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ChE102 Chemistry for Engineers

Final Exam Review package

Waterloo SOS

Fall 2010

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A gaseous compound known to contain only carbon, hydrogen, and nitrogen is mixed with exactly the

volume of oxygen required for its complete combustion to CO2, H2O, and N2. Burning 20 litres of the

gaseous reaction mixture (the unknown compound plus oxygen) produces 14 litres of CO2, 10 litres of

H2O vapour, and 2 litres of N2, all at the same temperature and pressure.(a) Of the 20 litres of reaction mixture, how many litres are oxygen?(b) What is the molecular formula of the unknown compound?

Basis: 20 litres of the reaction mixture containing (20-a) litres of CxHyNzplus avolumes of O2.

For an ideal gas at constant temperature and pressure, the volume of the ideal gas is directly proportional

to the number of moles of the ideal gas. Therefore, the volume ratios given in the problem statement can

also be interpreted as molar ratios.

The chemical equation for the reaction is therefore:

( ) 2222zyx N2OH01CO41ONHC20 +++ aa

Balancing this chemical equation for C, H, N, and O gives four equations:

( )

( )

( )

382:)(

420:)(

2020:)(

1420:)(

=

=

=

=

aOOxygen

zaNNitrogen

yaHHydrogen

xaCCarbon

The solution to these equations is:

4,20,14

19

===

=

zyx

a

Of the 20 litres of reaction mixture, 19 litres are O2, and the molecular formula of the unknown compound

is C14H20N4.

Dew point is defined as the temperature at which the vapour pressure of water is equal to the partial

pressure of water vapour in an air sample. Exactly 0.105 g of water was evaporated into 10.0 L of dry air

at 60.0C and 100 kPa. Using the vapour pressure data below, estimate the dew point of the moist air

sample if the total pressure is maintained at 100 kPa.

Data:

T[K] 280 285 290 295 298.15

Po[kPa] 0.9912 1.388 1.919 2.620 3.197

Basis: 0.105 g of H2O in 10.0 L of dry air at 60.0oC and 100 kPa.

Data: Molecular mass of H2O, molg01528.18OH2 =

Phases Q1

Phases Q2

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The number of moles of H2O evaporated is,

=OH2n 0.105 g H2O mol H2O = 0.0058284 mol H2O

18.01528 g H2O

The number of moles of dry air is,

=

= TR

VP

n airdry

100 kPa 10.0 L mol K

= 0.36101 mol dry air333.15 K 8.31451 kPa L

The partial pressure of H2O in the humid air is therefore,

kPa1.5888kPa100mol0.36101mol0058284.0

mol0058284.0totalOHOH 22

=

+== PyP

The dew point temperature of the humid air is the temperature at which this partial pressure of H2O

becomes equal to the vapour pressure of H2O. Inspection of the given vapour pressure data shows that

the dew point temperature should fall between 285 K and 290 K. Since the Clausius-Clapeyron equation

states that,

=

21

vap

2

1 11lnTTR

H

P

P

o

o

we find that between 285 K and 290K,

K734.5354

K290

1

K285

1

kPa919.1

kPa388.1ln

vap

vap

=

=

R

H

R

H

Therefore,

( )

K06.287K285

11K734.5354

kPa388.1

kPa5888.1ln

pointdew

pointdew

=

=

TT

The dew point temperature of the humid air is 287.06 K or 13.91oC.

At 25.0C, the vapour pressure of trichloroethene (C2HCl3) is 73.00 mm-Hg. At the same temperature,

the vapour pressure of chloroform (trichloromethane, CHCl3) is 199.1 mm-Hg.

a) What is the pressure exerted by a solution containing 32.05% by mass of C2HCl3 (the balancebeing CHCl3) at 25.0C?

b) What is the vapour phase mole fraction of CHCl3that is in equilibrium with the liquid solutionspecified in (a)?

Basis: 100 g of solution containing 32.05 g of C2HCl3and 67.95 g of CHCl3at 25.0C.

