Exercises

1. Physical behaviour of the states differs greatly even though chemical behaviour of the three states is identical because:

a.      Chemical properties depend on intermolecular non-bonding forces whereas physical properties depend on intermolecular bonding forces

b.     Chemical properties depend on non-bonding forces whereas physical properties depend on intermolecular bonding forces

c.      Chemical properties depend on intramolecular bonding forces whereas physical properties depend on intermolecular bonding forces

d.     Chemical properties depend on covalent bonding forces whereas physical properties depend on energy supplied

Solution: Chemical properties depend on intramolecular bonding forces whereas physical properties depend on intermolecular bonding forces.

Physical properties of an element depends on the intermolecular forces of that element which in turn changes on variations in temperature, pressure and volume, whereas, chemical properties depend upon intramolecular forces which remain the same for that element until another element is introduced.

 

2. Dipole-dipole forces act between molecules that have:

a.      An induced electrical dipole moment

b.     A permanent electrical dipole moment

c.      An electron cloud that causes temporary dipoles

d.     A permanent magnetic dipole moment

Solution: A permanent electrical dipole moment

Dipole moments depend upon the electron charge of the molecules.

 

3. Which of the following statements about Hydrogen bond incorrect?

a.      In hydrogen bonding H atom becomes partially positive and is attracted to the more negative O atom

b.     In hydrogen bonding H atom becomes partially positive and is attracted to the more negative F atom

c.      In hydrogen bonding H atom becomes partially negative and is attracted to the more positive N atom

d.     In hydrogen bonding H atom becomes partially positive and is attracted to the more negative N atom

Solution: In hydrogen bonding H atom becomes partially negative and is attracted to the more positive N atom

In hydrogen bonding the H atom becomes partially positive in order to create a hydrogen bond.

 

4. The intermolecular force primarily responsible for the condensed states of nonpolar substances is the:

a.      Dispersion force (or London force)

b.     Dipole – dipole force

c.      Hydrogen bond

d.     Ion-dipole forces

Solution: Dispersion force (or London force)

London or dispersion forces create temporary dipole moment for nonpolar molecules to form intermolecular bonds.

 

5. The average Kinetic energy and Thermal energy are proportional to the:

a.      Pressure

b.     Volume

c.      Specific heat

d.     Absolute temperature

Solution: Absolute temperature

            As increase or decrease in temperature leads to change in kinetic and thermal energy hence it is directly proportional.

 

6. Three states of matter are the result of:

a.      Balance between intermolecular forces and the thermal energy of the molecules

b.     Balance between Van der Waals forces and the London forces

c.      Balance between intermolecular forces and the intramolecular bonding energy of the molecules

d.     Balance between Hydrogen bonding forces and the London forces

Solution: Balance between intermolecular forces and the thermal energy of the molecules

The intermolecular forces between the molecules of matter change due to application of thermal energy from solid to liquid, and liquid to gas.

 

7. Gases have much lower density than the solids and liquids because:

a.      Gases contain lower specific enthalpy than solids

b.     Thermal energy overcomes the the attractive intermolecular forces

c.      Distance between the molecules is small compared to solids

d.     Gases do not have intermolecular forces that attract each molecule to another

Solution: Thermal energy overcomes the the attractive intermolecular forces

As thermal energy is applied to a gas it pushes the molecules further apart and reduces the intermolecular force between them, thus decreasing the density of the gas.

 

8. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?

a.      Decrease

b.     No regular pattern will be followed

c.      No effect

d.     Increase

Solution: Decrease

Intermolecular force of liquid decreases with increase in temperature, hence viscosity of liquid also decreases. However, some exceptions are there like liquid proteins and liquid sulphur.

 

9. Behaviour of the gas becomes more ideal when:

a.      Pressure is very low

b.     Pressure is very high

c.      Number of moles is very low

d.     Temperature is very low

Solution: Pressure is very low

            The real gases show ideal behaviour when pressure approaches zero.

 

10. Critical temperature (TC) of carbon dioxide is the highest temperature at which liquid carbon dioxide is observed. Above this temperature:

a.      It is gas

b.     It is liquid

c.      It is mixture of solid and liquid

d.     It is solid

Solution: It is gas

Liquid carbon dioxide first appears at 31 ͦ C, which is called critical temperature Tc.          

Above TC carbon dioxide is a gas and below it, it is a liquid.

