CHAPTER 1 THE SOLID STATE
Q1. What FeO is non-stoichiometric with the formula Fe0.95O ?1M
Ans. FeO has some Fe2+ ionsconverting into Fe3 ions, thereby leading to cation deficiency
in FeO. Therefore, it is non-stoichiometric with the formula.
Q2. Name the salt that can be added to AgCl to produce cation vacancy?1M
Ans. Cadmium chloride can be added to AgCl to produce cation vacancy.
Q3.What is the difference between phosphorous doped and gallium doped semiconductor? 1M
Ans. Phosphorous doped semiconductor is n-type whereas gallium doped is p-type semiconductor.
Q4. Why is an ionic crystal of Nacl a nonconductor?1M
Ans. ionic crystal of Nacl a nonconductor because the ion is fixed in their position and are not mobile.
Q5.Gold (atomic radius=0.144nm) crystallites in an fcc unit cell. What is the length of the unit cell? 2M
Ans. For fcc
Or
Q6. A unit cell consist of a cube in which there are anion () at each corner and one at the body Centre of the unit cell. The cation () are present at each face center .How many cation and anion are present in then unit cell and what is the formula of the compound? 2M
Ans. Number of anions
Contribution at the corner
Contribution to the Centre of unit cell
Total no. of anions=2
Contribution at face Centre
Formula of the compound is A3B
Q7. In a crystalline solid, anion ’c’ is arranged in ccp. Cation a occupy 50 % of the tetrahedral voids and cation ‘B’ occupy 50% of the octahedral voids .what is the formula of the solid? 2M
Ans.let anion present in the solid=C
The tetrahedral voids would be double
50% are occupied by ‘A’
Number of ‘B’ present in the solid
The octahedral voids would be same
50% are occupied by ‘B’
Number of ‘B’ present in the solid
Formula of the compound = AB1/2C
=A2BC2
Q8. Calculate the approximate number of unit cell present in 1g of ideal NaCl crystals which crystalizes in an fcc arrangement? 2M
Ans. 1 mole of NaCl=58.5 g
58.5 g of NaCl contain6.023×1023 formula units.
1g of NaCl contains formula units
4 NaCl formula units are present in one unit of NaCl.
Q9.ZnO is white but it turns yellow on heating and an exhibit enhances electrical conductivity. Explain. 2M
Ans. ZnO on heating loses oxygen irreversibly according to the equation:
ZnO(s) Zn2+ (aq) +O2(g) + 2e-
The Zn2+ ions are entrapped in the interstitial voids and electrons in neighboring interstitial voids to maintain electrical neutrality. Due to the presence of electrons, a yellow colour is associated with ZnO. These electrons enhance the electrical conductivity of ZnO.
Q10.Noble gases and metals crystallise with closed packed structures, yet the meeting points of noble gas crystals are exceptionally low. Why? 1M
Ans. Noble gases crystallise in close packed structures, but the forces of interaction between the atoms are weak dispersion forces and they therefore have low melting points. On the other hand, metals have strong metallic bonds (electrostatic forces of interaction) and therefore have high melting points.
Q11. An element crystallises into a structure which has a cubic structure with one atom at each corner and two atoms at one of its diagonals. If the volume of the
Unit cell is 24×10-24 cm3 and density us 7.2 g cm-3, calculate the number of atoms in 200 g of the elements? 2M
Ans. Number of atoms per unit cell =8× + 2 =3.
Now ρ
Or N= = =3.47 ×1024
Number of atoms in 200g of the element =3.47×1024 .
Q12. Pure silicon is an insulator. Silicon doped with phosphorus is a semiconductor. Silicon doped with gallium is also a semiconductor what is the difference between the two semiconductors? 2M
Ans. In pure silicon all electrons are involved in bonds formation. The bond formed is strong and cannot be broken easily. Therefore, there are no electrons for conduction, and pure silicon is an insulator.