Data: Molecular masses in g/mol:

377.119

388.131

3

32

CHCl

HClC

=

=

Phases Q3

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The composition of the C2HCl3-CHCl3solution is given in the following table:

Component Mass [g] Mole [mol] Mole fraction

C2HCl3 32.05 0.243934 0.29999

CHCl3 67.95 0.569205 0.70001

Total 0.813139 1.00000

The total vapour pressure is

Hgmm2712.161

Hgmm1.19970.0Hgmm00.7330.0

333232 CHClCHClHClCHClC

=

+=

+= oototal PxPxP

The total vapour pressure exerted by the solution is 161.271 mm-Hg.

Basis: 100 g of solution containing 32.05 g of C2HCl3and 67.95 g of CHCl3at 25.0C.

The vapour phase mole fraction of CHCl3is

=

==total

CHClCHCl

total

CHCl

CHCl333

3 P

Px

P

Py

o

0.70 199.1 mm-Hg

= 0.864208161.271 mm-Hg

The mole fraction of CHCl3in the vapour phase is 0.8642.

The vapour pressure of pure liquid CS2 is 0.3914 atm at 20C. When 40.0 g of rhombic sulphur is

dissolved in 1.00 kg of CS2, the vapour pressure of CS2 falls to 0.3868 atm. Determine the molecular

formula for the sulphur molecules dissolved in CS2.

Basis: 1.0 kg of CS2liquid.

Data: Initial vapour pressure of CS2, 0.3914 atm.

Mass of sulphur added, 40.0 g.

Final vapour pressure of CS2, 0.3868 atm.

Molecular mass of CS2, molg143.762CS =

Atomic mass of sulphur, molg066.32S =

We know what the change in pressure is due to the addition of sulphur to the CS2solution. Thus the mole

fraction of rhombic sulphur present is

=

=o

CS

rS

2P

Px

(0.3868 0.3914) atm= 0.011753

0.3914 atm

The solution contains 1.0 kg of CS2, which is

=2CS

n 1000 g CS2 mol CS2 = 13.13318 mol CS276.143 g CS2

Phases Q4

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From the expression for mole fraction we can find the number of moles of rhombic sulphur added,

rSmol15619.01 rS

CSrS

rSCSrS

rSrS

2

2

=

=

+=

x

nxn

nn

nx

Since we know the mass of rhombic sulphur added, we can find the molecular weight of rhombic sulphur,

=rS 40.0 g rS = 256.10533 g/mol rS0.15619 mol rS

To find the number of sulphur atoms in rhombic sulphur,

=S

rS

256.10533 g/mol= 7.9868

32.066 g/mol

Therefore rhombic sulphur is of the form S8.

The vapour pressure of benzene is 1.34 atm at 90oC and its boiling point at 1 atm is 80

oC. Assuming that

Hvapis constant, calculate the pressure at which benzene will boil at 0oC.

From the Clapeyron equation:

( ) ( )

( ) ( )

atmP

R

H

atmP

R

H

atm

atm

TTR

H

P

P

o

vap

o

vap

vap

o

o

04446.0

K15.27301

K15.273901

34.1ln

K15.27380

1

K15.27390

1

34.1

1ln

11ln

3

3

211

2

=

+

+

=

+

+

=

=

The pressure is 0.044476atm 760 mmHg atm-1

= 33.8 mmHg

Diethyl ether (C4H10O) is a volatile, highly flammable organic liquid that is used mainly as a solvent. The

vapour pressure, Pvap, of diethyl ether can be estimated using Antoines equation,

CKT

BAmmHgPvap

+=

)()(ln

where A= 16.0828, B= 2511.29 K, C= -41.95 K for diethyl ether. Calculate the enthalpy of vaporization

of diethyl ether at 25oC and 45oC and use these values to estimate the heat of vaporization for diethyl

ether.

