Chemistry Subjective_Phase VI_Questions and solutions

CHEMISTRY

Time:  Two Hours                                                                                                     Maximum Marks : 100

 

Note:

1. i) There are TEN 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. 1. 35 g of a salt ACl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of A⁺ and Cl^– if ACl forms CsCl type crystal having 2.2 g/cm³. Given K_b(AOH)  = 1.8 ´ 10^–5  = 0.732 for this cell unit.                                                    [10]

 

2. In a face centred lattice with all the positions occupied by A atoms the body centred octahedral hole in it is occupied by an atom B of an appropriate size for such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound. [10]

 

3. A solution containing 2.665g of CrCl₃×6H_­2O (M = 266.5 gms) is passed through a cation exchanger. The chloride ions obtained in solution react with AgNO₃ and give 2.87g of AgCl. Determine the structure of the compound. [10]

 

4. Determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in M/32 solution of salt at 298 K from the following cell data at 298 K.

Pt | H₂ (1 atm) | H⁺ (1M) || M/32 C₆H₅NH₂HCl (aq) | H₂ (1 atm) | Pt, E_cell = -0.188 V.        [10]

5. The EMF of the cell Pt |H_­2 (1 atm), HA (0.1 M, 30 ml) || Ag⁺ (0.8 M) | Ag is 0.9V. Calculate the EMF when 0.05 M NaOH (40 ml) is added. [10]
6. Calculate the cell potential of a cell having reaction: Ag₂S + 2e^– ® 2Ag + S^2- in a solution buffered at pH = 3 and which is also saturated with 0.1 M H₂S.

For H₂S: K₁ = 10^-8  and K₂ = 1.1 ´ 10^-13, K_sp(Ag₂S) = 2 ´ 10^-49,           [10]

 

7. The overall reaction of the discharge of storage battery is

Pb(s) + PbO₂(s) + 2H₂SO₄(l) ¾® 2PbSO₄(s) + 2H₂O(l)

If the standard free energies of formation of Pb₂O(s), H₂SO₄, PbSO₄ and water are – 217.38, – 909.27, –813.2 and – 237.18 kJ mol^–1 respectively.

1. a) Calculate DG⁰ for the reaction
2. b) Write the cell formed and the half cell reactions
3. c) Calculate the emf of the cell and predict whether the reaction is spontaneous or not.   [10]

 

8. The vapour pressure of pure liquid A at 300 K is 575 torr and that of pure liquid B is 390 torr. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of A in the vapour is 0.350. Calculate the total pressure of the vapour and the composition of liquid mixture. [10]

 

9. a) Sea water is found to contain 5.85% NaCl and 9.5% MgCl₂ by weight of solution. Calculate its normal boiling point assuming 80% ionisation of NaCl and 50% ionisation of MgCl₂ (K_b(H₂O) = 0.51 kg mol^–1K).

 

1. b) Two weak acid solutions HA₁ and HA₂ each with the same concentration and having pK_a values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25°C) and are inter connected through a salt bridge. Find emf of the cell.                                                                                          [8 +8]

 

10. An element crystallises into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of this unit cell is 24 ´ 10^–24 cm³and density is 7.2g cm^–3. Calculate the number of atoms present in 200 g of the element.     [4]

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CHEMISTRY

 

 

Solutions

 

 

1. ACl(s) ¾¾® A⁺(aq) + Cl^–(aq)

A⁺(aq) + H₂O  AOH(aq) + H⁺(aq)                       at equilibrium  C(1–a)

For salts of weak base and strong acid a =

H⁺ = Ca =

antilog (–4.827) =

M = 53.5

For CsCl type structure

r =

a = = 3.43Å

=  = 0.866 a = 0.866 ´ 3.43 Å  = 2.97 Å

Q  = 0.732

on solving  = 1.255 Å

= 1.715 Å

2. Let a be the edge of the cube so that

4r_A = a

a =

No since the atoms B has occupied the body central octahedral hole

 

 

= a

=

2r_B =

 

                = 1.414 – 1  = 0.414

Volume of the cube a³ =

Volume occupied by A and B=

 

Volume occupied per unit volume of unit cell

f = =  = 0.75

3. 143.45g of AgCl contains 35.45 g Cl^– ions

\ 2.87g of AgCl will contain  = 0.709 g Cl^– ions

266.35 g CrCl₃×6H₂O contains n ´ 35.45 g of ionisable Cl^– ions (where n = no. of Cl^– ions outside the coordination sphere).

Thus, 2.665g CrCl₃×6H₂O will contain g

Also  = 0.709 Þ n » 2

Keeping in view the octahedral geometry of the complex, its structure may be written as

[CrCl(H_­2O)₅]Cl₂×H₂O

 

4. E_cell = E_cathode – E_anode

\ E_cathode = E_cell – 0.188 V

=

Thus log [H⁺] =

Putting the various known values we get

[H⁺] = 6.63 ´ 10^–4 mol dm^–3.

