Ch E102 - Tutorial 6 Solutions PDF

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Tutorial 6 Solutions...

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ChE102 Chemistry for Engineers - Tutorial 6

Solve the following problems either individually or in small groups (3 students maximum). Submit one copy of the solutions per group. Ask the TA for help if you need assistance solving the problems.

Problem 1 Estimate the volume (in L) of pure methane gas (at STP) that will dissolve in 1.0 L of pure water (density = 0.998 g·cm–3) when the methane partial pressure is 1340 kPa. Data: The Henry’s Law constant for methane in water is 3.46 × 107 torr. STP is 1 bar and 0 °C First calculate the mole fraction of methane in solution using Henry’s Law: 𝑃𝑖 = 𝑘𝑥𝑖 𝑥𝑖 =

1340 kPa

3.46×107 torr

×

760 torr

101.325 kPa

= 2.896 × 10−4

mol methane mol solution

Calculate the amount of methane in solution. Since the mole fraction is much less than one assume that the change in volume when methane dissolves in 1.0 L water is negligible. 1 L solution ×

1000 cm3 solution 1 L solution

×

0.998 g solution 1 cm3

×

1 mol solution

18.01528 g solution

× 2.90 × 10−4

mol methane

mol solution

= 0.01604 mol methane

Use the ideal gas law to determine the volume of methane gas, measured at STP. 𝑉=

𝑛𝑅𝑇 𝑃

=

L atm

(0.01604 mol)(0.08206 mol K)(273.15 K) 1 atm 1 bar× 1.01325 bar

= 0.36 L

Problem 2 Four aqueous solutions of acetone, CH3COCH3, are prepared at different concentrations at 25 °C: (a) 0.100% CH3COCH3 by mass (b) 0.100 M CH3COCH3 (c) 0.100 mol CH3COCH3/kg H2O (d) 𝑥CH3 COCH3 = 0.100 Which of these solutions would have the highest partial pressure of water at 25 °C in the equilibrium vapour above the solutions? Data: Assume that the density of all solutions is approximately equal to the density of water, 1.00 g/mL. From Raoult’s law: 𝑃𝑖 = 𝑥𝑖 𝑃°𝑖 , therefore, the highest mole fraction of water (the most dilute solution of acetone) will result in the highest equilibrium vapour pressure. In order to compare the concentrations, we need each concentration to be given in the same units. The calculations below calculate the molarity of each solution. (a)

0.100 g acetone 100 g solution

×

1 mol acetone

58.08004 g acetone

×

1.00 g solution

1 mL solution

×

1000 mL solution 1 L solution

= 0.0172 M

(b) 0.100 M (c) The concentration will be very close to 0.100 M. Assume a basis of 1 kg = 1000 g H2O, then the mass of acetone is: 0.100 mol acetone ×

58.08004 g acetone 1 mol acetone

= 5.808004 g acetone

The mass of the solution is 1005.8 g solution. The concentration in mol/L is: 0.100 mol acetone 1005.8 g solution

×

1g

1 mL

×

1000 mL 1L

= 0.099 M

(d) Assume a basis of 1 mol solution 1 mol solution = 0.100 mol acetone + 0.900 mol H2O 𝑚solution = 0.100 mol acetone × 𝑉solution = 22.02 g ×

1 mL

1.00 g

×

58.08004 g acetone

1L

1000 mL

1 mol acetone

+ 0.900 mol H2 O ×

= 0.02202 L

18.01528 g H2 O 1 mol H2 O

= 22 .02 g

concentration =

0.100 mol acetone 0.02202 L

= 4.5 M

Solution (a) is the most dilute solution and will therefore have the highest equilibrium vapour pressure. Problem 3 Styrene, used in the manufacture of polystyrene, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains 38% of styrene (C8H8) and 62% of ethylbenzene (C8H10), by mass. The mixture is separated by fractional distillation at 90 °C. Determine the composition of the vapour in equilibrium with this mixture at 90 °C. Data: 𝑃°C8 H8 = 134 mmHg and 𝑃°C8 H10 = 182 mmHg at 90 °C.

In order to use Raoult’s law we first need to determine the mole fraction of each component in the solution. Basis: 100 g solution = 38 g C8H8 + 62 g C8H10 1 mol C H

38 g C8 H8 × 104.151528g C8 1 mol C H

8 H8

= 0.3649 mol C8 H8

62 g C8 H10 × 106.1674 8g C10H = 0.5840 mol C8 H10 𝑥C8 H8 =

0.3649 0.3649+0.5840

8 10

= 0.3846

𝑥C8 H10 = 1 − 𝑥C8 H8 = 0.6154

Now use Raoult’s law to determine the partial pressure of each component and the mole fraction in the vapour phase. 𝑃C8 H8 = 𝑥C8 H8 𝑃°C8 H8 𝑃C8 H8 = (0.3846)(134 mmHg) = 51.5 mmHg 𝑦C8 H8 =

