H 2O

What would happen to the level of liquid in the two arms if the semi-permeable membrane were permeable to a. H2O only? b. H2O, Na+, and Cl – [4]

Solution a. If the membrane is only permeable to water, there will be a net movement of water molecules to the solution side and the solution will rise up until the osmotic pressure of the two sides are equal. b. If the membrane is permeable to water and the ion solutes, the will be movement of both solutes and solvent until a new sate of equilibrium is attained and the levels of solution will be equal in both sides. 9. Explain the terms, isotonic solution, crenation, and hemolysis.

[6]

Solution Isotonic solutions are solutions with the same osmotic pressure. Crenation is the phenomenon which occurs when a biological cell is placed in a hypertonic solution and shrivels from the movement of water from the cell to the solution. Hemolysis is the reverse of crenation, when the cell is placed in a hypotonic solution and ruptures because of the movement of water from the solution to inside the cell. 10. What is ion-pairing?

[2]

Solution Not all the ions in concentrated solutions are moving freely; some get paired and thus count as one particle. This phenomenon is called ion-pairing. 11. Calculate the freezing point and boiling point of each of the following aqueous solutions. (Assume complete dissociation.). Kf = 1.86 o C.kg/mol, Kb = 0.51 o C.kg/mol [8] a. 0.050 m MgCl2 b. 0.050 m FeCl3

Data extraction Given: C(MgCl2) = 0.050 mol/kg, Kf = 1.86 o C· kg/mol, 4

R. T. C: Tf(solution) =?

Solution a.

∆ T =iK f m solute

[ T f ( H 2 O) −T f ( solution) ] =iK f msolute T f (solution )=T f (H 2 O )−iK f m solute =0 ℃ −3× 1.86 ℃ ∙kg /mol × 0.050 mol/kg=−0.28 ℃ ∆ T =iK b m solute

[ T b( solution )−T b ( H 2 O) ] =iK f msolute T b ( solution )=T b ( H 2 O) +iK b msolute =100 ℃+3 ×0.51 ℃ ∙ kg/mol × 0.050 mol / kg =100.08 ℃ b.

∆ T =i K f m solute

[ T f ( H 2 O) −T f ( solution) ] =iK f msolute T f (solution )=T f (H 2 O )−i K f msolute =0 ℃−4 × 1.86℃ ∙ kg / mol ×0.050 mol / kg=−0.37 ℃ ∆ T =iK b m solute

[ T b( solution)−T b( H 2 O ) ] =i K f msolute T b ( solution )=T b ( H 2 O) + iK b m solute =100 ℃ + 4 ×0.51 ℃ ∙ kg/ mol × 0.050 mol / kg=100.10 ℃

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