IGNOU BCHCT-135 Solved Assignment 2024 B.Sc. CBCS Chemistry cover page

IGNOU BCHCT-135 Solved Assignment 2024 | B.Sc. CBCS Chemistry

Solved By – Anjali Patel – Bachelor of Science (B.Sc) from Mumbai University

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IGNOU BCHCT-135 Assignment Question Paper 2024

bchct-135-solved-assignment-2024-qp-70f87f73-d018-4ab3-a508-bab17216331a

bchct-135-solved-assignment-2024-qp-70f87f73-d018-4ab3-a508-bab17216331a

BCHCT-135 Solved Assignment 2024
PART-(A)
  1. Discuss the fractional distillation for a mixture of benzene and toluene using a suitable diagram.
  2. (a) Give the thermodynamic derivation of distribution law.
    (b) Write any two applications of solvent extraction.
  3. (a) Derive the expression for distribution coefficient when the solute dissociates is one of the solvents.
    (b) Differentiate between true, metastable and unstable equilibria giving suitable examples.
  4. (a) State Gibbs phase rule and give its mathematical form.
    (b) How many components are there in the following systems? Briefly discuss.
    (i) C a C O 3 ⇌ C a O + C O 2 C a C O 3 ⇌ C a O + C O 2 CaCO_(3)⇌CaO+CO_(2)\mathrm{CaCO}_3 \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_2CaCO3⇌CaO+CO2
    (ii) C H 3 C O O H + C 2 H 5 O H ⇌ C H 3 C O O C 2 H 5 + H 2 O C H 3 C O O H + C 2 H 5 O H ⇌ C H 3 C O O C 2 H 5 + H 2 O CH_(3)COOH+C_(2)H_(5)OH⇌CH_(3)COOC_(2)H_(5)+H_(2)O\mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}CH3COOH+C2H5OH⇌CH3COOC2H5+H2O
  5. (a) Calculate the maximum number of phases and maximum number of degrees of freedom that
    (b) Can exist for a one-component system.
    When is a system called invariant? Illustrate with an example.
  6. Draw and explain the phase diagram of sulphur.
  7. (a) Define conductivity. Give its SI units.
    (b) List various factors on which the conductivity of electrolytic solutions depends.
  8. Draw and explain the conductometric titration curves for the titration of the following:
    (i) H C l H C l HCl\mathrm{HCl}HCl vs N a O H N a O H NaOH\mathrm{NaOH}NaOH
    (ii) C H 3 C O O H C H 3 C O O H CH_(3)COOH\mathrm{CH}_3 \mathrm{COOH}CH3COOH vs N a O H N a O H NaOH\mathrm{NaOH}NaOH
  9. (a) List the functions of a salt bridge.
    (b) What are the conditions which a reversible cell should satisfy.
  10. (a) What is an electrolyte concentration cell? Write the expression for E cell E cell E_(“cell “)E_{\text {cell }}Ecell for such a
    (b) cell.
    Write the reactions occurring at electrodes in the electrolysis of water. Why are a few drops of conc. H 2 S O 4 H 2 S O 4 H_(2)SO_(4)\mathrm{H}_2 \mathrm{SO}_4H2SO4 added in this process?
PART-(B)
  1. Discuss the two methods of decarboxylation of carboxylic acids and comment on the nature of products formed.
  2. How can you propane the following compound starting from ethanoyl chloride?
original image
(ii) C H 3 C O O C H 2 C H 3 C H 3 C O O C H 2 C H 3 CH_(3)COOCH_(2)CH_(3)\mathrm{CH}_3 \mathrm{COOCH}_2 \mathrm{CH}_3CH3COOCH2CH3
(iii) C H 3 C H O C H 3 C H O CH_(3)CHO\mathrm{CH}_3 \mathrm{CHO}CH3CHO
13. Briefly explain Hofmann elimination. Also give the importance of this reaction.
14. Differentiate between Sandmeyer reaction and Gattermann reaction giving suitable examples.
15. Discuss the Hinsberg test for distinguishing primary, secondary and tertiary amines.
16. (a) Discuss the preparation of 2-aminobutanoic acid using Strecker synthesis.
(b) What is cope elimination? Give reaction.
17. (a) Briefly explain this general structure and classification of peptides.
(b) What is bradykinin? Give its role?
18. How is C C CCC-terminal indentified in a peptide or a protein? Discuss.
19. (a) Explain the cyctic hemiacetal formation by glucose.
(b) Explain mutarotation in glucose.
20. Discuss the important features of structure of cellulose giving suitable diagram.
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

BCHCT-135 Sample Solution 2024

bchct-135-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

bchct-135-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

BCHCT-135 Solved Assignment 2024
PART-(A)
  1. Discuss the fractional distillation for a mixture of benzene and toluene using a suitable diagram.
Answer:
Fractional distillation is a technique used to separate mixtures of liquids that have different boiling points. It’s particularly effective for components whose boiling points are close to each other, like benzene and toluene. Let’s discuss this process step-by-step, focusing on a mixture of benzene and toluene.