On doping
(i)With phosphorus which is pentavalent, four of its electrons are involved in bound formation with silicon and the fifth valence electron is free to conduct for conduct electricity. This type of semiconductor is called n-type semiconductor.
(ii)With gallium which is trivalent, three bounds are formed with neighboring Si atoms and one electron deficient bound is formed .It gives rise to a hole, which moves in the crystals and enhances the conduction. This type of semiconductor is called p-type semiconductor.
Q13 the compositions of a sample of wustile is Fe0.93O1.00. What percentage of iron is present in the form of Fe (III)? 2M
Ans. Let the number of Fe3+ = x
Then the number of Fe2+ =0.93-x
Since in a neutral species
Total positive charges = Total negative charges
3x +2(0.93-x) =2
Or 3x+1.86 – 2x =2
X=0.14
Fraction of Fe (III) =
Q14. An element of atomic mass of 40 gmol-1 occurs in fcc structure with cell edge of 540 pm .calculate the Avogadro number if its density is 1,7 gcm-3 ? 2M
Ans. For fcc, Z=4
N= =
=
=59.9 ×1023mol-1
Q15.Explain with suitable examples:3M
(i)Ferromagnetism (ii) Ferrimagnetism (iii) 12-16 and 13-15compound
Ans. Ferromagnetism: is shown by compounds by which the domains are aligned in the direction of the external magnetic field .They are strongly attracted by the magnet and retains magnetism even in
The absence of a magnetic field eg.,CrO2
Ferrimagnetism: is shown by compounds by which the domains are aligned in parallel and antiparallel direction but in unequal numbers. They have small magnetic momentum associated with them e.g. Fe3O4.
(i) 12-16 and 13-15compound are formed by the combination of group 12 and group16, group 13-15 elements. These substances have high electrical conductivity. eg.ZnS, GaAs.
CHAPTER 2-SOLUTIONS
1.Does the mixture of a gas always represent a solution?1M
Ans.Yes, since the mixture is homogeneous.
2.What is the molarity of an aqueous solution of NaOH containing 0.5g in 500cm3 of the solution? 1M
Ans. Molarity
3.What will happen to the boiling point of a solution on mixing two miscible liquids showing positive deviation from Raoults law? 1M
Ans.The boiling point will decrease as the vapour pressure of the solution will decrease.
4.What type of azeotrope will be formed upon mixing chloroform and acetone?1M
Ans.A maximum boiling azeotrope will be formed as the vapour pressure of the solution will decrease.
5.On mixing 20 ml of ‘A’ and 20 ml of ‘B’ heat is liberated during the process. Explain.
1M
Ans.Liberation of heat during the process indicates the formation of stronger bonds than presentbetween the reactants and therefore, the components on mixing show negative deviation from ideal behavior.
6.Two liquids A and B boil at 80°C and 125°C respectively.Which out of them has a higher vapour pressure? 1M
Ans.’A’ has higher vapour pressure.
7.What will be the Van’t Hoff factor for compound which undergo trimerization? 1M
Ans. Van’t Hoff factor i=
8.Out of 0.1 m sucrose solution and 0.1 m KCl solution,which one will have a higher boiling point? 1M
Ans.0.1 m KCl solution will have a higher boiling point than a solution of 0.1 m sucrose as KCl dissociates in solutions.
9.You are provided with one litre each of 0.5 M NaOH and 0.25 M NaOH.What maximum volume of 0.3 M NaOH can be obtained from these solutions without using water at all? 3M
Ans.Volume of 0.25M NaOH used = 1L
Volume of 0.5M NaOH =
Total volume of NaOH = (1
Molarity of solution
M =
M = 0.3 M
M1 =- 0.25 , V2= 1 L
M2 = 0.5 M , V2 = L
0.3 =
0.3 (1 + x) = 0.25 + 0.5x
0.3 + 0.3x = 0.25 + 0.5x
x = = 0.25 L
, maximum volume of 0.3 NaOH solution is 0.25 L
10. Explain why are aquatic species more comfortable in cold water rather than warm water . 2M
ans. Aquatic species get oxygen for their survival from water containing dissolved oxygen .
according to Henry’s constant = .
with increase in temperature , for the same pressure the value of increases and mole fraction of gas decreases . The solubility of oxygen in water thus decreases and hence causes discomfort to the aquatic species. The aquatic species are more comfortable in cold water as more oxygen is dissolved in water at a lower temperature.
