Chemistry Subjective_Phase I to III_Questions & 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. 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?                                                                                                                              [7]
2. 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.                                                                     [5]

 

2. Calculate the time (in minutes) required for the solution to be optically inactive (following first order kinetics), from the following data.

t/min	0.0	7.2	36.8	46.0	¥
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 r_o is independent of temperature)  [10]

3. 3. a) The value of K_p for the reaction

2H₂O(g) + 2Cl₂(g) 4HCl(g) + O₂(g)  is 0.035 atm at 400^oC, when the partial pressures are expressed in atmosphere. Calculate K_c for the reaction,

Cl₂(g) + H₂O(g)                                                                              [5]

1. 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).                                            [8]

 

4. Explain the following.
5. a) CH₂F₂ involves sp³ hybridised C atom. Still none of the angles is 109.5° but is different from this. Why?
6. b) The boiling point of CH₄ is less than that of SiH₄ but reverse is true for CCl₄ and SiCl₄ Why?
7. 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?
8. 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]

 

5. Identify A,B,C,D,E and F

 

 

 

[2 ´ 6]

 

6. Bring out the following conversion

 [5]

7. By referring to the equilibrium constants listed below, find the equilibrium for each of the following:
8. a) Fe(OH)₃(s) + 3H₃O⁺  Fe³⁺ + 6H₂O
9. b) PbSO₄(s) +   PbCrO₄(s) +

K_sp[Fe(OH)₃] = 4  ´ 10^–38 , K_w = 10^–14, K_sp(PbSO₄) = 2 ´ 10^–8, K_sp[PbCrO₄] = 2 ´ 10^–14.  [10]

 

8. Arrange the following in the order of the property mentioned in brackets and explain your answer.
9. a) NH₃, CF₃ NH₂, CH₃NH₂, (CH₃)₂NH (basicity)
10. b) SiO₂, SO₃, N₂O₅, CO₂ (acidity)
11. c) Ag₂O, V₂O₅, CO, N₂O₅ (acidity)
12. d) CH₄, NH₃, H₂O & HF (strength of conjugate base)                                                  [2.5 ´4]

 

9. 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?                                                                                           [4]
10. 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.                                                                                                 [6]
11. 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 K_b of the base and pH
12. a) at equivalence point
13. b) on addition of 20 ml HCl more after equivalence point.                                                  [10]

vvv

 

 

CHEMISTRY

SOLUTIONS

 

1. 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, M_x = 1.00

i.e., x = hydrogen

K.E. of ejected electron from the metal atom Y

K.E. = hn – hn_o =  =

=  [13.32 – 1.08] eV (hn_o = 1.08eV)

K.E. = 12.24 eV

= 12.24 ´ 1.6 ´ 10^–19  = 1.95 ´ 10^–18 Joules

 

1. b) l =

 

=

9.33 ´ 10^–8‑ =

933Å =

V =

V = 1.723 ´ 10^–16 V

 

2. = = =

= 0.0112

log

= 0.0149

The time at which, the solution becomes optically inactive at 40°C can be calculated as follows,

 

r_t = 0

 

t =

t = 79.16 minutes

i.e., after 79.16 minutes , the solution becomes optically inactive

 

3. a) K_P = K_C (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 = K_C (0.082 ´ 673)

K_C = 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 (K_C)ⁿ or (K_P)ⁿ of the original reaction.

\ K_C for O₂(g) + 2HCl(g)  Cl₂(g) + H₂O(g)

is 39.7 (mol.l^-1)^-½

 

1. b) Let us assume that we start with C moles of N₂O₄(g) initially.

N₂O₄(g)    2NO₂(g)

Initially                   C                           0

At equilibrium        C(1-a)                     2Ca

where a is the degree of dissociation of N₂O₄(g)

Since,

Initial vapour density =

 

Since vapour density and actual density are related by the equation,
V.D.     =  = =  26.25

\ 1 + a =

\  a = 0.752

\  K_p =

K_p=   5.2 atm.

 

 

4. 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°.
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.
6. 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.
7. 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.

 

5.

 

6.

 

7. a) 2H₂O  H₃O⁺ + OH^–

K_w = [H₃O⁺][OH^–]  —— (1)

Fe(OH)₃(s)  Fe³⁺ + 3OH^–

K_sp = [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

K_c =  =  = 4 ´ 10⁴

1. b) PbSO₄(s) +   PbCrO₄(s) +

K_c =  —— (4)

PbSO₄(s)  Pb²⁺ +

\ K_sp[PbSO₄] = [Pb²⁺]  ——- (5)

PbCrO₄(s)  Pb²⁺ +

\ K_sp[PbCrO₄] = [Pb²⁺] —— (6)

Dividing equations (5) and (6), we get,

 

K_c =  = 10⁶

 

8. a) CF₃NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH
9. b) SiO₂ < CO₂ < N₂O₅ < SO₃
10. 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.
11. 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^–‑

 

9. 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

 

2. 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%

10. 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 = pK_b + log

4.16 = pK_b + log               …(1)

Similarly after addition of 25 ml HCl

4.76 = pK_b + log           …(2)

From (2) – (1)

0.6 =

3.98 = Þ N = 0.1

putting N = 0.1 in equation (1)

4.16 = pK_b + log

K_b= 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 = (pK_w – pK_b  – 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