1. Introduction to the Components: Benzene and Toluene

  • Benzene (C₆H₆): Boiling Point ≈ 80.1°C
  • Toluene (C₇H₈): Boiling Point ≈ 110.6°C

2. Principle of Fractional Distillation

Fractional distillation works on the principle that different liquids boil at different temperatures. When a mixture is heated, the component with the lower boiling point vaporizes first. The vapor is then cooled and condensed back into a liquid, which can be collected separately.

3. Setup for Fractional Distillation

The setup typically includes:
  • A distillation flask containing the mixture.
  • A fractionating column, which provides a surface area for repeated vaporization and condensation.
  • A thermometer to monitor the temperature.
  • A condenser where vapors are cooled and condensed back into liquid.
  • Collection flasks for the separated components.

4. Process of Fractional Distillation for Benzene and Toluene

  1. Heating the Mixture: The mixture of benzene and toluene is heated. Benzene, with its lower boiling point, starts to vaporize first.
  2. Vapor Rises Through Fractionating Column: The vapor enters the fractionating column. As it rises, it cools and condenses on the packing material.
  3. Re-Vaporization and Condensation: The condensed vapor may re-vaporize as it absorbs heat from the rising vapors. This process of repeated vaporization and condensation separates the more volatile component (benzene) from the less volatile one (toluene).
  4. Collection of Benzene: When the temperature at the top of the column reaches the boiling point of benzene, the vapor at this stage is almost pure benzene. It is then condensed and collected in a separate flask.
  5. Collection of Toluene: After all benzene has been distilled off, the temperature rises to the boiling point of toluene. The process is continued to collect toluene in a different flask.

5. Diagram of Fractional Distillation

A diagram would typically show the distillation flask connected to a fractionating column, which is in turn connected to a condenser. The condenser leads to separate collection flasks. The thermometer is positioned at the top of the fractionating column to monitor the temperature accurately.
original image

6. Conclusion

Fractional distillation is an efficient method to separate liquids like benzene and toluene. The key to its effectiveness lies in the careful control of temperature and the efficiency of the fractionating column, which allows for the separation based on the differing boiling points of the components.
  1. (a) Give the thermodynamic derivation of distribution law.
Answer:
The distribution law, also known as the partition law or Nernst distribution law, is a principle in thermodynamics and chemistry that describes how a solute distributes itself between two immiscible solvents. It’s based on the concept of chemical potential and equilibrium. Let’s derive this law thermodynamically.

1. Introduction to the Concept

The distribution law states that at equilibrium, a solute distributes itself between two immiscible solvents in a constant ratio, independent of the total amount of solute. Mathematically, it’s given by:
K d = [ A ] 1 [ A ] 2 K d = [ A ] 1 [ A ] 2 K_(d)=([A]_(1))/([A]_(2))K_d = \frac{[A]_1}{[A]_2}Kd=[A]1[A]2
where K d K d K_(d)K_dKd is the distribution coefficient, and [ A ] 1 [ A ] 1 [A]_(1)[A]_1[A]1 and [ A ] 2 [ A ] 2 [A]_(2)[A]_2[A]2 are the concentrations of the solute in solvents 1 and 2, respectively.

2. Thermodynamic Basis

The derivation is based on the concept of chemical potential ( μ μ mu\muμ). At equilibrium, the chemical potential of the solute in both solvents is the same. The chemical potential of a substance can be expressed as:
μ = μ 0 + R T ln ⁡ a μ = μ 0 + R T ln ⁡ a mu=mu^(0)+RT ln a\mu = \mu^0 + RT \ln aμ=μ0+RTln⁡a
where μ 0 μ 0 mu^(0)\mu^0μ0 is the standard chemical potential, R R RRR is the gas constant, T T TTT is the temperature, and a a aaa is the activity of the solute, which is proportional to its concentration.