11. A mixture of ethanol and acetone shows positive deviation from ideal behaviour whereas that of chloroform and acetone shows negative deviation.Explain 1M
Ans. A mixture of ethanol and acetone shows positive deviation from Raoult’s law because on adding acetone to ethanol breaks the intermolecular hydrogen bonds between ethanol and therefore the escaping tendency of the components increases thereby increasing the vapour pressure.
On the other hand, when chloroform and acetone are mixed, strong hydrogen bonds are formed between chloroform and acetone, thereby dercreasing the escaping tendency of the components and therefore lowering the vapour pressure of the solution.
12.The vapour pressure of water is 12.3 kPa at 300K. Calculate the vapour pressure of one molal solution containing a solute in it. 3M
Ans. 1 molal solution =
Number of moles of water =
Assuming = 1g/ml
Mass of
= = 55.5 mol
= = 0.0177
= P = 12.3 ( 1 – 0.0177)
= 12.3 × 0.9823 = 12.08 kPa
13. A solution is prepared by dissolving 8 g of BaCl2 in 100g of water raises the boiling point of water by 0.52 K. Calculate the percentage (%) dissociation of BaCl2.Kb for water = 0.52 K kg mol-1 3M
Ans. We know that,
=
0.5 =
= = 2.6
α =
α =
α = 0.80
% dissociation = 0.80 × 100 = 80%
14.0.90 g of non- electrolyte was dissolved in 87.9 g of benzene. This raised the boiling point of benzene by 0.25K. If the molecular mass of the non-electrolyte is103.0 g mol-1,Calculate the molar elevation constant for benzene. 2M
Ans. Kb=
Kb =
= = 2.52K kg mol-1
15.The following two graphs depict the conductivity change with temperature.Predict, giving reasons,the type of substance which they represent. 2M
temperature→
temperature→
Ans. Graph(a) represents a semiconductor as the conductivity increases with temperature.In semiconductors,on heating,the electrons gain energy and overcome the energy gap to jump from the valence band to the conduction band,showing enhanced conductivity.
Graph(b) represents metallic conductor,as in metallic conductors,on increasing the temperature the positively charged kernels start vibrating and therefore hinder the movement of the kernels.This leads to a decrease in conductivity of metallic conductors with increase in temperature.
CHAPTER 3 ELECTROCHEMISTRY
- The resistance of a 0.01 M NH4OH solution was found to be 3000 Ω in a conductivity cell with cell constant 0.345 cm-1 . Calculate
(i)the conductivity of NH4OH
(ii)the degree of dissociation of NH4OH
(iii)the dissociation constant of NH4OH
given: =73.4 S cm2 mol-1
= 197.6 S cm2 mol-1 3M
ans. (i) we know that , K=C G*
where G is the conductance of the cell and G* is cell constant
given , C = = Ω-1 ; G* = 0.345 cm-1
K = = 1.1510-4 S cm-1 .
(ii) degree of dissociation α =
=
= 73.4 197.6
= 271.0 S cm2 mol-1
now , = =
= 11.5 S cm2 mol-1
α = = 0.042
(iii) dissociation constant =
=
=
= 1.83 10-5
2. How does the molar conductivity of strong and weak electrolyte vary with its concentration in solution ? 2M
ans. For a strong electrolyte the molar conductance varies with increase in dilution. It increases linearly as the interionic interactions between the completely dissociated electrolyte decreases on dilution.