Phases Q5

Phases Q6

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mmHgPvap 12.53195.41)27325(

29.25110828.16ln

1,=

+=

mmHgPvap 93.108095.41)27345(

29.25110828.16ln

2,=

+=

Use Clausius-Clapeyron equation to estimate Hvap:

=

211,

2, 11ln

TTR

H

P

P vap

vap

vap

mol

kJ

KKKmol

J

TTP

PRH

vap

vap

vap 23.27994318

1

298

1

12.531

93.1080ln

.31451.8

11ln

11

211,

2,=

=

=

A salt AB2dissociates partially in water. The dissociate occurs as,+ + B2AAB 22

A 0.10 molal aqueous solution of AB2 has a freezing point of 0.250C. Calculate the degree of

dissociation of the salt, assuming that it is independent of temperature.

Data: OHformolCkg86.1 2o=fK

Basis: 0.10 molal aqueous solution of AB2containing 1 kg of H2O.

From the freezing point depression we can calculate the effective molality of the solution.

=

=f

f

effK

Tm

0.250oC mol

= 0.13441mol

1.86 kg oC kg H2O

Let be the degree of dissociation of AB2 in the solution (i.e., is equal to the fraction of AB2 thatdissociates in solution). Thus, we have in the solution,

( )

-

B

2

A

2AB

Bmol20.0

Amol10.0

ated)(undissociABmol110.0

-

2

2

=

=

=

++

n

n

n

The effective molality is,

( )[ ] [ ]OHkg1

mol13441.0

OHkg1

mol20.010.0

OHkg1

mol20.010.0110.0

OHkg1 2222

BAAB -22 =+

=++

=++ + nnn

Solving this equation

Phases Q7

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mol20.0

mol10.0OHkg1OHkg1

mol13441.02

2

=

The degree of dissociation of AB2in water is 0.17205.

A saturated solution is prepared at 70C containing 32.0g CuSO4per 100.0 g solution. A 335 g sample of

this solution is then cooled to 0C and CuSO45H2O precipitates. If the concentration of a saturated

solution at 0C is 12.5 g CuSO4/100 g solution, what mass of CuSO45H2O precipitate would be obtained?(Hint: Note that the solution composition is stated in terms of CuSO4 and that the solid that precipitates is

the copper sulphate hydrate, CuSO45H2O).

Basis: 335 g of solution containing 32.0 g of CuSO4per 100 g

Data: Molecular masses in g mol-1

:

681.249

607.159

O5HCuSO

CuSO

24

4

=

=

CuSO4dissolved initially:

=4CuSOin,

m

335 g soln 32.0 g CuSO4= 107.2 g CuSO4

100 g soln

CuSO4dissolved at end:

=4CuSOf,

m

(335 x) g

soln12.5 g CuSO4 =

0.125 * (335 m)

g CuSO4100 g soln

Where m is the mass (g) of CuSO45H2O precipitate.

The mass fraction of CuSO4in the CuSO45H2O precipitate is

=x 159.607 g CuSO4 1 mol CuSO4 1 mol precipitate =

0.639 244 g

CuSO4/g precipitatemol CuSO4 1 mol precipitate 248.681 g precipitate

Mass balance on CuSO4:

Mass dissolved at

start- mass dissolved at end = mass precipitated

107.2 - 0.125 * ( 335 m) = 0.639 244 * m

Solving, mass of precipitate, m = 127 g.

Phases Q8

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The vapour pressure of pure water water is 17.54 mm-Hg at 20oC. Calculate the vapour pressure

lowering of aqueous solutions that are 0.25 molality in

a) NaCl,

b) CaCl2,c) sucrose (C12H22O11).

Basis: 0.25 mol of NaCl in 1000 g of H2O.

Data: Molecular mass of H2O, molg01528.18OH2 =

The number of moles of H2O is1000 g H2O 1 mol H2O = 55.50844 mol H2O

18.01528 g H2O

Assuming complete dissociation of NaCl in the solution, the numbers of moles of Na+ and Cl- in the

solution are both 0.25 mol. The vapour pressure lowering of the solution is,

Hg-mm0.15658

Hg-mm55.17mol50844.55mol25.0mol25.0

mol25.0mol25.0

oOH

OHClNa

ClNaOH 2

2

2

=

++

+=

++

+=

+

+

Pnnn

nnP

The vapour pressure lowering of the aqueous solution is 0.1566 mm-Hg.