3. Derivation

Let’s consider a solute distributed between two solvents, 1 and 2. At equilibrium, the chemical potential of the solute in both solvents is equal:
μ 1 = μ 2 μ 1 = μ 2 mu_(1)=mu_(2)\mu_1 = \mu_2μ1=μ2
Substituting the expression for chemical potential, we get:
μ 1 0 + R T ln ⁡ a 1 = μ 2 0 + R T ln ⁡ a 2 μ 1 0 + R T ln ⁡ a 1 = μ 2 0 + R T ln ⁡ a 2 mu_(1)^(0)+RT ln a_(1)=mu_(2)^(0)+RT ln a_(2)\mu^0_1 + RT \ln a_1 = \mu^0_2 + RT \ln a_2μ10+RTln⁡a1=μ20+RTln⁡a2
Rearranging and isolating the terms, we have:
R T ln ⁡ a 1 − R T ln ⁡ a 2 = μ 2 0 − μ 1 0 R T ln ⁡ a 1 − R T ln ⁡ a 2 = μ 2 0 − μ 1 0 RT ln a_(1)-RT ln a_(2)=mu_(2)^(0)-mu_(1)^(0)RT \ln a_1 – RT \ln a_2 = \mu^0_2 – \mu^0_1RTln⁡a1−RTln⁡a2=μ20−μ10
Using the properties of logarithms, this can be simplified to:
R T ln ⁡ ( a 1 a 2 ) = μ 2 0 − μ 1 0 R T ln ⁡ a 1 a 2 = μ 2 0 − μ 1 0 RT ln((a_(1))/(a_(2)))=mu_(2)^(0)-mu_(1)^(0)RT \ln \left( \frac{a_1}{a_2} \right) = \mu^0_2 – \mu^0_1RTln⁡(a1a2)=μ20−μ10
Now, taking the exponential of both sides:
a 1 a 2 = e μ 2 0 − μ 1 0 R T a 1 a 2 = e μ 2 0 − μ 1 0 R T (a_(1))/(a_(2))=e^((mu_(2)^(0)-mu_(1)^(0))/(RT))\frac{a_1}{a_2} = e^{\frac{\mu^0_2 – \mu^0_1}{RT}}a1a2=eμ20−μ10RT
Since the activities are proportional to the concentrations, we can write:
[ A ] 1 [ A ] 2 = K d = e μ 2 0 − μ 1 0 R T [ A ] 1 [ A ] 2 = K d = e μ 2 0 − μ 1 0 R T ([A]_(1))/([A]_(2))=K_(d)=e^((mu_(2)^(0)-mu_(1)^(0))/(RT))\frac{[A]_1}{[A]_2} = K_d = e^{\frac{\mu^0_2 – \mu^0_1}{RT}}[A]1[A]2=Kd=eμ20−μ10RT

4. Conclusion

The distribution law shows that the ratio of the concentrations of a solute in two immiscible solvents at equilibrium is constant at a given temperature. This constant is the distribution coefficient, K d K d K_(d)K_dKd, and it depends on the difference in standard chemical potentials of the solute in the two solvents and the temperature.
This derivation highlights the thermodynamic underpinnings of the distribution law, emphasizing the role of chemical potential and equilibrium in the distribution of substances between phases.
(b) Write any two applications of solvent extraction.
Answer:
Solvent extraction, a widely used technique in chemical and biochemical processes, involves the separation of compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. Here are two significant applications of this technique:

1. Metal Purification and Recovery

Solvent extraction is extensively used in the metallurgical industry for the purification and recovery of metals from ores and secondary sources (like electronic waste). The process is particularly important in the extraction and purification of rare and precious metals, as well as in the production of high-purity metals.
  • Example: Copper Extraction
    • In the hydrometallurgical processing of copper ores, solvent extraction is used to separate copper from other metal ions in a leach solution. The copper is selectively extracted into an organic solvent, leaving behind impurities in the aqueous phase. After extraction, the copper is transferred from the organic solvent into an aqueous solution from which it can be electrowon.
    • This process is known as the solvent extraction-electrowinning (SX-EW) process and is used to produce high-purity copper from low-grade oxide ores, which are not suitable for direct smelting.

2. Pharmaceutical and Biochemical Applications

In the pharmaceutical and biochemical industries, solvent extraction is employed for the isolation and purification of natural products, the preparation of pharmaceuticals, and in various stages of drug development.
  • Example: Extraction of Natural Products
    • Many natural products, such as essential oils, alkaloids, and antibiotics, are extracted from plant or microbial sources using solvent extraction. The choice of solvent is crucial and depends on the solubility of the target compound and the matrix of the source material.
    • For instance, the extraction of essential oils from plant materials (like lavender or eucalyptus) often involves using a non-polar solvent to dissolve the oil, followed by separation from the aqueous phase. This method allows for the recovery of concentrated extracts without the degradation of heat-sensitive components, which is a risk in distillation methods.

Conclusion

Solvent extraction is a versatile and efficient method for separating and purifying compounds in various industries. Its applications range from metal recovery and purification in the mining and metallurgical sectors to the extraction of valuable compounds in pharmaceutical and natural product chemistry. The technique’s effectiveness lies in its ability to selectively dissolve and separate specific components based on their differing solubilities in two immiscible solvents.
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