 

11. Above the critical temperature (TC) of carbon dioxide:

a.      Solidification of the gas is possible

b.     Liquefaction of the gas is possible

c.      Increase of pressure makes

d.     Liquefaction of the gas is not possible

Solution: Liquefaction of the gas is not possible

Critical temperature of a gas is highest temperature at which liquefaction of the gas first occurs and for further liquefaction the gas has to be cooled.      

                 

12. The three states of matter of H2O are in equilibrium at:

a.      Steam point

b.     Ice point

c.      Critical point

d.     Triple point

Solution: Triple point

Triple point is the point on a phase diagram at which the three states of matter: gas, liquid, and solid coexist or are in equilibrium.

 

13. Surface tension decreases as the temperature is raised because:

a.      Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction increases

b.     Increase in temperature decreases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases

c.      Increase in temperature decreases the kinetic energy of the molecules and effectiveness of intermolecular attraction is not changed

d.     Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases

Solution: Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases

The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

 

14. Water has high surface tension and high capillarity because of:

a.      Ionic bonds

b.     Hydrogen bonds

c.      Dispersion forces

d.     Covalent bonds

Solution: Hydrogen bonds

            Hydrogen bonds are very strong. As the attractive forces are large, the surface tension and capillarity of water is high.

 

15. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 ͦ C?

Solution:

We have,

p1 = 1 bar

p2 = ?

V1 = 500 dm3

V2 = 200 dm3

  T = is constant at 30  ͦC

According to Boyle’s law,

           =

p2 =

     =

     = 2.5 bars

 

16. A vessel of 120 mL capacity contains a certain amount of gas at 35 ͦ C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 ͦ C. What would be its pressure?

Solution:

We have,

p1 = 1.2 bar

p2 = ?

V1 = 120 mL

V2 = 180 mL

 T = is constant at 35  ͦC

According to Boyle’s law,

 =

p2 =

     =

     = 0.8 bars

 

17. At 0 ͦ C the density of a certain oxide of a gas at 2 bar is same as that of di-nitrogen at 5 bar. What is the molecular mass of the oxide?

Solution:

We have,

       p1 = 2 bar

       p2 = 5 bar

        M = Relative atomic mass × Molar mass constant (1 g/mol)

Molar mass of dinitrogen

      M1 = 2 × 14 × 1

 = 28 g/mol

We know that

         d =  

For dinitrogen,

       d1 =  

For unknown oxide,

       d2 =

Since

       d1 = d2

 =

   140 = 2 × M

       M =

            = 70 g/mol

 

18. Pressure of 1 g of an ideal gas A at 27 ͦ C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Relationship between their molecular masses is:

Solution:

We have,

pA = 2 bar

      pTotal = 3 bar

pB = 3 – 2

      = 1 bar

           mA = 1 g       

           mB = 2 g

Substituting in equation,

           pV = nRT

           pV =  RT

              p =  

              p =  

             2 =                             ------ (1)

 1 =                             ------ (2)

Equating (1) and (2),

   =

   =

Therefore we get,

          MB = 4MA

 

19. The drain cleaner, drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20  ͦC and one bar will be released when 0.15 g of aluminium reacts?

Solution:

            Aluminium in drainex reacts with caustic soda to produce dihydrogen.

            2Al + 2NaOH + 2H2O     →   2NaAlO2 + 3H2

            2 moles of aluminium   →   3 moles of dihydrogen        

            ⇒ 2 × 27 g = 54 g                    ⇒ 3 × 22.40 L

            Therefore,

                 =  

                 = 0.1867 L

                 = 187 ml

            We have,

            p1 = 1 bar

            p2 = 1 bar

            T1 = 273 K

            T2 = 20 + 273

     = 293 K

            V1 = 187 ml

            V2 = ? ml

            We know that,

           V2 =

                 =

                 = 200.699 ml

                 = 201 ml approximately 202.5 ml

           

20. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27  ͦ C?

Solution:

            We know that,

            Mass of carbon dioxide              m1 = 4.4 g

            Molar mass of carbon dioxide M1 = 44 g/mol

            Mass of methane                          m2 = 3.2 g

            Molar mass of methane             M2 = 16 g/mol

            Now,

            Amount of carbon dioxide        n1 =

                                                                            = 0.1 mole

            Amount of methane                    n2 =

                                                                            = 0.2 mole

            Therefore total moles                   n = n1 + n2

                     = 0.1 + 0.2

                  = 0.3 mole­

            Ideal gas equation is,

            pV = nRT

            We know that,

              R = 0.083 bar dm3 K−1 mol−1

  T = 27  ͦC

      = 27 + 273

      = 300 K

   V = 9 dm3

   p =

      =       

      = 0.83 bar

      = 0.83 × 105 Pa (1 bar = 105 Pa)

      = 8.3 × 104 Pa

The pressure exerted by both gases is 8.3 × 104 Pa.