It is given by, Λm = ΛmA .
where A is a constant depending upon the type of the electrolyte. For a weak electrolyte, the degree of dissociation α increases with dilution according to
α =
(where k = dissociation constant, C = concentration and α = degree of dissociation )
since more ions are obtained upon dilution, therefore there is a steep rise in molar conductance for a weak electrolyte at infinite dilution.
3. What is a fuel cell ? Give the working of a fuel cell. 3M
ans: Fuel cells are those cells in which the energy produced from the combustion of fuels such as H2 , CO , CH4 etc. is directly converted into electrical energy.
Working of a fuel cell. Anode and cathode are graphite electrodes containing suitable NaOH of KOH is placed between the electrodes as electrolyte. H2 gas and O2 gas are bubbled through the porous electrodes into the electrolyte.
Reactions involved.
At anode : 2 (g) + 4(aq) → 4(l) + 4
At cathode :(g) + 2(l) + 4 → 4(aq)
Overall cell reaction : 2 (g) + (g) → 2(l)
4. For the electrode Pt, H2 (1 atm) | H+(aq) (x M), the reduction potential at 25 is -0.34 V. Write the electrode reaction equation and calculate the value of ‘x’. Also, calculate the pH of the solution. 3M
ans: The equation is :
H+(aq) + → (g)
Nernst equation for electrode potential is
= log
= 0 log
0.34 = log
= log
log x = = 0.1738
x = antilog (0.1738) = 1.49 M
5. What is corrosion ? What are the factors which affect corrosion ? Give the mechanism involved in rusting. 3M
ans: The process of slowly eating away of the metal due to the attack of atmosphere gases on the surface of the metal resultibg into the formation of oxides, sulphides, etc. is called corrosion.
Factors affecting corrosion :
(i)Reactivity of the metal
(ii)Presence of impurities
(iii)Presence of air and moisture
(iv)Strains in metals
(v)Presence of electrolytes.
Mechanism of corrosion :
At anode : Fe(s) → (aq) + 2
At cathode : + →
2 +
2 + (g) + 2 →(l)
2(aq) + (g) + (l) → + 4
+ (l) → (aq)
6. Three iron sheets have been coated separately with three metals A, B, C the standard reduction potentials of which are V, and respectively. The = Identify in which case rusting will take place faster when coating is damaged. 1M
ans: Rusting will occur faster if coating of ‘C’ is damaged as ‘C’ has a lower tendency to undergo oxidation as compared to Fe (based on the values of the reduction potential).
7. How many moles of copper will be deposited from a solution of copper sulphate by passing 24, 125 C of electricity ? 1M
ans: (aq) + 2 →
2 96500 C deposit 1 mole of Cu
24125 C will deposit 24125 = 0.125 mole of Cu
8. Why is rusting inhibited in an alkaline medium ?1M
ans: The ions from the alkalies react with ions, amd form unionized water. Since ions take part in the process of rusting, the removal of these ions in an alkaline medium inhibits rusting.
9. Predict whether and Na will react with one another. Give reason.
Given = +2.87 V and = .2M
ans: The reaction between and Na will be 2Na + → 2NaF
, 2Na → 2 + 2 (oxidation)
and F2 → 2 (reduction)
2F + 2 → 2
, Na is anode and F2 acts as cathode
, =
= + 2.87 ()
= + 5.58 V
, is +ve, so the reaction between Na and F2 occurs.
10. Calculate the equilibrium constant of the reaction :
Cu(s) + 2(aq) →(aq) + 2Ag(s);
2M
ans: we know that,
= log
, 0.46 = log
log = = = 15.5668
= antilog (15.5668)
= 3.6881
11. The conductivity of 0.1 M KCl solution at 298 K is 0.0129 S . The resistance of this solution is found to be 58 Ω. What is the cell constant of the cell? A 0.1 M AgNO3 solution at 298 K in the same conductivity cell offered a resistance of 60.5 Ω. What is the conductivity of 0.1 M AgNO3 solution? 2M
ans: Given KKCl = 0.0129 S
R = 58 Ω , C = =
(cell constant) = ?
