the mole concept

New Way Chemistry for Hong Kong A-Level Book 11

The Mole ConceptThe Mole Concept

2.12.1 The MoleThe Mole

2.22.2 Molar Volume and Avogadro’s LawMolar Volume and Avogadro’s Law

2.32.3 Ideal Gas EquationIdeal Gas Equation

2.4 2.4 Determination of Molar MassDetermination of Molar Mass

2.52.5 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

22


2.2.11 The MoleThe Mole


2.1 The mole (SB p.18)

What is “mole”?What is “mole”?

Item Unit used to count

No. of items per unit

Shoes pairs 2

Eggs dozens 12

Paper reams 500

Particles in Chemistry

moles 6.02 1023

for counting particles like atoms, ions, molecules

for counting common objects


6.02 1023

= 602 000 000 000 000 000 000 000

Avogadro constant(the amount in 1 mole)

How large is the amount in 1 mole?How large is the amount in 1 mole?



$ 6.02 1023All the people in the world

so that each get:

$ 1000 note

count at a rate of 2 notes/sec

?2000 years



1

or

number of particles

= number of moles (6.02 1023)

number of particles

= number of moles (6.02 1023)

How to find the number of How to find the number of moles?moles?


Number of moles = consant Avogadroparticles of Number


Why defining 6.02 x 10Why defining 6.02 x 102323 as the as the amount for one mole?amount for one mole?

12 g carbon contains 6.02 1023 12C atoms

The mole is the amount of substance containing as many particles as the number

of atoms in 12 g of carbon-12.



Relative mass12

1

C atom

…….

6.02 1023

H atom

…….

6.02 1023

1 g

Molar mass

Molar mass is the mass, in grams, of 1 mole of a substance, e.g. the molar mass of H atom is 1 g.

Relative atomic masses



Molar mass is the same as the relative atomic mass in grams.

Molar mass is the same as the relative molecular mass in grams.

Molar mass is the same as the formula mass in grams.


Example 2-1AExample 2-1A Example 2-1BExample 2-1B Example 2-1CExample 2-1C

Example 2-1DExample 2-1D Example 2-1EExample 2-1E Check Point 2-1Check Point 2-1


2.2.22Molar Volume anMolar Volume and Avogadro’s Lad Avogadro’s La

ww


What is molar volume of What is molar volume of gases?gases?

at 25oC & 1 atm

(Room temp & pressure / R.T.P.)

2.2 Molar volume and Avogadro’s law (SB p.24)


Avogadro’s LawAvogadro’s Law2.2 Molar volume and Avogadro’s law (SB p.24)

Equal volumes of all gases at the same temperature and pressure contain the

same number of molecules.


same number of molecules.


same number of moles of gases.


same number of moles of gases.


Avogadro’s LawAvogadro’s Law


So 1 mole of gases should have the same volume at the same temperature and pressure.

V n where n is the no. of moles of gas V n where n is the no. of moles of gas



Interconversions involving Interconversions involving number of molesnumber of moles


Example 2-2DExample 2-2D Check Point 2-2Check Point 2-2


2.2.33 Ideal Gas Ideal Gas

EquationEquation


Boyle’s lawBoyle’s law

2.3 Ideal gas equation (SB p.27)

At constant temperature, the volume of a given mass of a gas is inversely proportional to the pressure exerted on it

PV = constant



Schematic diagrams explaining Boyle’s law



A graph of volume against the reciprocal of pressure for a gas at constant temperature


At a constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.


Charles’ lawCharles’ law



Schematic diagrams explaining Charles’ law



A graph of volume against absolute temperature for a gas at constant pressure


PV = nRTPV = nRT


Ideal gas equationIdeal gas equation

P1

Combining:

V n (Avogadro’s Law)

V (Boyle’s Law)

V T (Charles Law)

V

V = where R is a constant (called the universal gas constant)

PnT

PRnT


For one mole of an ideal gas at standard temperature and pressure,

P = 760 mmHg, 1 atm or 101 325 Nm-2 (Pa)

V = 22.4 dm3 mol-1 or 22.4 10-3 m3 mol-1

T = 0 oC or 273K




273K

1mol3m3-1022.42Nm 101325

By substituting the values of P, V and T in S.I. Units into the equation, the value of ideal gas constant can be found.

