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Chemistry Subjective_Phase IV to VI_Questions and solutions

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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. Predict the products in the following reactions [5 ´ 2 = 10]
            i)

 

            ii)

 

            iii)

 

            iv)

 

            v)

 

  1. Identify the missing products (or) reagents in the following sequence of reactions:
            i)                                [6]

 

            ii)                                                   [4]

 

            iii)                                [5]
  1. An organic compound (A) having molecular formula C6H10O on treatment with methyl magnesium iodide followed by acid treatment for a long time gave (B). (B) on ozonolysis yielded (C) and (C) on treatment with a dilute basic solution gave 1-acetyl cyclopentene. (B) on reaction with HCl gave (D). Identify (A) to (D) and show the mechanism involved in the conversion of (C) to 1-acetyl cyclopentene. [10]

 

  1. An organic compound A (C7H6O) gives positive test with Tollens reagent. On treatment with alcoholic potassium cyanide, A yields the compound B (C14H12O2). Compound B on reduction with amalgamated zinc and concentrated hydrochloric acid yield an unsaturated compound C, which adds a molecule of bromine. The compound B can be oxidised with nitric acid to a compound D, having molecular formula C14H10O2. Compound D on heating with potassium hydroxide undergoes rearrangement and subsequent acidification of rearranged products yields an acidic compound E (C14H12O3). Identify compounds A to E giving the relevant reactions. [10]

 

  1. A sample of lead weighing 1.05 g was dissolved in a small quantity of nitric acid to produce aqueous solution of Pb2+ and Ag+ (which is present as impurity). The volume of the solution was increased to 300 ml by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be 0.503 V at 25°C. What was the
    % of Ag in the lead metal? Given  = 0.799 V. Neglect amount of Ag+
    converted to Ag.                                                                                                                     [10]

 

  1. Vapour pressure of Benzene and Toluene mixture at 50°C is given by P(mm Hg) = 180xB + 90, where xB is the mole fraction of benzene. A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be the mole fraction of C6H6 in the vapour phase?                                  [10]

 

  1. 1 g of monobasic acid in 100g of water lowers the freezing point by 0.168°C. if 0.2 g of same acid requires 15.1 ml of alkali for complete neutralization, calculate degree of dissociation of acid (Kf for H2O = 1.86 K mol–1 Kg)                                                                                                                                [10]

 

  1. Heat of combustion of ethylene (C2H4) is –337 kcal/mol under standard state. Assuming 70% efficiency, how many Kg of water at 20°C can be converted into steam at 100°C burning 1m3 of C2H4 gas at STP. Specific heat of water is 1kcal Kg–1 K–1 and its latent heat of vapourisation is 540 kcal kg–1. [10]

 

  1. Standard heat of formation of hydrazine (), H2O (), and H2O() are –50.4 , –193.2 and
    – 242.7 kJ mol–1 respectively. What is the heat of reaction under standard state for the following reaction; N2H4(l) + 2H2O2(l) ¾® N2(g) + 4H2O(l) [7]

 

  1. The edge length of a cubic unit cell of an element (atomic mass = 95.54) is 313 pm and its density is 10.3 g/ml calculate the atomic radius.                                                                                              [8]

6

 

PHASE – IV to VI

CHEMISTRY

 

 

Solutions

 

1.         i)             ii)
            iii)             iv)
            v)    

 

  1. i) A = HO – CH2 – C º C – CH2 – OH

B = HO – CH2 – CH2 – CH2 – CH2 – OH

C = Br – CH2 – CH2 – CH2 – CH2 – Br

D = CN – CH2 – CH­2 – CH2 – CH­2 – CN

E = H2N – CH2 – CH­2 – CH2 – CH­2 – CH2 – CH2 – NH2

F = HO2C – CH2 – CH2 – CH2 – CH2 – CO2H

 

            ii)
 
  • K = CH3I/Base

L = PCC

 

N = (i) BrCH2CO2Et/Zn (ii) H3O+

 

 

3.
 

 

 

  1. The compound A is benzaldehyde (C6H5CHO). The various given reactions are as follows.

 

  1. E = E° – log

0.503 = 0.799 –

[Ag+] = 9.62 ´ 10–6 M

Moles of Ag+ = 9.62 ´ 10–6 ´

= 2.89 ´ 10–6

Mass of Ag = 2.89 ´ 10–6 ´ 108 = 3.11 ´ 10–4 g

\ Percentage of Ag = 0.0296%

 

  1. P = 180xB + 90

If xB = 1,  = 270 mm Hg

If xB = 0,  = 90 mm Hg

Mole fraction of Benzene  =  = 0.6

Mole fraction of Toluene  = 1– 0.6  = 0.4

Mole fraction of benzene in the vapour over the first solution

= 0.82

and that of toluene   = 1– 0.82 = 0.18

Hence the new solution will have the composition

= 0.82,  = 0.18

P = 0.82 ´ 270 + 0.18 ´ 96  = 237.6 mm Hg

Now mole – fraction of benzene in the vapour over the second solution  =  = 0.93

  1. DT =

DT = 0.168, w  = 1g, W = 100 g, Kf = 1.86

\ m­acid = 110.71         (This is exp. Mol.wt.)

Now meq of acid  = Meq of Alkali

´ 1000  = 15.1 ´

Eq. wt  = 132.45  = Mol. wt. (Monobasic acid)

HA         H+        +          A

Before dissociation     1                      0                      0

After dissociation        1–a                  a                      a

= 1  + a Þ a  = 19.6%

 

  1. 1m3 = 1000 L = mol

Enthalpy change due to burning of 1m3 ethylene = – 337 ´  Kcal

Due to 70% efficiency, useful heat is  = 337 ´  ´  Kcal

= 10531. 25 Kcal

H2O(l) is converted into H2O(g) in two stages

H2O(l), (at 25°C) ¾¾® H2O (l, at 100°C) DH = –80 Kcal Kg–1 (rise in temp. = 80°C)

H2O (l, 100°C) ¾¾®  H2O(g, 100°C), DH = 540 Kcal Kg–1

DH total   = 620 Kcal Kg–1

Thus water converted to steam  =  =  = 16.98 Kg

 

  1. Taking the facts that heat of formation of a substances is its enthalpy and enthalpy of the element is zero we have

DH° (298K) = SDH products  – SDHreactants

 

[ 0 + 4 ´ (–242.7)] –  [ –50.4 + 2(–193.2)]

= – 534 kJ mol–1

 

  1. r =

Z =

Z » 2

Therefore the unit cell is b.c.c

\ 4r =  a

r =  313  = 135.5 pm

vvv

 

 

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