Category: Basic Chemistry K-Scheme Lab Manual Answer

 Practical No. 15: Determination of Thinner Content in Oil Paint

• Conclusion: Oil paints contain significant volatile thinner which evaporates during drying.
• Recommendation: Preheat oven, use desiccator for cooling, and accurately weigh samples for precise results.

• Answer: The weight of paint decreases after heating because the volatile components (thinner/solvent) present in the paint evaporate due to heat. Paint is a mixture of pigments, binders, and volatile solvents. When heated at 120°C, the volatile thinner evaporates completely, leaving behind only the non-volatile solid components (pigments and binders), resulting in weight loss.

• Answer: A desiccator is used to cool heated samples in a moisture-free environment. It prevents the absorption of moisture from the atmosphere by the dried sample during cooling. This ensures accurate weight measurement as any absorbed moisture would increase the weight, leading to incorrect calculations of thinner content. It contains a desiccant like CaCl₂ or silica gel to absorb moisture.

• Answer: The oven should be preheated to the required temperature (120°C) before placing the paint sample so that the sample immediately reaches the desired temperature. This ensures consistent heating conditions and complete evaporation of thinner. If the oven is not preheated, the sample would heat gradually, giving variable results and incomplete drying of the paint.

• Answer: CaCl₂ (Calcium chloride) , not CaCO₃, is commonly used as a desiccant. Calcium chloride is a hygroscopic substance that absorbs moisture from the air inside the desiccator, maintaining a dry environment. This prevents the cooled sample from absorbing atmospheric moisture, ensuring accurate weight measurement after heating.

• Answer: The paint sample is heated in the electric oven at 120°C for one hour. This temperature is sufficient to evaporate all volatile thinner content from the paint without decomposing the non-volatile components. The temperature is maintained constant throughout the heating process to ensure complete evaporation of thinner.

• Conclusion: Closed cup method is more suitable for oils with low flash points (below 49°C).
• Recommendation: Use correct thermometer, maintain 1-1.5°C per minute heating, and conduct test in dark room for better visibility.

• Answer: Precautions include:
  • (i) Fill oil carefully to avoid bubbles and ensure oil is not above the filling mark,
  • (ii) Conduct test in a dark room for better visibility of flash,
  • (iii) Use correct range of thermometer,
  • (iv) Maintain temperature increase at 1 to 1.5°C per minute,
  • (v) Stir oil continuously except during introduction of test flame,
  • (vi) Ensure sliding shutter operates smoothly,
  • (vii) Standard test flame should be of prescribed size.

• Answer: Limitations include:
  • (i) Suitable only for oils with flash point below 49°C,
  • (ii) Not suitable for viscous oils,
  • (iii) Closed cup method may not reflect actual fire hazard in open conditions,
  • (iv) Results may vary if the apparatus is not cleaned properly,
  • (v) Requires careful temperature control and dark room conditions for accurate results.

• Answer: Flash point indicates the minimum temperature at which oil gives sufficient vapours to ignite momentarily, helping determine storage and transport temperature limits. Fire point indicates the temperature at which oil burns continuously for at least 5 seconds. Both parameters are crucial for fire safety, detecting volatile contaminants, identifying oils, and selecting appropriate lubricants for specific working temperatures.

• Answer: Apparatus used:
  • (i) Cleveland Open Cup Apparatus – for flash and fire point of high flash point oils,
  • (ii) Abel’s Closed Cup Apparatus – for flash point of oils below 49°C,
  • (iii) Pensky-Martens Closed Cup Apparatus – for flash point of oils,
  • (iv) Tagliabue Open Cup Apparatus – for flash and fire point, and
  • (v) Koehler Open Flash Tester.

 Practical No. 13: Flash and Fire Point by Cleveland’s Open Cup Apparatus

• Conclusion: The lubricating oil can be safely used up to its flash point temperature.
• Recommendation: Conduct test in draft-free area, avoid breathing over oil, and heat at 3-5°C per minute for accurate results.

• Answer: Precautions include:
  • (i) Perform the test in a draft-free laboratory space.
  • (ii) Avoid breathing directly over the open oil surface to prevent inhalation of vapours.
  • (iii) Maintain an accurate heating rate of 3 to 5°C per minute.
  • (iv) Introduce the test flame from the standard specified distance.
  • (v) Use the correct temperature range thermometer.
  • (vi) Fill the oil cup exactly up to the filling mark.
  • (vii) Ensure the test flame is kept at a standard bead size.
  • (viii) Note the temperature carefully the exact moment the first flash appears.

• Answer: Cleveland open cup apparatus is generally used for the determination of flash points of fuel oils and other heavy oils having flash points above 79°C. It is highly suitable for oils with higher flash points and is commonly used for industrial lubricating oils, heavy petroleum products, and other high-viscosity industrial fluids. It is not suitable for volatile oils with very low flash points.

• Answer: The thermometer is mounted vertically by means of a firm laboratory clamp in such a way that the bottom of the bulb is about 1 cm above the absolute bottom of the cup. The thermometer should be positioned centrally and must not touch the metallic sides of the cup. The bulb should be completely immersed in the oil sample without touching the container to ensure accurate measurement of oil temperature.