51.5

51.5+112

= 0.31

𝑃C8 H10 = 𝑥C8 H10 𝑃°C8 H10 𝑃C8 H10 = (0.6154)(182 mmHg) = 112 mmHg 𝑦C8 H10 = 1 − 𝑦C8 H8 = 0.69

Problem 4 An aqueous solution is 15.0 % NaBr by mass. Assuming that the solution behaves ideally and that NaBr completely dissociates, what would be the expected boiling point of the solution at a total pressure of 1 atm? Data: For water Kb = 0.512 kg·°C·mol–1 Sodium bromide dissociates as: NaBr(s)  Na+(aq) + Br-(aq) Assuming complete dissociation, the van ‘t Hoff factor, i = 2. ∆𝑇𝑏 = 𝑖𝐾𝑏 𝑏 = 𝐾𝑏 𝑏𝑒𝑓𝑓 To determine the change in boiling point, we must first determine the molality of the solution. Assume a basis of 100. g solution. 15.0 g NaBr

100. g solution × 100.g solution ×

1 mol NaBr 102.8938 g NaBr

= 0.1458 mol NaBr

Since the solution is 15% NaBr by mass the remaining 85% must be water. 𝑚solvent = 85 g = 0.085 kg 𝑏=

0.1458 mol NaBr 0.085 kg solvent

∆𝑇𝑏 = 2 × 0.512

= 1.715 mol/ kg solvent ℃ kg mol

× 1.715

mol kg

𝑇𝑏 = 100 + 1.76 ℃ = 101.76 ℃

= 1.76 ℃

Problem 5 A salt AB2 dissociates partially in water. The dissociate occurs as, AB2  A2   2 B

A 0.10 molal aqueous solution of AB2 has a freezing point of –0.250 °C. Calculate the degree of dissociation of the salt, assuming that it is independent of temperature. Data: For water Kf = 1.86 kg·°C·mol–1

From the given information calculate the van ‘t Hoff factor, i: ∆𝑇𝑓 = −𝑖𝐾𝑓 𝑏 = 𝐾𝑓 𝑏𝑒𝑓𝑓 𝑏𝑒𝑓𝑓 = −

∆𝑇𝑓 𝐾𝑓

=

0.250

1.86

℃ kg mol

= 0.134 mol/kg solvent

Assume a basis of 1 kg. Therefore, a 0.10 molal solution contains 0.10 mol AB2. Let x represent the fraction of AB2 that dissociates:

initial (mol) change (mol) final (mol)

AB2 0.10 -0.10x 0.10 – 0.10x



From the effective molality we know that after dissociation: 0.134 = 0.10 − 0.10𝑥 + 0.10𝑥 + 0.20𝑥 = 0.10 + 0.20𝑥 Solve for x:

𝑥=

0.134−0.10 0.20

= 0.17 = 17%

A2+ 0 +0.10x 0.10x

2B0 +0.20x 0.20x

Problem 6 A 4.0 g mixture consisting of sucrose (C12H22O11) and zinc nitrate (Zn(NO3)2) is dissolved in 150.0 g of water. If the resulting solution freezes at –0.768 °C, what is the mass fraction of sucrose in the mixture? Assume that Zn(NO3)2 dissociates completely in water. Data: For water Kf = 1.86 kg·°C·mol–1 ∆𝑇𝑓 = ∆𝑇𝑓

sucrose

+ ∆𝑇𝑓 zinc nitrate = −𝐾𝑓 [(𝑖𝑏)sucrose + (𝑖𝑏)zinc nitrate ]

Sucrose does not dissociate, therefore i = 1. Zinc nitrate dissociated as follows: ZnNO3 2  Zn2   2 NO3

Therefore, i = 3. 𝑛sucrose

∆𝑇𝑓 = −𝐾𝑓 [𝑏sucrose + 3𝑏zinc nitrate ] = −𝐾𝑓 [

𝑚H2 O

+

3𝑛zinc nitrate 𝑚H2 O

]=−

𝐾𝑓

𝑚H2 O

[

𝑚sucrose

𝑀sucrose

+

Let x represent the mass of sucrose, then 4.0 – x represents the mass of zinc nitrate. ∆𝑇𝑓 = −

𝐾𝑓

𝑚H2 O

𝑥

[ 342.30008 +

−0.768 ℃ = −

3(4.0−𝑥) ] 189.3998

1.86 ℃ kg /mol 0.150 kg

𝑥

[ 342.30008 +

3(4.0−𝑥)

189.3998

]

Solve for x: (0.768 ℃)(0.150 kg) 1.86 ℃ kg/mol

𝑥=

=

𝑥

342.30008

(0.768 ℃)(0.150 kg) 12.0 − 189.3998 1.86 ℃ kg/mol 1 3 − 342.30008 189.3998

0.11 g sucrose 4.0 g mixture

+

12.0−3𝑥

189.3998

1

= 𝑥 ( 342.30008 −

= 0.11 g sucrose

= 0.028 = 2.8% sucrose by mass

3

189.3998

)+

12.0

189.3998

3𝑚zinc nitrate 𝑀zinc nitrate

]...