Basis: 0.25 mol of CaCl2in 1000 g of H2O.Data: Molecular mass of H2O, molg01528.18OH2 =

Assuming complete dissociation of CaCl2 in the solution, the numbers of moles of Ca2+ and Cl- in the

solution are 0.25 mol and 0.50 mol respectively. The vapour pressure lowering of the solution is,

Hg-mm23383.0

Hg-mm55.17mol50844.55mol50.0mol25.0

mol50.0mol25.0

oOH

OHClCa

ClCaOH 2

22

2

2

=

++

+=

++

+=

+

+

Pnnn

nnP

The vapour pressure lowering of the aqueous solution is 0.2338 mm-Hg.

Basis: 0.25 mol of C12H22O11in 1000 g of H2O.

Data: Molecular mass of H2O, molg01528.18OH2 =

Phases Q9

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Assuming no dissociation of C12H22O11in the solution, the number of moles of C12H22O11in the solutionis 0.25 mol. The vapour pressure lowering of the solution is,

Hg-mm0.078643

Hg-mm55.17mol50844.55mol25.0

mol25.0

oOH

OHOHC

OHC

OH 22112221

112221

2

=

+=

+= P

nn

nP

The vapour pressure lowering of the aqueous solution is 0.07864 mmHg.

An aqueous solution is 15.0% by weight with respect to NaBr. Assuming that the solution behaves

ideally and that dissociation is complete, what would be the expected boiling point of the solution at a

total pressure of 1 atm?

Data: OHformolCkg512.0 2=bK

Basis: 100.0 g of NaBr solution containing 15.0g of NaBr.

Data: Molecular mass of NaBr, molg8938.102NaBr =

Sodium bromide dissociates in aqueous solution as,+ + (aq)(aq)(s) BrNaNaBr

The effective molality of the solution is,

=effm 15 g NaBr mol NaBr 2 mol Na+and Br-ions 1000 g H2O

= 3.4302mol ions

85 g H2O 102.8938 g NaBr 1 mol NaBr 1 kg H2O kg H2O

The boiling point elevation of the solution is

== beffb KmT 3.4302 mol 0.512 kg

oC

= 1.756oCkg mol

The boiling point of the NaBr solution is 101.76oC at 1 atm.

At 90.0oC, the vapour pressures of benzene (C6H6) and toluene (C7H8) are 1.34 atm and 0.534 atm,

respectively. Benzene and toluene form a nearly ideal solution at 90.0oC.a) What is the mole fraction of toluene in a benzene-toluene solution that begins to boil at 90.0oC

under 1 atm?

b) What is the mass percentage of benzene in the vapour over this solution at 90.0oC and 1 atm?

Basis: 1 mol of benzene-toluene solution boiling at 1 atm and 90.0oC.

Phases Q10

Phases Q11

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( )

( )( )

C68.7atkPa38.561

Hg-mm23.289

Hg-mm200

K85.341

o

K85.331

1

K85.341

1K07.4185

11

11K85.341

167

=

=

=

=

e

ePP TTR

H

ooHC

vap

The liquid phase mole fractions can be determined using Raoult's law.

( ) kPa701167146146146167146 HCHCHCHCHCHC

=+=+= oototal PxPxPPP

Since kPa,561.38andkPa325.101 o HCo

HC 167146== PP solving the above equation gives .50091.0

146HC =x The

mole fraction of hexane in the vapour phase is then

=

==total

o

total P

Px

P

Py 146146146

146

HCHCHC

HC 0.50091 101.325 kPa

= 0.7250670 kPa

The mole fractions of hexane in the liquid and vapour phases are 0.5009 and 0.7251, respectively.

A 4.0 g mixture consisting of sucrose (C12H22O11) and zinc nitrate (Zn(NO3)2) is dissolved in 150.0 g ofwater. If the resulting solution freezes at 0.768C, what is the mass fraction of sucrose in the mixture?

Assume that Zn(NO3)2dissociates completely in water.

Data: OHformolCkg86.1 2o=fK

Basis: 4.0 g of the C12H22O11-Zn(NO3)2mixture containing x g of C12H22O11 dissolved in 150.0 g of

H2O.