R =

=

= 8.314 J K-1mol-1

TPV



Relationship between the ideal gas equation and the individual gas laws



Example 2-3AExample 2-3A Example 2-3BExample 2-3B

Example 2-3CExample 2-3C Check Point 2-3Check Point 2-3


2.2.44DeterminatioDeterminatio

n of Molar n of Molar MassMass


Mass of volatile liquid injected

= 26.590 - 26.330 = 0.260 g

2.4 Determination of molar mass (SB p.32)


Volume of trichloromethane vapour

= 74.4 - 8.2 = 66.2 cm3



Temperature = 273 + 99 = 372 K



Pressure = 101325 Nm-2




PV = nRT

101325 Nm-2 66.2 10-6 m3

= n 8.314 J K-1 mol-1 372 K

n = 2.169 10-3 mol


Molar mass

=

=

= 119.87 g mol-1

ethanetrichlorom of moles of Numberethanetrichlorom of Mass

mol 3-102.169

g 0.260



PV = nRT………..(1)

n = ………..(2)

Where

m is the mass of the volatile substance

M is the molar mass of the volatile substance

Combing (1) and (2), we obtain

PV = RT

M =


Mm

Mm

PVmRT

Example 2-4AExample 2-4A

Example 2-4BExample 2-4B

Check Point 2-4Check Point 2-4


2.2.55 Dalton’s Law Dalton’s Law

of Partial of Partial PressuresPressures


In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the component gases (the sum that each gas would exert as if it is present alone under the same conditions).

In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the component gases (the sum that each gas would exert as if it is present alone under the same conditions).

PT = PA + PB + PC

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

2.5 Dalton’s law of partial pressures (SB p.35)


Consider a mixture of gases A, B and C occupying a volume V. It consists of nA, nB and nC moles of each gas.

The total number of moles of gases in the mixture

ntotal = nA + nB + nC

If the equation is multiplied by RT/V, then

ntotal (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)

i.e. Ptotal = PA + PB + PC

(so Dalton’s Law is a direct consequence of the Ideal Gas Equation)



Besides, the partial pressure of each component gas can be calculated from the Ideal gas law.

PA = nA(RT/V) and Ptotal = ntotal(RT/V)

i.e. PA= (nA/ntotal) Ptotal

PA = xA Ptotal


Check Point 2-5Check Point 2-5


Example 2-5DExample 2-5D


The END


What is the mass of 0.2 mol of calcium carbonate?


Back

The chemical formula of calcium carbonate is CaCO3.

Molar mass of calcium carbonate = (40.1 + 12.0 + 16.0 3) g mol-1

= 100.1 g mol-1

Mass of calcium carbonate = Number of moles Molar mass

= 0.2 mol 100.1 g mol-1

= 20.02 g

Answer


Calculate the number of gold atoms in a 20 g gold pendant.


Back

AnswerMolar mass of gold = 197.0 g mol-1

Number of moles =

= 0.1015 mol

Number of gold atoms

= 0.1015 mol 6.02 1023 mol-1

= 6.11 1022

1mol g 197.0g 20


It is given that the molar mass of water is 18.0 g mol-1.

(a)What is the mass of 4 moles of water molecules?

(b)How many molecules are there?

(c)How many atoms are there?


Answer



(a) Mass of water = Number of moles Molar mass

= 4 mol 18.0 g mol-1

= 72.0 g

(b) There are 4 moles of water molecules.

Number of water molecules

= Number of moles Avogadro constant

= 4 mol 6.02 1023 mol-1

= 2.408 1024


Back


(c) 1 water molecule has 3 atoms (i.e. 2 hydrogen atoms and 1 oxygen atom).

1 mole of water molecules has 3 moles of atoms.

Thus, 4 moles of water molecules have 12 moles of atoms.

Number of atoms = 12 mol 6.02 1023 mol-1

= 7.224 1024


A magnesium chloride solution contains 10 g of magnesium chloride solid.

(a) Calculate the number of moles of magnesium chloride in the solution.


Answer

(a) The chemical formula of magnesium chloride is MgCl2.

Molar mass of MgCl2 = (24.3 + 35.5 2) g mol-1 = 95.3 g mol-1

Number of moles of MgCl2 =

= 0.105 mol

1mol g 95.3g 10


(b) Calculate the number of magnesium ions in the solution.


Answer

(b) 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Cl- ions.