We know that, K = C =
= 0.0129 58 = 0.7482
Since, AgNO3 solution is taken in the same conductivity cell therefore, (cell constant) remains same.
Now, K = ? , = 0.7482 , C = =
,K = C = 0.7482 = 0.0124
12. A solution of Ni( is electrolysed between platinum electrodes using a current of 0.5 A for 30 minutes. What weight of Ni will be produced at cathode ? (atomic weight of Ni = 58.9 u) 2M
ans: we know that,
Q = I t
Given, I = 5.0 A
t = 30 minutes = 30 60 = 1800 s
, Q = 5 1800 = 9000 C
Also, (aq) + 2 → Ni(s)
2 96500 C deposits 1 mole i.e. 58.9 g of Ni
, 9000 C deposits 9000 = 27.5 g of Ni .
13. Calculate the standard electrode potential of the electrode for a cell in which the following cell reaction takes place 3M
Mg(s) + 2(aq) → (aq) + 2Ag(s)
Given : [] = 0.1 M , [] = 0.01 M
= +0.80 V ; = +2.90 V.
ans: the cell is
Mg(s) │║ (0.01 M) │Ag(s)
= log
2.90 = log
2.90 = 0.02955 log 103
2.90 = 0.02955 3
= 0.8865
, = 2.90 + 0.08865 = 2.989 V
Now, =
= = 0.80 2.989
= 2.189 V
14. Given Cu2++ 2Cu , = + 0.34 V ,
+ , = + 0.80 V
(a)Construct a galvanic cell.
(b)For what concentration of will the emf of the c ell be zero at 25, if the concentration of Cu2+ is 0.01M ? 3M
ans: (a) Cu(s) │ Cu2+(aq) ║ Ag+(aq) │ Ag(s)
(b) we know that,
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)
= log
, = 0
, = log
(0.80 0.34) = log
log = = 15.5668
= antilog(15.5668) = 3.688 1015
= = 2.17
= = 1.64 M.
15. Calculate the molar conductance at infinite dilution of ethanoic acid from the following data : 2M
(HCl) = 425.9 cm2 mol-1
(NaCl) = 126.4 cm2 mol-1
ans: = +
= 91 + 425.9 126.4
= 390.5 cm2 mol-1
CHAPTER 4 – CHEMICAL KINETICS
QUESTIONS:-
1) For the reaction: A+2B AB2 , the rate constant is 1.26 x 10-3 Lmol-1s-1. What is the order of the reaction? 1M
Answer: The order of the reaction is 2.
2) One gram of pulverized wood burns faster than one gram piece of wood. Why?1M
Answer: pulverized wood has a larger surface area which enhances the rate of the reaction and therefore pulverized wood burns faster than a solid piece of wood .
3) 87.5% of a substance e following first order kinetics disintegrated in 45 minutes. What is the half life of the substance? 1M
Answer: ( ½)n = (as 87.5% sample has disintegrated.)
=> (½)n= 1/8
=> (½)n=(½)³
Number of half lives = 3
Therefore t1/2 =45/3 = 15 minutes.
4) Can a reaction have activation energy equal to zero ?1M
Answer: If Ea=0 , then according to the Arrhenius equation k=Ae-E/RT , k=Ae0=A .This would mean that the rate constant is equal to the collision frequency , i.e. each collision would result in product formation which cannot be true , and thus Ea for any reaction cannot be zero.
5) The half life for radioactive decay of c14 is 5730 years . An archeological artifact contained wood that had only 80% of the c14 found in the living tree. Find the age of the wooden artifact. 2M
Answer: radioactive decay follows first order kinetics .