 

21. What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of O2 at 0.7 bar are introduced in a 1 L vessel at 27 ͦ C?

Solution:

We have,

V1 = 0.5 L

V2 = 2.0 L

p1 = 0.8 bar

p2 = 0.7 bar

We know that,

           pV = nRT

So for H2,

          0.8 × 0.5 = n1RT

n1 =

So for O2,

          0.7 × 2.0 = n2RT

            n2 =

Using Dalton’s law of partial pressures,

When the mixture is put in a 1 L vessel,

               p × 1 L = (n1 + n2) × RT

  p =  × RT

  p = 1.8 bars

The pressure of the mixture of gases is 1.8 bars.

 

22. Density of a gas is found to be 5.46 g/dm3 at 27 ͦ C at 2 bar pressure. What will be its density at STP?

Solution:

We have,

       d1 = 5.46 g/dm3

       T1 = 27  ͦC

= 27 + 273 K

= 300 K

= 2 bars

At standard temperature and pressure,

       d2 = ?

       T2 = 0  ͦC

 = 273 K

        p2 = 1 bar

Using Molar mass formula from ideal gas equation,

       M =

Therefore,

 =

         = 

       d2 =

= 3 g/mol

At standard temperature and pressure density is 3 g/mol.

 

23. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 ͦ C and 0.1 bar pressure. What is the molar mass of phosphorus

Solution:

            We have,

            V         = 34.05 mL

                        = 0.03405 L

      T         = 546  ͦC

                        = 546 + 273 K

                        = 819 K

            p          = 0.1 bars

            m        = 0.0625 g

            R         = 0.0821 L atm K-1 mol -1

            From ideal gas equation we can find molar mass,

            M        =

                   = 

                        =

                        = 1234.212 mol-1

            The molar mass of Phosphorous is 1234.212 mol-1.

 

24. Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K−1mol−1).

Solution:

            We have,

                                    V         = 5 dm3

                                    p         = 3.32 bar

                                    R         = 0.083 bar dm3 K−1 mol−1

                                    n          = 4 mol

            Using ideal gas equation,

                                    pV       = nRT

                                     T        =

                                                =

                                                = 50 K

The temperature of the gas is 50 K.

 

25. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Solution:

            We know that,      

            Mass of N2                m       = 1.4 g

            Molar mass of N2    M      = 28 g/mol

            Moles of N2                n       =

                                                            = 0.05 mol

            The number of nitrogen molecules in 0.05 mol of N2,

                                                            = 0.05 × 6.023 × 1023    

                                                                        = 0.30115 × 1023

                                                            = 3.0115 × 1022

            Number of electrons in 1 molecule of N2 = 14 electrons

            Therefore,

                                                            = 3.0115 × 1022 × 14

                                                            = 4.22 × 1023  electrons

The number of electrons in 1.4 g of dinitrogen is 4.22 × 1023.

 

26. The Van der Waals Equation adjusts the measured volume:

a.      Up by subtracting a factor from the entire container volume that accounts for the molecular volume

b.     Down by adding a factor from the entire container volume that accounts for the molecular volume

c.      Down by subtracting a factor from the entire container volume that accounts for the molecular volume

d.     Up by subtracting a factor from the entire container volume that accounts for the container volume

Solution: Down by subtracting a factor from the entire container volume that accounts for the molecular volume

Van der Waals proposed that we correct the volume   of a real gas as it is too large at high pressures by subtracting a term from the volume of the real gas before we substitute it into the ideal gas equation. He therefore introduced a constant constant (b) into the ideal gas equation that was equal to the volume actually occupied by a mole of gas particles.

 

27. The Van der Waals Equation adjusts the measured pressure:

a.      Up by adding a factor that accounts for intermolecular attractions

b.     Up by subtracting a factor that accounts for intermolecular attractions

c.      Down by adding a factor that accounts for intermolecular attractions

d.     Up by adding a factor that accounts for intramolecular attractions

Solution: Up by adding a factor that accounts for intermolecular attractions

Van der Waals added a term to the pressure as it was too small. This term contained a second constant (a) and has the form: .