Therefore, 0.105 mol of MgCl2 contains 0.105 mol of Mg2+ ions.

Number of Mg2+ ions

= Number of moles of Mg2+ ions Avogadro constant

= 0.105 mol 6.02 1023 mol-1

= 6.321 1022


(c) Calculate the number of chloride ions in the solution.


Answer

(c) 0.105 mol of MgCl2 contains 0.21 mol of Cl- ions.

Number of Cl- ions

= Number of moles of Cl- ions Avogadro constant

= 0.21 mol 6.02 1023 mol-1

= 1.264 1023


(d) Calculate the total number of ions in the solution.


Answer(d) Total number of ions

= 6.321 1022 + 1.264 1023

= 1.896 1023

Back


What is the mass of a carbon dioxide molecule?


AnswerThe chemical formula of carbon dioxide is CO2.

Molar mass of CO2 = (12.0 + 16.0 2) g mol-1 = 44.0 g mol-1

Number of moles = =

=

Mass of a CO2 molecule =

= 7.31 10-23 g

constant Avogadromolecules of Number

mass MolarMass

1-

2

mol g 44.0

molecule CO a of Mass1-23 mol 10 6.02

1

1-23

-1

mol 10 6.02mol g 44.0

Back



(a)Find the mass in grams of 0.01 mol of zinc sulphide.

(a) Mass = No. of moles Molar mass

Mass of ZnS = 0.01 mol (65.4 + 32.1) g mol-1

= 0.01 mol 97.5 g mol-1

= 0.975 g

Answer



(b)Find the number of ions in 5.61 g of calcium oxide.

Answer

(b) No. of moles of CaO =

= 0.1 mol

1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.

No. of moles of ions = 0.1 mol 2

= 0.2 mol

No. of ions = 0.2 mol 6.02 1023 mol-1

= 1.204 1023

1mol g 16.0)(40.1g 5.61



(c) Find the number of atoms in 32.05 g of sulphur dioxide.

Answer(c) Number of moles of SO2 =

= 0.5 mol

1 SO2 molecule contains 1 S atom and 2 O atoms.

No. of moles of atoms = 0.5 mol 3

= 1.5 mol

No. of atoms = 1.5 mol 6.02 1023 mol-1

= 9.03 1023

1-mol g 2) 16.0 2.13(g 32.05



(d)There is 4.80 g of ammonium carbonate. Find the

(i) number of moles of the compound,

(ii) number of moles of ammonium ions,

(iii) number of moles of carbonate ions,

(iv) number of moles of hydrogen atoms, and

(v) number of hydrogen atoms. Answer



(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1

(i) No. of moles of (NH4)2CO3 = = 0.05 mol

(ii) 1 mole (NH4)2CO3 gives 2 moles of NH4+ ions.

No. of moles of NH4+ ions = 0.05 mol 2 = 0.1 mol

(iii) 1 mole (NH4)2CO3 gives 1 mole of CO32- ions.

No. of moles CO32- ions = 0.05 mol

(iv) 1 (NH4)2CO3 formula unit contains 8 H atoms.

No. of moles of H atoms = 0.05 mol 8

= 0.4 mol

(v) No. of H atoms = 0.4 mol 6.02 1023 mol-1 = 2.408 1023

1mol g 96.0g 4.80

Back


What is the difference between a theory and a law?

Back


A law tells what happens under a given set of circumstances while a theory attempts to explain why that behaviour occurs.

Answer


Find the volume occupied by 3.55 g of chlorine gas at room temperature and pressure.

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)Answer

Molar mass of chlorine gas (Cl2) = (35.5 2) g mol-1 = 71.0 g mol-1

Number of moles of Cl2 =

= 0.05 mol

Volume of Cl2 = Number of moles of Cl2 Molar volume

= 0.05 mol 24.0 dm3 mol-1

= 1.2 dm3

1mol g 71.0g 3.55


Back


Find the number of molecules in 4.48 cm3 of carbon dioxide gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogadro constant = 6.02 1023 mol-1)

Answer

Molar volume of carbon dioxide at S.T.P. = 22.4 dm3 mol-1

= 22400 cm3 mol-1

Number of moles of CO2 =

= 2 10-4 mol

Number of CO2 molecules = 2 10-4 mol 6.02 1023 mol-1

= 1.204 1020

13

3

mol cm 22400cm 4.48


Back


The molar volume of nitrogen gas is found to be 24.0 dm3 mol-1 at room temperature and pressure. Find the density of nitrogen gas.

