Time: Two Hours Maximum Marks : 100
- i) There are TEN questions in this paper. Attempt ALL
- ii) Answer each question starting on a new page. The corresponding question number must be written in the left margin. Answer all parts of a question at one place only.
iii) Use of Logarithmic tables is permitted.
- iv) Use of calculator is NOT PERMITTED
Gas Constant R = 8.314 J mol–1K–1
= 0.0821 lit atm mol–1 K–1
= 2 Cal mol–1
Avogadro’s Number N = 6.023 ´ 1023
Planck’s constant h = 6.625 ´ 10–34 J sec.
Velocity of light c = 3 ´ 108 m sec–1
1 electron volt ev = 1.6 ´ 10–19 J
F = 96500 C
Atomic Masses Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39,
Cl = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75,
Fe = 56, Ag = 108
Enrolment No. :
- a) An electron from an atom X de-excites from a state in which its s-function has 6 radial nodes to some other state, where its associated de-Broglie’s wavelength is minimum. Moreover, 20 ml of the gaseous atom X to pass through an orifice takes 5.0 secs. The diffusion of same ml oxygen gas through the same orifice takes 28.28 secs. If the frequency of light emitted is utilised in removing an e– from a metal atom Y whose work function is 1.08 eV, what is the kinetic energy of the ejected electron from the metal
atom Y? 
- b) How much electric field strength should be applied in between two electric plates so as to produce an electron beam of same wavelength as that of the wavelength of light emitted from the atom X due to de-excitation from the states mentioned above. 
- Calculate the time (in minutes) required for the solution to be optically inactive (following first order kinetics), from the following data.
|Rotation of polarized light/degree (at 25°C)||24.1||21.4||12.4||10.0||–10.7|
If the activation energy of this reaction is 15K cals/mole, at what time the solution become optically inactive, when the optical rotation is carried out at 40°C. (assume ro is independent of temperature) 
- 3. a) The value of Kp for the reaction
2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) is 0.035 atm at 400oC, when the partial pressures are expressed in atmosphere. Calculate Kc for the reaction,
Cl2(g) + H2O(g) 
- b) The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 gdm-3. Calculate the equilibrium constant of the reaction, N2O4(g) 2NO2(g). 
- Explain the following.
- a) CH2F2 involves sp3 hybridised C atom. Still none of the angles is 109.5° but is different from this. Why?
- b) The boiling point of CH4 is less than that of SiH4 but reverse is true for CCl4 and SiCl4 Why?
- c) The bond energy of C=C bond is greater than that of –N=N– but the bond energy of
–CºC– is less than that of Nº Why?
- d) Naphthalene undergoes oxidation more readily than benzene but only to the stage where a substituted benzene is formed. Further oxidation requires more vigrous conditions. Why? [4´4]
- Identify A,B,C,D,E and F
[2 ´ 6]
- Bring out the following conversion
- By referring to the equilibrium constants listed below, find the equilibrium for each of the following:
- a) Fe(OH)3(s) + 3H3O+ Fe3+ + 6H2O
- b) PbSO4(s) + PbCrO4(s) +
Ksp[Fe(OH)3] = 4 ´ 10–38 , Kw = 10–14, Ksp(PbSO4) = 2 ´ 10–8, Ksp[PbCrO4] = 2 ´ 10–14. 
- Arrange the following in the order of the property mentioned in brackets and explain your answer.
- a) NH3, CF3 NH2, CH3NH2, (CH3)2NH (basicity)
- b) SiO2, SO3, N2O5, CO2 (acidity)
- c) Ag2O, V2O5, CO, N2O5 (acidity)
- d) CH4, NH3, H2O & HF (strength of conjugate base) [2.5 ´4]
- a) The police often use a device called a breath analyser to test drivers suspected of being drunk. For this process a sample of driver’s breath or blood is drawn into the breath analyser where it is treated with an acidic solution of K2Cr2O7. The limit of blood alcohol content is 0.1% by mass. If a 60 gm sample of blood from a driver required 28.64 ml of 0.07654 M K2Cr2O7 for titration, should the police prosecute the individual for drunken driving? 
- b) Octane is a component of gasoline. Complete combustion of octane leads to CO2 and H2 Incomplete combustion leads to CO and H2O which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 2.000 gallon of octane is burned in an engine The total mass of CO, CO2 and H2O produced is 23.06 Kg. Calculate the fraction of octane converted to CO2. The density of octane is 2.650 Kg/ gallon. 
- A weak base 50 ml was titrated with 0.1 M HCl. The pH of the solution after addition of 12 ml and 25 ml were found to be 9.84 and 9.24 respectively. Calculate Kb of the base and pH
- a) at equivalence point
- b) on addition of 20 ml HCl more after equivalence point. 
- a) From the problem, we can infer that n2 = 7 (7s has 6 radial nodes (n-l-1) and n1 = 1 (wavelength of electron is minimum). From the rate of diffusion data, we can find out the molecular weight of the gaseous molecule.
on solving, Mx = 1.00
i.e., x = hydrogen
K.E. of ejected electron from the metal atom Y
K.E. = hn – hno = =
= [13.32 – 1.08] eV (hno = 1.08eV)
K.E. = 12.24 eV
= 12.24 ´ 1.6 ´ 10–19 = 1.95 ´ 10–18 Joules
- b) l =
9.33 ´ 10–8‑ =
V = 1.723 ´ 10–16 V
- = = =
The time at which, the solution becomes optically inactive at 40°C can be calculated as follows,
rt = 0
t = 79.16 minutes
i.e., after 79.16 minutes , the solution becomes optically inactive
- a) KP = KC (RT)Dn
Dn = moles of product – moles of reactants = 5 – 4 = 1
R = 0.082 L atm/mol K, T = 400 + 273 = 673 K
\ 0.035 = KC (0.082 ´ 673)
KC = 6.342 ´ 10-4 mol l-1
\ for the reverse reaction would be
\ = = 1576.8 (mol l-1)-1
When a reaction is multiplied by any number n (integer or a fraction) the or becomes (KC)n or (KP)n of the original reaction.