 

28. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27  ͦC. (Density of air = 1.2 kg m−3 and R = 0.083 bar dm3 K−1 mol−1).

Solution:

            Payload = mass of displaced air − mass of balloon

            We have,

            Radius of balloon               = 10 m

            Mass of balloon                 = 100 kg

            Volume of balloon/sphere

 =

                                                            =

                                                            = 4188.79 m3

Density of air                       = 1.2 kg m-3

            Mass of air displaced by the balloon,

                                                            = 1.2 ⨯ 4188.79

                                                            = 5026.548 kg

            Ideal gas equation is,

                   pV = nRT

            We know that,

                                                     R    = 0.083 bar dm3 K−1 mol−1

                                               T    = 27  ͦC

                                                           = 27 + 273

= 300 K

            Molar mass of Helium    = 4 g/mol

                                                      N =

                                         pV = RT

                                          m =

                                          m =           

                                                          = 1117.010 kg

            Total mass of balloon     = 1117.010 + 100

                                                          = 1217.010 kg

            Payload of balloon          = mass of displaced air – mass of balloon

                                                          = 5026.548 – 1217.010

                                                          = 3809.538 kg

The payload of the balloon is 3809.538 kg.

 

29. Calculate the volume occupied by 8.8 g of CO2 at 31.1 ͦ C and 1 bar pressure. R = 0.083 bar dm3 K−1 mol−1.

Solution:

            We have,

            Mass of CO2 ,           m        = 8.8 g

                                          T         = 31.1 ͦ C

                                                            = 31.1 + 273

                                                            = 304.1 K

                                                p          = 1 bar

                                                R         = 0.083 bar dm3 K−1 mol−1

            Molar mass of CO2, M      = 44 g/mol

            Using ideal gas equation,

                pV   = nRT

                   n   =

            Therefore,

                pV   = RT

                  V   =

                        =

                        = 5.04806 L

The volume of CO2 is 5.04806 L.

 

30. 2.9 g of a gas at 95 ͦ C occupied the same volume as 0.184 g of dihydrogen at 17 ͦ C, at the same pressure. What is the molar mass of the gas?

Solution:

            We have,

            For gas 1,

            m1       = 2.9 g

      T1        = 95 ͦ C

                        = 95 + 273

                        = 368 K

            m2       = 0.184 g

      T2        = 17 ͦ C

                        = 17 + 273

                        = 290 K

            p1        = p2

            V1        = V2

            R         = 0.083 bar dm3 K−1 mol−1

            Molar mass of Dihydrogen,

M2       = 2.01588 g/mol

            Using ideal gas equation,

            p1V1    = n1RT1

            p2V2    = n2RT2

                   n2 =

                        =

                        = 0.0913 moles

              V2       =

                         =

              V2       =  L

            Since p1 = p2 and V1 = V2

              V1       =  L

            We can find n1 by substituting V1

                        p1V1    = n1RT1

p1 ×         = n1 × 0.083 × 368

               n1     =

               n1     = 0.072 moles    

                  =

                M1   =

                   =

                        = 40.3 g/mol

The payload of the balloon is 40.3 g/mol.

 

31. A mixture of dihydrogen and dioxygen at one bar pressure contains 20 % by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Solution:

            We know from Dalton’s law of partial pressures,          

                   pi =  × pTotal

            where,

        =

            Assume total mass is 100 g,

            Mass of dihydrogen m1                         = 20 % of 100g

                                                                                  = 20 g

            Molar mass of dihydrogen M1             = 2 g/mol

            Number of moles of dihydrogen n1   =

                                                                                   =

                                                                                   = 10 moles

            Mass of dioxygen m2                               = 100 – 20

                                                                                   = 80 g

            Molar mass of dioxygen M2                    = 32 g/mol

            Number of moles of dioxygen n2          =

                                                                                    =

                                                                                    = 2.5 moles

            Total pressure of both gases ptotal       = 1 bar

            Total number of moles n                         = 10 + 2.5

                                                                                    = 12.5 moles

            Mole fraction of dihydrogen x1             =

                                                                                    =

                                                                                    = 0.8

            Partial pressure of dihydrogen p1        =  × ptotal

                                                                                    = 0.8 × 1

                                                                                    = 0.8 bar

The partial pressure of dihydrogen in the mixture is 0.8 bar.