Answer


Back

Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1

Density = =

Density of N2

=

=1.167 g dm-3

VolumeMass

volume Molarmass Molar

1-3

-1

mol dm 24.0mol g 28.0


1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the relative molecular mass of the gas?

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)Answer


Number of moles of the gas =

= 0.05 mol

Molar mass of the gas =

= 32 g mol-1

Relative molecular mass of the gas = 32 (no unit)

13

3

mol dm 24.0dm 1.2

mol 0.05g 1.6

Back



(a)Find the volume occupied by 0.6 g of hydrogen gas at room temperature and pressure.

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)

Answer

(a) No. of moles of H2 = = 0.3 mol

Volume = No. of moles Molar volume

= 0.3 mol 24.0 dm3 mol-1

= 7.2 dm3

1mol g 2)(1.0g 0.6



(b) Calculate the number of molecules in 4.48 dm3 of hydrogen gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1)Answer

(b) No. of moles of H2 = = 0.2 mol

No. of H2 molecules = 0.2 mol 6.02 1023 mol-1

= 1.204 1023

13

3

mol dm 2.42dm 4.48



(c)The molar volume of oxygen gas is 22.4 dm3 mol-1 at standard temperature and pressure. Find the density of oxygen gas in g cm-3 at S.T.P.

Answer

(c) Density = =

Molar mass of O2 = (16.0 2) g mol-1 = 32.0 g mol-1

Molar volume of O2 = 22.4 dm3 mol-1 = 22400 cm3 mol-1

Density =

= 1.43 10-3 g cm-3

VolumeMass

volume Molarmass Molar

13

-1

mol cm 22400mol g 32.0



(d) What mass of oxygen has the same number of moles as that in 3.2 g of sulphur dioxide?

Answer

(d) No. of moles of SO2 =

No. of moles of O2 = 0.05 mol

Mass = No. of moles Molar mass

Mass of O2 = 0.05 mol (16.0 2) g mol-1

= 1.6 g

1mol g 2)16.0(32.1g 3.2

Back



A 500 cm3 sample of a gas in a sealed container at 700 mmHg and 25 oC is heater to 100 oC. What is the final pressure of the gas?

Answer

As the number of moles of the gas is fixed, should be a constant.

=

=

P2 = 876.17 mmHg

The final pressure of the gas at 100 oC is 876.17 mmHg.

Note: All temperature values used in gas laws are on the Kelvin scale.

TPV

1

11

T

VP

2

22

T

VP

K 25)(273cm 500 mmHg 700 3

K )001(273

cm 500 P 3

2

Back



A reaction vessel of 500 cm3 is filled with oxygen gas at 25 oC and the final pressure exerted on it is 101 325 Nm-2. How many moles of oxygen gas are there?

(Ideal gas constant = 8.314 J K-1 mol-1)Answer

PV = nRT

101325 Nm-2 500 10-6 m3 = n 8.314 J K-1 mol-1 (273 + 25) K

n = 0.02 mol

There is 0.02 mol of oxygen gas in the reaction vessel.

Back



A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles of nitrogen gas are pumped into the vessel, what is the highest temperature it can be safely heated to?

Answer

Back

Applying the equation,

T = = = 1523.4 K

The highest temperature it can be safely heated to is 1250.4 oC.

1-1-

3-3-2

mol K J 8.314 mol 2m 105 Nm 10132550

nRPV



(a)State Boyle’s law, Charles’ law and ideal gas equation.

Answer(a) Boyle’s law states that at constant temperature, the volume of a giv

en mass of a gas is inversely proportional to the pressure exerted on it.

Charles’ law states that at constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.

The ideal gas equation, PV = nRT, shows the relationship between the pressure, volume, temperature and number of moles of a gas.



(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to? Answer

(b)

=

=

T2 = 586 K

1

1

T

P

2

2

T

P

K 20) (273atm 5 2T

atm 10



(c) A balloon is filled with helium at 25 oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon?Answer

(c) PV = nRT

1.5 101325 Nm-2 450 10-6 m3

= n 8.314 J K-1 mol-1 (273 + 25) K

n = 0.0276 mol



(d)25.8 cm3 sample of a gas has a pressure of 690 mmHg and a temperature of 17 oC. What is the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K?

