Chemistry _Subjective_Phase I to VI_Questions & Solutions

CHEMISTRY

 

 

Time:  Two Hours                                                                                                     Maximum Marks : 100

 

 

Note:

1. i) There are TWELVE questions in this paper. Attempt ALL
2. 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.

1. iv) Use of calculator is NOT PERMITTED

 

Useful Data:

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

 

 

Name                    :

Enrolment No.  :

 

1. Provide plausible mechanism for each of the following:

a)
b)	[4 + 4]

 

2. An organic compound A (MF – C₁₂H₁₃N) does not decolourise Br₂/CCl₄ readily though it can be nitrated and sulphonated. Upon boiling with dil. acid it is hydrolysed with evolution of NH₃ gas. The organic product of hydrolysis upon permanganate oxidation gives a dicarboxylic acid B (MF – C₈H₆O₄) of which only one mononitroisomer is possible and which upon decarboxylation gives benzene. The same compound (hydrolysis product) upon treatment with NBS and subsequent treatment with hot alc. KOH gives a compound C
   (MF – C₁₂H₁₂O₂). The compound C upon reductive ozonolysis with only one equivalent of O₃ gives a compound D(MF – C₁₂H­₁₂O₄). In the presence of dil. NaOH the compound D undergoes intramolecular aldol condensation. Identify A, B, C, D and two probable aldol condensation products of D which are expected to be formed after acidification and
   subsequent heating.                                                                                                               [10]

 

3. a) Given at 25°C:

Cu(s) ¾® Cu²⁺(aq) + 2e,            DG⁰ = + 65.04 kJ

Ag(s) ¾® Ag⁺(aq) + e,                DG⁰ = +77.200 kJ

What would be the concentration of Ag⁺ ion in a solution containing 0.06 M of Cu²⁺ ion such that both metals can be deposited together? Assume that Ag and Cu do not dissolve in one another.

1. b) Write the IUPAC nomenclature of the following:
2. i) K₃[Fe(NC)₅NO]×2H₂O
3. ii) [Pt(NH₃)₂Cl₂] [ZnCl₄]                                                                                             [5 + 5]

 

4. a) An organic liquid, A , immiscible with water, when boiled together with water, has a boiling point of 90°C at which the partial vapour pressure of water is 526 mm Hg. The superincumbent pressure is 736 mm Hg. If mass ratio of liquid and water collected is
   2.5 : 1, what is the molecular mass of the liquid?

 

1. b) Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He⁺ ion, emitting a single photon in this process. He⁺ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV range that are likely to be emitted during and after the combination (h = 4.14 × 10^–15 eVs).     [5 + 5]

 

5. a) 100 ml of water containing soluble Ca(HCO₃)₂ required 30 ml. N/50 HCl solution for the titration using methyl orange indicator. Calculate the temporary hardness of water in ppm unit.

[Density of water is 1 gm/cc]

94. b) Calculate the heat of oxidation of ethyl alcohol to CO₂ when the heats of formation of CO₂ and H₂O(l) are given as – 94.05 and – 68,32 kcal and that of liquid ethyl alcohol is
    – 66.4 kcal.
95. c) Calculate the no. of moles of electrons in 18 ml of water water at 4°C temperature.

[4 + 4 + 2]

6. Explain the followings:
7. a) The bleaching action of Cl₂ is permanent while that of SO₂ is temporary.
8. b) PbCl₄ is available in the form of unstable liquid and gradually decomposes to PbCl₂ and Cl₂ while PbBr₄ and PbI₄ couldn’t be prepared.
9. c) F₂O₂ is more stable than H₂O₂.                                                                             [3 + 3 + 3]
10. A metal sulphate ‘A’ gives sky blue colour in oxidising flame and opaque red in reducing flame during borax bead test.When ‘A’ reacts with BaCl₂ it gives a white ppt. ‘B’ which is insoluble in HCl. Into the solution of ‘A’ when excess KI is added two compound ‘C’ and ‘D’ are produced and the gas liberated when passed through starch solution it becomes violet. When potassium ferrocyanide is added to a solution of ‘A’ a chocolate coloured ppt. E appeared along with formation of ‘D’. Identify A to E. [5]

 

8. Write down the products in the following

a)
b)	[3 + 3]

 

9. Find out the missing alphabets in the following sequence

CH₃ – – CO₂Et

[8]

 

10. The compound Co(en)₂(NO₂)₂Cl has been prepared in a number of isomeric forms. One form undergoes no reaction with either AgNO₃ or ethylene diamine. A second form reacts with AgNO₃ but not with ethylenediamine and the third form reacts with both AgNO₃ and ethylenediamine. Identify each of the these forms by their IUPAC names and discuss the above reaction in light of Werner’s theory. [6]

 

11. After solid SrCO₃ was equilibrated with a pH = 8.60 buffer, the solution was found to have [Sr²⁺] = 2.2 ´ 10^–4. Determine the solubility product of SrCO₃? K₂ for H₂CO₃ = 4.7 ´ 10^–11.