C₆H₅NH + H₂O   C₆H₅NH₂ + H₃O⁺

[H₃O⁺] = h.c.

 

  =  1.43 ´ 10^-5 mol dm^-3

 

5. Since E = E⁰_C – E⁰_A –

0.9 = 0.77 – 0 –

\ K_a = 2.5 ´ 10^–4

When 40 ml NaOH is added, the [H⁺] is given by

pH = pK_a + log

pH = 4 – log 2.5 + 0.30120 = 3.9

[H⁺] = 1.25 ´ 10^–4

E = 0.77 =  = 0.994 V

 

6. Ag₂S 2Ag⁺ + S^2-       K_sp = 2 ´ 10^-49

2Ag⁺ + 2e^– ® 2Ag       E^o  = 0.8 V

————————————————

Ag₂S + 2e^– ® 2Ag + S^2-

= -RTln K_sp

=  2.7 ´ 10⁵ J/mol

=  -nFE^o = -2 ´ 96500 ´ 0.8

= -1.54 ´ 10⁵ J/mol

=   +  = 1.156 ´ 10⁵ J/mol

= -nFE^o¢

\ E^o¢   =

=  -0.598V

\ E     = E^o¢ –

K₁K₂    = 1.1 ´ 10^-21 =

\ [S^2-] =

\ E     = -0.598 – log 1.1 ´ 10^-16

            E    = –0.127 V                                               

 

7. We know,
8. a) D_reaction = G⁰_product – G⁰_reactant

                        = 2 (– 813.2) + 2(–237.18)

= – 1(–217.38) – 2(–909.27)

                  = – 64.84 kJ

 

1. b) Cell reaction is

Pb | PbSO₄ | H­₂SO₄ | PbO₂

The half cell reactions are

At anode                Pb(s) + SO₄^2– (aq) ¾® PbSO₄ + 2e^–

At cathode             PbO₂(s) + SO₄^2–(aq) + 4H⁺ + 2e^– ¾® PbSO₄ + 2H₂O

–––––––––––––––––––––––––––––––––––––––––––––

                                                Pb(s) + PbO₂(s) + 2H₂SO₄(aq) ¾® 2PbSO₄(s) + 2H₂O(l)

 

1. c) We know   DG⁰ = – nFE⁰

E⁰ = – 64.84 kJ

-–––––––––––––

– 2(96.5 kJ mol^–1V^–1)

                  = 0.336

 

8. A and B are volatile liquids

Given P⁰_A = 575 torr

P⁰_B = 390 torr

Let mole fraction of A in solution = X_A

Hence P_Total = P⁰_AX_A + P⁰_B (1 – X_A)

Also X₁¹ = mole fractionof A in vapour = 0.35

X₁¹ =  = 0.35

=

this gives X_A = 0.24

Hence total presure = 575 ´ 0.24 + 390 ´ 0.76

= 434.4 torr

Composition of liquid mixture

A = 24 mol %

B = 76 mol %

 

9. a) 100 g solution contains = 5.85 g NaCl = 0.1 mol NaCl

100g solution contains = 9.5 g MgCl₂ = 0.1 mol MgCl₂

Hence weight of solvent (H₂O) = 100 – (5.8 + 9.5) = 84.65 g

NaCl ionises 80%

NaCl  Na⁺ + Cl^–

i = 1 + (y + 1)x = (1 + x) = 1 + 0.8 = 1.8

Here y is the number of ions per mol of solute and X is degree of ionisation

Hence no. of mole of NaCl from 0.1 mol due to ionisation = 1.8 ´ 0.1 = 0.18 mol

For MgCl₂, it comes out to be 0.2

Total mol of NaCl and MgCl₂ in solution = 0.18 + 0.2 = 0.38

(n₁ + n₂)R⁰ = 0.38

DT_b =

=  = 2.29

Hence boiling point of solution = 100 + 2.29 = 102.29°C

 

1. b) The cell is

Pt H_{2(1 atm)} | HA₂ || HA₁ | H_{2 (1 atm)} Pt

At LHS                                                                                         = log[H⁺]

=  + 0.059 (pH)₁

At RHS       =  – 0.059 (pH)₁

For Acid HA₁

[H⁺] = Ca =

pH =

For HA₂

PH =  log C

\ E_cell = E_OP + E_RP

= 0.059  [Q C are same] =

E_cell = + 0.059

Since E_cell is positive the reaction is spontaneous.

 

10. d =

From this M = 104.02

No. of mol in 200 g =  = 1.9226

No. of atoms in 200 g = 1.922 ´ N₀ = 11.57 ´ 10²³ atoms

 

 

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