(1 atm = 760 mmHg)Answer

(d)

V2 = 15.06 cm3

2

22

1

11

T

VP

T

VP

K 345

V atm 1.85

K 17) (273

cm 25.8atm 760690

2

3

Back



A sample of gas occupying a volume of 50 cm3 at 1 atm and 25 oC is found to have a mass of 0.0286 g. Find the molar mass of the gas.

(Ideal gas constant = 8.314 J K-1 mol-1; 1 atm = 101325 Nm-2) Answer

Back

M = 13.99 g mol-1

Therefore, the molar mass of the gas is 13.99 g mol-1.

RTMm

PV

K 25) (273 mol K J 8.314M

g 0.0286m 10 50 Nm 101325 11362-



The density of a gas at 450 oC and 380 mmHg is 0.0337 g dm-3. What is its molar mass? (1 atm = 760 mmHg = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer

Back

The unit of density of the gas has to be converted to g m-3 for the calculation.

0.0337 g dm-3 = 0.0337 103 g m-3 = 33.7 g m-3

PM = RT

M =

= = 4.0 g mol-1

Therefore, the molar mass of the gas is 4.0 g mol-1.

PRT

2

-1-1-3

Nm 101325760380

K 450) (273 mol K J 8.314 m g 33.7


(a)0.204 g of phosphorus vapour occupies a volume of 81.0 cm3 at 327 oC and 1 atm. Determine the molar mass of phosphorus.

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer


(a) PV = RT

101325 Nm-2 81.0 10-6 m3

= 8.314 J K-1 mol-1 (273 + 327) K

M = 123.99 g mol-1

The molar mass of phosphorus is 123.99 g mol-1.

Mm

Mg 0.204


(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3 measured at 97 oC and 1.62 atm. Calculate the molar mass of the gas.

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer


(b) PV = RT

1.62 101325 Nm-2 4.16 10-3 m3

= 8.314 J K-1 mol-1 (273 + 97) K

M = 54.06 g mol-1

The molar mass of the gas is 54.06 g mol-1.

Mm

Mg 12.0


(c) A sample of 0.037 g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium.

(1 atm = 760 mmHg = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer




(c) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

PV = nRT

101325 Nm-2 38.2 10-6 m3

= n 8.314 J K-1 mol-1 (273 + 25) K

n = 1.52 10-3 mol

No. of moles of H2 produced = 1.52 10-3 mol

No. of moles of Mg reacted = No. of moles of H2 produced

= 1.52 10-3 mol

Molar mass of Mg = = = 24.34 g mol-1

The relative atomic mass of Mg is 24.34.

moles of No.Mass

mol 10 1.52g 0.037

3-

760740

Back


Air is composed of 80 % nitrogen and 20 % oxygen by volume. What are the partial pressures of nitrogen and oxygen in air at a pressure of 1 atm and a temperature of 25 oC?

Answer

Back2.5 Dalton’s law of partial pressures (SB p.3

6)

Mole fraction of N2 =

Mole fraction of O2 =

Partial pressure of N2 =

= 81060 Nm-2

Partial pressure of O2 =

= 20265 Nm-2

10080

10020

2Nm 10132510080

2Nm 10132510020


The valve between a 5 dm3 vessel containing gas A at a pressure of 15 atm and a 10 dm3 vessel containing gas B at a pressure of 12 atm is opened.

(a) Assuming that the temperature of the system remains constant, what is the final pressure of the system?

(b) What are the mole fractions of gas A and gas B?

Answer




(a) Total volume of the system = (5 + 10) dm3 = 15 dm

By Boyle’s law, P1V1 = P2V2

Partial pressure of gas A (PA)

=

= 5 atm

Partial pressure of gas B (PB)

=

= 8 atm

By Dalton’s law of partial pressures, Ptotal = PA + PB

Final pressure of the system = (5 + 8) atm = 13 atm

3

3

dm 15dm 5 atm 15

3

3

dm 15dm 10 atm 12



(b) Mole fraction of gas A =

=

= 0.385

Mole fraction of gas B =

=

= 0.615

total

A

P

P

atm 13atm 5

total

B

P

P

atm 13atm 8

Back


0.25 mol of nitrogen and 0.30 mol of oxygen are introduced into a vessel of 12 dm3 at 50 oC. Calculate the partial pressures of nitrogen and oxygen and hence the total pressure exerted by the gases.

