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]

2. Identify the missing products (or) reagents in the following sequence of reactions:

3. 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]

4. 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]

5. 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]

6. Vapour pressure of Benzene and Toluene mixture at 50°C is given by P(mm Hg) = 180x_B + 90, where x_B 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]

7. 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 (K_f for H₂O = 1.86 K mol^–1 Kg)                                                                                                                                [10]

8. 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]

9. Standard heat of formation of hydrazine (), H₂O_2­ (), 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]

10. 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

2. 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

• K = CH₃I/Base

L = PCC

N = (i) BrCH₂CO₂Et/Zn (ii) H₃O⁺

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

5. 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%

6. P = 180x_B + 90

If x_B = 1,  = 270 mm Hg

If x_B = 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

7. DT =

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

\ m_­acid = 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%

8. 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

9. 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  – SDH_reactants

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

= – 534 kJ mol^–1

10. r =

Z =

Z » 2

Therefore the unit cell is b.c.c

\ 4r =  a

r =  313  = 135.5 pm

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