Therefore; k=log
Given: t1/2=5730 years
=100 , [R]=80
We know: k =
... k= = 1.21 x 10-4 years
And t= log
t= 1.903 x 104(log 100 - log 80)
t= 1.903 x 104(2-1.903)
=1845.91 years .
6) For a reaction A + B C, it is found that : (i) rate becomes double when concentration of A is doubled . (ii) Rate becomes 16 times when concentration of both A and B is doubled . Write the rate law for the reaction and calculate the overall order of the reaction . 3M
Answer: let order with respect to A = x let order with respect to B=y
Then, Rate(r) = k[A]x[B]y
Now rate doubles when [A] doubles
... 2r = k[2A]x[B]y
... 2r = 2xk[A]x[B]y =2xr
2 = 2x =>x=1.
Now rate becomes 16 times when concentration of Both and B doubled.
... 16r = k[2A]x[2B]y
... 16r = 2x2yk[A]x[B]y = 2x2yr
16r = 212y
8r = 2y => y=3
Order with respect to A = x=1 Order with respect to B= y=3
The overall order of reaction =1+3=4.
7) The following data were obtained for the reaction : 2NO2(g) + F2(g) 2NO2F(g)
Experiment / Initial Concentration / Initial Rate (MolL-1min-1)[NO2] / [F2]
1 / 0.20 / 0.05 / 6.0 x 10-3
2 / 0.40 / 0.05 / 1.2 x 10-2
3 / 0.80 / 0.10 / 4.8 x 10-2
Determine the order of the reaction and write the rate law of this reaction . 5M
Answer: we know , Rate = k[NO2]m[F2]n
Using experiments 1 and 2 , we get, = =
Dividing the two we get , =
= => m=1
Using experiments 1 and 3, we get , = =
Dividing the two we get, =
=
= => n=1
The order of the reaction is m+n = 1+1 = 2 Rate law is , Rate = k[NO2][F2]
8) For a reaction: A + B products, the rate law is given by : Rate = k[A]1/2[B] .What is the order of the reaction? 1M
Answer: the order of the reaction is 1/2 + 2 = 5/2 = 2.5.
9) What are pseudo unimolecular reactions ? Give an example . 2M
Answer: pseudo unimolecular reactions are those reactions which seem to be of higher order but are actually first order reactions. Example : hydrolysis of an ester . CH3COOC2H5 + H2O CH3COOH + C2H5OH .
10) (a) Define activation energy of a reaction . (b) An increase in 10k of temperature rarely doubles the kinetic energy of the particles but this increase in temperature is , may be enough , to double the rate of the reaction . Explain. 2M
Answer: (a) activation energy is the energy which is to be provided to the reactants to reach the threshold energy level.
(b) an increase in 10k in temperature rarely doubles the kinetic energy of the particle but this increase in temperature may be enough to double the rate of the reaction because the molecules possessing threshold energy double themselves when the temperature is raised by 10k.
11) For a reaction : 2A + B + C A2B + C , the rate = k[A][B]2 with k = 2.0 x 10-6M-2s-1. Calculate the initial rate of the reaction when [A]=0.1M , [B]=0.2M, and [C]=0.8M.If the rate of the reverse reaction is negligible then calculate the rate of the reaction after[A] is reduced to 0.06M. 2M
Answer: The rate law is Rate= k[A][B]2 At[A] = 0.1M and [B] = 0.2M , the initial rate of the reaction is : = 2.0 x 10-6 x (0.2)2 = 8 x 10-9mol L-1s-1.
After [A] is reduced to 0.06 M , [B] = 0.2 – = 0.18M .
... Rate = 2.0 x 10-6 x 0.06 x (0.18)2 = 3.88 x 10-99mol L-1s-1.
12) (a) Give any one example of (i) Zero order reaction . (ii) first order reaction . (b)The rate constant of a reaction is 1.9 x 10-5mol-2 L-2s-1. What is the order of the reaction ? 2M