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1) Answer




Let the partial pressure of nitrogen be PA.

Using the ideal gas equation PV = nRT,

PA 12 10-3 m3 = 0.25 mol 8.314 J K-1 mol-1 (273 + 50) K

PA = 55946 Nm-2 (or 0.552 atm)

Let the partial pressure of oxygen be PB.


PB 12 10-3 m3 = 0.30 mol 8.314 J K-1 mol-1 (273 + 50) K

PB = 67136 Nm-2 (or 0.663 atm)



Total pressure of gases

= (55946 + 67136) Nm-2

= 123082 Nm-2

Or

Total pressure of gases

= (0.552 + 0.663) atm

= 1.215 atm

Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and 0.663 atm respectively, and the total pressure exerted by the gases is 1.215 atm.

Back


4.0 g of oxygen and 6.0 g of nitrogen are introduced into a 5 dm3 vessel at 27 oC.

(a) What are the mole fraction of oxygen and nitrogen in the gas mixture?

(b) What is the final pressure of the system?





(a) Number of moles of oxygen =

= 0.125 mol

Number of moles of nitrogen =

= 0.214 mol

Total number of moles of gases = (0.125 + 0.214) mol

= 0.339 mol

Mole fraction of oxygen = = 0.369

Mole fraction of nitrogen = = 0.631

1gmol 32.0g 4.0

1gmol 28.0g 6.0

mol 0.339mol 0.125

mol 0.339mol 0.214



(b) Let P be the final pressure of the system.


P 5 10-3 m3 = 0.339 mol 8.314 J K-1 mol-1 (273 + 27) K

P = 169 107 Nm-2 (or 1.67 atm)

Back



(a)Define mole fraction.

(b)State Dalton’s law of partial pressures.

Answer(a) Mole fraction of a substance is the ratio of the number of moles of

that substance to the total number of moles in the mixture.

(b) Dalton’s law of partial pressures states that in a mixture of gases which do not react with each other, the total pressure exerted is the sum of the pressure that each gas would exert as if it it present alone under the same conditions.

Ptotal = PA + PB + PC + … …



(c)The valve between a 6 dm3 vessel containing gas A at a pressure of 7 atm and an 8 dm3 vessel containing gas B at a pressure of 9 atm is opened. Assuming that the temperature of the system remains constant and there is no reaction between the gases, what is the final pressure of the system?

Answer



(c) By Boyle’s law: P1V1 = P2V2

Partial pressure of gas A =

= 3.00 atm

Partial pressure of gas B =

= 5.14 atm

By Dalton’s law: Ptotal = PA + PB

Final pressure of the system = 3.00 atm + 5.14 atm = 8.14 atm

3

3

dm 8) (6dm 6

atm 7

3

3

dm 8) (6dm 8

atm 9



(d)2 g of helium, 3 g of nitrogen and 4 g of argon are introduced into a 15 dm3 vessel at 100 oC.

(i) What are the mole fractions of helium, nitrogen and argon in the system?

(ii) Calculate the total pressure of the system, and hence the partial pressures of helium, nitrogen and argon.




(d) (i) No. of moles of He =

= 0.50 mol

No. of moles of N2 =

= 0.11 mol

No. of moles of Ar =

= 0.10 mol

Total no. of moles of gases = 0.50 mol + 0.11 mol + 0.10 mol

= 0.71 mol

1-mol g 4.0g 2

1-mol g 2) (14.0g 3

1-mol g 39.9g 4



Mole fraction of He = = 0.704

Mole fraction of N2 = = 0.155

Mole fraction of Ar = = 0.141

(ii) Let the total pressure of the system be P.

PV = nRT

P 15 10-3 m3 = 0.71 mol 8.314 J K-1 mol-1 (273 + 100) K

P = 146786 Nm-2

Partial pressure of He = 146786 Nm-2 0.704 = 103337 Nm-2

Partial pressure of N2 = 146786 Nm-2 0.155 = 22752 Nm-2

Partial pressure of Ar = 146786 Nm-2 0.141 = 20697 Nm-2

mol 0.71mol 0.50

mol 0.71mol 0.11

mol 0.71mol 0.10

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