Data: Molecular masses in g/mol:

( ) 40.189

30.342

23

112212

NOZn

OHC

=

=

Zinc nitrate dissociates in aqueous solution as

( ) + + 32

23NO2ZnNOZn

The effective molality of the solution is (eqn. 1),

( ) ( ) ( )

( ) ( ) ( ) kgmol08612.042239.0

NOZnmol1

ionsmol3

NOZng40.189

NOZnmolNOZng0.4

OHCg30342OHCmolOHCg

OHkg150.01

2323

2323

112212

112212112212

2

xx

.xmeff

=

+

=

But the effective molality can also be calculated from the freezing point depression as (eqn. 2)

T

0.768 oC mol = 0.412903226 mol/kg

Phases Q13

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1.86 kg oC

Therefore equating eqn. 1and 2.

(0.42239 0.08612x) mol/kg = 0.412903226 mol/kg

Solving for x gives x = 0.110116114 g.

The mass fraction of sucrose in the mixture is therefore x g/4.0 g = 0.0275.

A solution of 0.07265 g of a human hormone in 100 mL of solution has an osmotic pressure of 12.60

mmHg at 21.6oC. What is the molecular mass of the hormone in g /mol?

Basis: 0.07265 g of hormone in 100 mL.

The osmotic pressure is given by:

V

TRm

V

TRn

==

Solving for the molecular mass

=

=

V

TRm

0.07265 g 0.082 Latm 294.6 K= 1058.5 g/mol

1.65810-2atm molL 0.1 L

The molecular mass of the hormone is 1058 g/mol.

At 100oC the vapour pressure of toluene is 557.2 mmHg. Toluene and benzene form nearly ideal

solutions. A particular benzene-toluene solution has a normal boiling point of 100oC, and the vapour in

equilibrium with the liquid contains 58.26% toluene by mass. What is the composition of the liquid,

expressed as a mass percentage of toluene? Solve this without looking up the vapour pressure of benzene.

Basis: 100 g of benzene-toluene vapour in equilibrium with its liquid solution at 100oC.

Data: Molecular masses in g/mol:

1402.921134.78

87

66

HC

HC

==

The following table shows the composition of the vapour phase:

Component mass [g] mol y

toluene 58.26 0.6323 0.5420benzene 41.74 0.5344 0.4580

Total 100.00 1.1666 1.0000

Phases Q14

Phases Q15

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The mole fraction composition of the liquid phase is

=

==o

HC

totalHC

oHC

HC

HC

87

87

87

87

87P

Py

P

Px

0.5420 760 mmHg= 0.7392

557.2 mmHg

The mass percentage of toluene in the liquid phase is therefore

( )7697.0

187876687

8787

87

HCHCHCHC

HCHC

HC =+

=

xx

xw

The liquid phase contains 76.97% toluene by mass.

Consider the homogeneous gas phase reaction

)()()( RBA ggg +

occurring at 400.0 K in a 1.0 L vessel. Initially there are 0.070 mol of A, 0.030 mol of B and 0.150 mol

of R. What is the equilibrium concentration of R (in mol/L)?

Data: for the above reaction at 400.0 K: 1atm9749.0 =PK

Basis: 0.070 mol of A, 0.030 mol of B, and 0.150 mol of R in a 1.0 L vessel at 400.0 K.

The net change in the number of moles in the reaction (n) is 1. Therefore, conversion of KPinto Kcgives

( ) == nPc TRKK 0.9749 0.0820578 atm L 400.0 K = 31.9993 L/mol

atm mol K

The stoichiometry of the reaction is summarised in the following table:

Component initial [mol] change [mol] final [mol] concn. final [mol/L]

A 0.070 x 0.070x 0.070x

B 0.030 x 0.030x 0.030x

R 0.150 +x 0.150+x 0.150+x

When equilibrium is reached,[ ]

[ ] [ ]==

BA

RQ

0.150+x= Kc= 31.9993 L/mol

(0.070-x)(0.030-x)

Rearranging this gives the quadratic equation:

( ) ( ) 008280147.019993.49993.31 2 = xx

Solving for x:

( )