[10]

 

12. A proportional counter is filled in succession with the same amount of 3 samples of CO₂ at a fixed pressure and the following counts were noted.
13. a) Dead CO₂ from coal : 10,800 counts in 900 minutes.
14. b) CO₂ from contemporary wood : 33,000 counts in 200 minutes.
15. CO₂ from test sample : 18,800 counts in 400 minutes

If half life of ¹⁴C = 5570 year, then estimate the age of sample?                                           [8]

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CHEMISTRY

 

 

Solutions

 

1.         a)		[4]
b)		[4]
2.

3. a) DG⁰ = –

So, 65040 = – 2 ´ 96500 ´

\  = – 0.337V

Similarly, 77200 = – 1 ´ 96500 ´                                                                           [3]

\  = – 0.80V

According to Nernst equation

= log[Cu²⁺]

= 0.337 + log(0.06) = 0.301

Similarly,

= 0.80 + log[Ag⁺]                                                                                          [2]

Two metals will be deposited together when their electrode potentials are equal

0.301 = 0.80 + 0.059 log[Ag⁺]

[Ag⁺] = 10^–8.47 = 0.348 ´ 10^–8M                                                                                          [2]

1. b) i)    Potassium pentacyano(N) nitrosyl ferrate (II)dihydrate
2. ii) Diamminedichloroplatinum(IV) tetrachlorozincate(II)                                      [2.5 + 2.5]

 

4. a) At boiling point P_mix = 736 mm

Thus at boiling point,

= 526 mm

= 210 mm

Also, × mole fraction in vapour phase                                                        [1]

Let ‘a’ g of liquid and water is collected or this is the amount of vapours at equilibrium

Thus, Mass of liquid vapours =

Mass of water vapours =                                                                                            [1]

Now for liquid

210 = 736 ´                             …(1)                                                     [1]

Now for H₂O

526 = 736 ´                             …(2)                                                     [1]

From (1) and (2)

Þ m = 112.7                                                                                             [1]

1. b) The energy of the incident photon is

E₁ =  = 3.1 eV

The maximum kinetic energy of the emitted electrons is

E_max = E₁ – W = 3.1 – 1.9 eV = 1.2 eV                                                                              [1]

It is given that

emitted electrons of maximum energy + ₂He²⁺ ¾®  + Photon

The 4^th  excited state implies that the electron enters in the n = 5 electronic state. In this state the energy is

E₅ =  = – 2.18 eV

The energy of the emitted photon in the above combination reaction is

E = E_max + (–E₅) = 1.2 + 2.1 = 3.3 eV                                                                                [1]

After recombination reaction, the electron may undergo transitions from a higher level to a lower level thereby emitting photons.

The energies in the lower electronic levels of He⁺ are

E₄ =  = – 3.4eV

E₃ =  = – 6.04 eV

E_­2 =  = – 13.6 eV

The possible transitions are

n = 5 ¾® n = 4    DE = E₄ – E₅ = –3.4 –(–2.1) =  –1.3 eV                                               [1]

n = 5 ¾® n = 3    DE = E₃ – E₅ = –6.04–(–2.1) =  –3.94 eV                                            [1]

n = 4 ¾® n = 3    DE = E₃ – E₄ = –6.04 – (–3.4)= –2.64 eV                                            [1]

Hence the photons that are likely to be emitted in the range of 2eV to 4 eV are 3.3 eV, 3.94 eV and 2.64 eV.

 

5. a) Ca(HCO₃)₂ + 2HCl = CaCl₂ + 2H₂O + 2CO₂

No. of ml eq. of HCl used       = 30 ´  = no. of ml eq. of CaCl₂ produced

= no. of ml eq. of CaCO₃ present in 100 ml of water.