 

32. The vapour pressure of different substances at the same temperature increases if:

a.      The molecules of the substance are covalent bonded

b.     The intermolecular forces of the substance are weak

c.      The intermolecular forces of the substance are strong

d.     The molecules of the substance are ionic bonded

Solution: The intermolecular forces of the substance are weak

As temperature increases, intermolecular forces become weak. Different substances have different boiling points and the vapour pressure increases as temperature increases.

 

33. A person living in ooty observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude:

a.      Pressure increases

b.     Temperature increases

c.      Pressure decreases

d.     Temperature decreases

Solution: Pressure decreases

When a liquid is heated, it will turn into vapour at a certain temperature. In an open container, these molecules will wander off (diffuse) into the room and out into the atmosphere. Eventually all the liquid will evaporate. Pressure decreases and no cooking takes place. But in a pressure cooker there will be repeated evaporation and condensation leading to pressure being maintained till a certain point. Hence cooking is easier in a pressure cooker.

 

34. Which of the following property of water can be used to explain the spherical shape of rain drops?

a.      Surface tension

b.     Critical phenomena

c.      Pressure

d.     Viscosity

Solution: Surface tension

Liquid drops will be spherical as a spherical shape has the minimum surface energy for a given volume.

 

35. The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon:

a.      Charge of interacting particles

b.     Polarisability of interacting particles

c.      Mass of interacting particles

d.      Strength of permanent dipoles in the particles

Solution: Polarisability of interacting particles

Large electron clouds get easily polarised. Increase in polarisation leads to an increase in the magnitude of the London force.

 

36. The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?

Solution:

Let the number of moles of dihydrogen and dioxygen be 1 and 4.

Mole fraction of O2            =

                                                            =

                                                            =

                                                            = 0.8

          Partial pressure of dioxygen

                                                            = Mole fraction × Total Pressure

                                                            = 0.8 × 1

                                                            = 0.8 atm

                                                            = 0.8 × 105 Nm−2

 = 8 × 104 Nm−2

The partial pressure of dioxygen is 8 × 104 Nm−2.

 

37. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?

a.      No regular pattern will be followed

b.     Increase

c.      No effect

d.     Decrease

Solution: Decrease

Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

 

38. With regard to the gaseous state of matter which of the following statements is correct?

a.      Complete order of molecules and fixed position of molecules

b.     Complete disorder of molecules and Random motion of molecules

c.      Fixed position of molecules and complete order of molecules

d.     Random motion of molecules and fixed position of molecules

Solution: Complete disorder of molecules and Random motion of molecules

In gaseous state, molecules are in a state of random motion, i.e., it is the state in which molecules are disorderly arranged. Gaseous state has higher entropy than the liquid as well as solid.

 

39. Under which of the following two conditions, a gas deviates most from the ideal behaviour?

a.      High temperature and Low pressure

b.     High pressure and Low temperature

c.      Low temperature only

d.     Low pressure only

Solution: High temperature and Low pressure

Real gases show ideal behaviour when temperature is high and pressure is low and the intermolecular forces are practically negligible. Thus to deviate from ideal behaviour temperature has to below and pressure high.

 

40. Heat of vaporization is always larger than the heat of fusion because:

a.      It takes more energy to separate particles completely than just to free them from their fixed positions in the solid

b.     It takes less energy to separate particles completely than just to free them from their fixed positions in the solid

c.      It takes more energy to separate particles completely than just to free them from their random positions in the solid

d.     It takes no energy to separate particles completely than just to free them from their oscillating positions in the solid

Solution: It takes more energy to separate particles completely than just to free them from their fixed positions in the solid

The energy required to completely separate the molecules, moving from liquid to gas, is much greater that if you were just to reduce their separation, solid to liquid. 

 

41. The ease with which the electron cloud of a particle can be distorted is called its:

Solution: Polarizability

Neutral nonpolar species have spherically symmetric arrangements of electrons in their electron clouds. When in the presence of an electric field, their electron clouds can be distorted. The ease of this distortion is defined as the polarizability of the atom or molecule.      