\ KC for O2(g) + 2HCl(g) Cl2(g) + H2O(g)
is 39.7 (mol.l-1)-½
- b) Let us assume that we start with C moles of N2O4(g) initially.
Initially C 0
At equilibrium C(1-a) 2Ca
where a is the degree of dissociation of N2O4(g)
Initial vapour density =
Since vapour density and actual density are related by the equation,
V.D. = = = 26.25
\ 1 + a =
\ a = 0.752
\ Kp =
Kp= 5.2 atm.
- a) In this compound two types of angles are there one is HCH and another is FCF. The FCF angle is smaller than 109.5° because of higher electronegativity of F and due to this s – character decreases. On the other hand in HCH bond the s – character increases hence angle also increases from 109.5°.
- b) The electronegativity of Si is less than that C and due to this in SiCl4 the Cl atom acquires more than twice as much negative charge and give rise to higher electrostatic repulsion among SiCl4 than among CCl4 molecule hence boiling point of SiCl4 is lower than that of CCl4.
- c) Because of the lone pair repulsion the -N=N- is weaker than C=C bond but the NºN bond is stronger than CºC bond because the lone pairs on the two nitrogen atoms are on the opposite sides and too far apart to interact strongly with each other.
- d) Resonance energy of naphthalene is 61 Kcal / mol and for benzene it is 36 Kcal/mol so energy required to destroy the first ring is (61-36)=25 Kcal and if once one ring is destroyed 36 kcal of energy is required which is greater than that for the 1st ring hence more vigrous conditions are required.
- a) 2H2O H3O+ + OH–
Kw = [H3O+][OH–] —— (1)
Fe(OH)3(s) Fe3+ + 3OH–
Ksp = [Fe3+][OH–]3 ————— (2)
cubing equation (1) and dividing equation (2), we get,
= ——— (3)
The left hand side of equation (3) is the equilibrium constant of equilibrium (a), hence
Kc = = = 4 ´ 104
- b) PbSO4(s) + PbCrO4(s) +
Kc = —— (4)
PbSO4(s) Pb2+ +
\ Ksp[PbSO4] = [Pb2+] ——- (5)
PbCrO4(s) Pb2+ +
\ Ksp[PbCrO4] = [Pb2+] —— (6)
Dividing equations (5) and (6), we get,
Kc = = 106
- a) CF3NH2 < NH3 < CH3NH2 < (CH3)2NH
- b) SiO2 < CO2 < N2O5 < SO3
- c) The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states of V in V2O5 and N in N2O5 are both +5. But the electronegativity of N is higher, making N2O5 the most acidic oxide.
- d) The size of central atom decreases from C to F, at the same time volume available to electron on the central atom also decreases because in CH3– ion only 1/4th of the space is available for electrons. Thus increasing basic character of these conjugate bases is
F– < OH– < NH2– < CH3–‑
- a) 3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 ¾¾® 3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
Moles of K2Cr2O7 = 28.64 ´ 10–3 ´ 0.07654
Moles of CH3CH2OH= ´ 28.64 ´ 10–3 ´ 0.07654 = 3.288 moles
= 3.228 ´ 46 ´ 10–3 gm of C2H5OH = 0.1512 gm
% by mass = = 0.25%
Since 0.25 > 0.1 the police should prosecute
- b) Mass of 2 gallon of octane = 2.650 ´ 2 ´ 1000 gm
Moles of octane = in 2 gallon
+ O2 ¾¾® +
+ ¾¾® +
8 ´ 44x + 9 ´ 18x + 8y ´ 28 + 9y ´ 18 = 23.06 ´ 103 gm
x + y = 46.48 Þ x = 40.22
% = = ´ 100 = 86.5%
- Let BOH is base with normality N
BOH + HCl ¾® BCl + H2O
Initial Meq. 50 N 10 ´ 0.1 0
Meq. of after titration 50 N – 1 0 1
Medium is still basic as clear from pH. So acid acts as limiting reagent. Further mixture above forms a basic buffer solutions.
pOH = pKb + log
4.16 = pKb + log …(1)
Similarly after addition of 25 ml HCl
4.76 = pKb + log …(2)
From (2) – (1)
3.98 = Þ N = 0.1
putting N = 0.1 in equation (1)
4.16 = pKb + log
Kb= 1.73 ´ 10–5
At equivalence point we calculate pH by salt hydrolysis
B+ + H2O BOH + H+
C(1 – a) Ca Ca
C = = 0.05 (Total volume is doubled)
pH = (pKw – pKb – log C) = 5.12 = (14 – 4.76 – log 0.05) = 5.27
After addition of HCl, the pH is calculated from HCl left unreacted only
Meq. of HCl left = 20 ´ 0.1 = 2
Normality of HCl = = 0.0166; pH = – log (0.0166) = 1.78