\no. of ml eq. of CaCO₃ in 10⁶ ml or 10⁶ gm of water = 30 ´ ´ 10⁴                          [2]

= ´ 10 gm eq. of CaCO₃ = ´ 10 ´ 50 gm of CaCO₃

= 300 gm of CaCO₃

\ Temporary hardness of water is 300 ppm.                                                                    [2]

1. b) C₂H₅OH(l)       +    3O₂ = 2CO₂         +    3H₂O(l)

DH_f⁰      – 66.4              0          2 ´(–94.05)      3 ´ (–68.32)

DH for the reaction      =

= 2 ´ (–94.05) + 3 ´ (–68.32) – (–66.4)

= – 326.6 kcal

\ 326.6 kcal heat will be evolved in the oxidation of ethanol.                                           [4]

1. c) 18 ml of water = 18 gm water = 1 mole water

as density of water is 1 gm/cc at 4°C temperature

\ in 18 gm of water, no. of molecules of water = 6.023 ´ 10²³

In one molecule of water no. of moles of electron is 8

\ Total moles of electron in 18 gm of water

= 8 ´ 6.023 ´ 10²³ = 4.818 ´ 10²⁴                                                                                      [2]

 

6. a) Chlorine bleaches substance by oxidation while SO₂ bleaches by reduction. So reduced bleached product after sometimes undergoes oxidation in presence of air or oxygen and SO₂ bleaching is only a temporary bleaching.                    [3]
7. b) ‘Pb’ is the bottom element of group IV A with electronic configuration…….. 4f¹⁴5d¹⁰6s²6p². Due to presence of so many ‘d’ and ‘f’ electrons the effective nuclear charge of Pb is very high and two electrons of ‘6s’ orbital remain inert and don’t behave as valence cell electrons (inert pair effect). So (+4) oxidation state of ‘Pb’ is not stable.

Now Pb⁺⁴ being hard cation (having high effective nuclear charge) a considerable percentage of covalency is incorporated in PbCl₄ and it is liquid and Pb⁺⁴ is hard, hence it is to some extent stable with hard base like Cl^–. But due to inert pair effect it dissociates into PbCl₂ & Cl₂. Br^– and I^–‑ being soft base and Pb⁺⁴ being hard and unstable PbBr₄ and PbI₄ is not at all stable.                                                                                               [3]

1. c) H₂O₂ and F₂O₂ both have peroxy linkage which is not so much stable due to lone pair – lone pair repulsion

 

Due to higher electronegativity of ‘F´atom the electron density on oxygen atoms decreases and interelectronic repulsion in peroxy linkage also decreases.                                               [3]

 

7. A = CuSO₄, B = BaSO₄, C = Cu₂I₂, D = K₂SO₄, E = Cu₂[Fe(CN)₆]. [5 ´ 1]

 

8.         a)

[3]

 

b)

[3]

 

9.

 

10. One form does not undergo reaction with AgNO₃ indicating absence of free Cl^– i.e. Cl^– in the outer sphere and with ethylene diamine indicating that the Cl^– and NO₂^– are anti to each other. Had they been cis, they should have been replaced by ethylene diamine which is a bidentate ligand.

Second form reacts with AgNO₃ but not with ethylene diamine which indicates availability of free Cl^–. The third compound reacts with AgNO₃ and ethylene diamine indicating Cl^– in the outer sphere and presence of NO₂^– in the cis-position.                                                                                                           [3]

A = [Co(en)₂ (NO₂) Cl]NO₂

B = [Co(en)₂(NO₂)₂]Cl

C = [Co(en)₂(NO₂)₂]Cl

 

[3]

11. K_sp = [Sr²⁺] [CO­₃^2–]

CO₃^2– + H₂O H CO₃^– + OH^–

K_h =                                                                                     [2]

[H₃O⁺] = antilog (– 8.6) = 2.51 ´ 10^–9

\ [OH^–] =

\                                         [4]

The carbonate ion which dissolves forms HCO₃^– in a 1 : 1 ratio or remains unreacted. Thus

[Sr²⁺]   = [HCO₃^–] + [CO₃^2–]

= 53.4 [CO₃^2–] + [CO₃^2-–]

= 54.4 [CO₃^2–]

\ [CO₃^2–] =  \ K_sp = 2.2 ´ 10^–4  = 8.9 ´ 10^–10                                  [4]

 

12. Counting rate of dead CO₂ = = 12 counts/min.

This gives the background counting rate counting rate of contemporary wood

=  = 165 cpm (counts/min)

\ Actual counting rate = 165 – 12 = 153 cpm

Counting rate of test sample =  cpm

True counting rate of test sample = 47 – 12 = 35 cpm                                                            [4]

\ Age =

= = 11858.19 years                                                                           [4]

 

 

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