 

42. Capillarity results from a competition between the:

a.      Intramolecular forces within the liquid (cohesive forces) and those between the liquid and the tube walls (adhesive forces)

b.     Intermolecular forces within tubewall and intramolecular forces within tubewall

c.      Intermolecular forces within the liquid (cohesive forces) and those between the liquid and the tube walls (adhesive forces)

d.     Intermolecular forces within liquid and intramolecular forces within liquid

Solution: Intermolecular forces within the liquid (cohesive forces) and those between the liquid and the tube walls (adhesive forces)

The rise of water in a thin tube inserted in water is caused by forces of attraction between the molecules of water (intermolecular forces) and the glass walls and among the molecules of water themselves.

 

43. Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water:

CaC2 (s) + 2H2O (l) → C2H2 (g) + Ca(OH)2 (aq)

For a sample of acetylene collected over water, total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas (23 ͦ C), the vapour pressure of water is 21 torr. How many grams of acetylene are collected?

Solution:

            We have,

            P         = Total gas pressure 738 torr − Vapor pressure of the water 21 torr

                        = 717 torr of C2H2

            T         = 23 C + 273

                        = 296 K

            V         = 523 ml

                        = 0.523 L

            As per Ideal Gas equation

                                    pV       = nRT

                      717 × 0.523   = n × 62.36 × 296 K

                                     n         = 0.02032 moles of C2H2

            Molar mass of C2H2  M    = 26.04 g/mol

            Moles of C2H2            m     = n × M

                                                            = 0.02032 × 26.04

            Mass of N2                   m     = 0.529 g of C2H2

There are 0.529 g of C2H2.

 

44. A scale model of a blimp rises when it is filled with helium to a volume of 55.0 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.

Solution:

            We have,

            Initial readings of blimp,

                                    n          = 1.10 mol

                                    V         = 26.2 dm3

                                    p          = T = k

                                    R         = 0.083 bar dm3 K−1 mol−1

            Using ideal gas equation,

                                    pV       = nRT

                                              =

                                                =

                                                = 0.0035 atm/K

            Final readings of blimp,

                                    n          = ? mol

                                    V         = 55.0 – 26.2

                                                = 28.8 dm3

                                    p          = T = k

                                    R         = 0.083 bar dm3 K−1 mol−1

            Using ideal gas equation,

                                    pV       = nRT

                                    n          =  × k

                                                =  × 0.0035

                                                = 1.214 mol

            Mass of Helium

m         = 1.214 × 4 (Molar mass of He = 4 g/mol)

                                                = 4.856 g

The mass of Helium needed to fill up the blimp is 4.856 g.

 

45. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium?

Solution:

            The reaction is,

            2K + Cl2 → 2KCl

            We know that,

                                    T         = 293 K

                                    V         = 5.25 L

                                    p         = 0.950 atm

                                    m        = 17.0 g

                                    R         = 0.083 bar dm3 K−1 mol−1

            Using ideal gas equation,

                                    pV       = nRT

                                    n          =

                                                =

                                                = 0.205 mol

            Mass of Potassium chloride,

                                     m       = 0.205 × 74.55 (Molar mass of KCl = 74.55 g/mol)

                                                = 15.28275 g

            Since there are 2 moles,

                                     m       = 2 × 15.28275

                                                = 30.5655 g

The mass of KCl formed is 30.5655 g.

 

46. Which of the following is correct for critical temperature?

a.      At critical temperature the gas and the liquid phases have different critical densities

b.     It is the lowest temperature at which liquid and vapour can coexist

c.      Beyond the critical temperature, there is no distinction between the two phases and a gas cannot be liquefied by compression

d.     At critical temperature (Tc), the surface tension of the system is not zero

Solution: Beyond the critical temperature, there is no distinction between the two phases and a gas cannot be liquefied by compression

Critical temperature of a substance is a temperature at and above which gases can't be liquid fied, no matter how much pressure is applied.

 

47. If a gas is heated at constant pressure, its density

a.      Will decrease

b.     Will increase

c.      May increase or decrease

d.     Will remain unchanged

Solution: Will decrease

When gas is heated at constant pressure the molecules gain energy and thus disperseand amount of gas at a particular place decreases so density decreases.

 

48. One gram mole of a gas at NTP occupies 22.4 L as volume. This fact was derived from:

a.      Dalton's theory

b.     Avogadro's hypothesis

c.      Berzelius hypothesis

d.     Law of gaseous volumes

Solution: Will decrease

Avagadro gave a hypothesis stating that the volume of gas is directly proportional to the number of molecules of gas. As one moles of substance always contains same number of molecules i.e., 6.022 × 1023. So the volume is always 22.4 h at NTP. NTP is normal temperature.