• Answer: The oil should be heated at a steady rate of about 3 to 5°C per minute during the general determination of flash and fire points using Cleveland’s open cup apparatus. As the temperature approaches the expected flash point, the heating rate is carefully controlled, and the test flame is introduced periodically after every degree rise in temperature.

 Practical No. 12: Steam Emulsification Number

• Conclusion: An oil with the lowest steam emulsification number is best for industrial applications as it breaks the emulsion quickly.
• Recommendation: Shake all tubes vigorously for the exact same duration and accurately record the separation time using a precise stopwatch.

• Answer: A good lubricating oil has a low steam emulsification number because it does not form stable emulsions with water. A low emulsion tendency prevents the formation of oil-water mixtures that can trap dirt, grit, and other foreign matter. If trapped, these dirt particles cause abrasion and wear of lubricated machine parts. A low number ensures that if water enters the system, the mixture breaks quickly into distinct layers, allowing smooth lubrication.

• Answer: No, an emulsion cannot be formed by mixing two miscible liquids. An emulsion is a colloidal dispersion of two immiscible liquids (like oil and water) where one liquid is dispersed in the other as fine droplets. Miscible liquids completely dissolve in each other to form a single homogeneous solution, not an emulsion. Emulsions strictly require two immiscible liquids and an emulsifying agent to stabilize the mixture.

 Practical No. 11: Determination of Effect of Temperature on Viscosity using Redwood Viscometer-I

• Conclusion: Viscosity decreases with an increase in temperature, which increases the flow rate of the oil.
• Recommendation: Filter the oil thoroughly before testing, level the viscometer properly, and ensure accurate temperature control for reliable measurements.

• Answer: Clean the viscometer by following these steps:
  • (i) Removing all residual oil from the cup.
  • (ii) Using a suitable solvent (like petroleum ether or benzene) to dissolve any remaining oil.
  • (iii) Rinsing thoroughly with the solvent and then with distilled water.
  • (iv) Passing a stream of air to remove all traces of the solvent.
  • (v) Ensuring the jet is completely clean and unobstructed.
  • (vi) Drying the viscometer completely before its next use.

• Answer: The water bath provides a constant and uniform temperature for heating the oil sample. It ensures that the oil reaches the desired temperature uniformly without any localized overheating. Maintaining a constant temperature during the experiment is essential because the viscosity of oil is highly dependent on temperature.

• Answer: Precautions include:
  • (i) Filter the oil to remove any solid particles or impurities.
  • (ii) Level the viscometer perfectly using the leveling screws.
  • (iii) Place the receiving flask correctly to avoid foaming.
  • (iv) Maintain an accurate temperature by constant stirring of the water bath.
  • (v) Start and stop the stopwatch precisely when the oil reaches the markers.
  • (vi) Ensure the oil stream strikes the neck of the receiving flask.
  • (vii) Completely drain the oil after each reading.

• Answer: The receiving flask should be placed immediately below and in line with the discharging jet. The oil stream from the jet should strike the neck of the receiving flask to prevent foaming. The flask must be kept steady and not moved during the experiment, and the 50 ml mark should be clearly visible at eye level for accurate timing.

• Answer: Types of viscometers include:
  • (i) Redwood Viscometer (No. I and No. II)
  • (ii) Saybolt Viscometer
  • (iii) Engler Viscometer
  • (iv) Ostwald Viscometer
  • (v) Brookfield Viscometer (rotational)
  • (vi) Falling Ball Viscometer
  • (vii) Capillary Viscometer
  • (viii) Rotational Viscometers

 Practical No. 10: Determination of Rate of Corrosion at Different Temperatures for Aluminium

• Conclusion: Metal deterioration is maximum in acidic medium at higher temperatures.
• Recommendation: Handle acids carefully, dry aluminium strips completely before weighing, and maintain accurate temperature control for reliable results.

• Answer: Maximum change in weight is observed in Hydrochloric acid (HCl) among all the acids tested. This is because HCl is a strong acid and aluminium reacts vigorously with HCl, displacing hydrogen and forming aluminium chloride. The reaction rate is maximum in HCl, leading to the highest weight loss.

• Answer: The gas liberated is Hydrogen (H₂).
• Chemical Reaction: 2Al + 6HCl → 2AlCl₃ + 3H₂↑. The hydrogen gas is released as bubbles during the reaction between aluminium and hydrochloric acid.

• Answer: The compound formed is Aluminium Chloride (AlCl₃). The reaction produces aluminium chloride which dissolves in the solution, and the aluminium strip loses weight as it dissolves.

• Answer: No protective film is formed when aluminium is dipped in HCl. In fact, the oxide layer (Al₂O₃) which normally protects aluminium from corrosion is also attacked by HCl, exposing the metal surface to further corrosion. However, in HNO₃, a protective passive oxide film is formed.

 Practical No. 09: Preparation of Corrosive Medium for Aluminium

• Conclusion: Corrosive media of desired concentrations can be prepared using normality equation.
• Recommendation: Always add acid to water (never water to acid) with stirring, wear safety equipment, and label all solutions properly.

• Answer: When metal comes in contact with acids/bases, Wet corrosion (also called Electrochemical corrosion) takes place. In acidic medium, hydrogen gas is liberated, and the metal dissolves forming metal salts. In alkaline medium, metals like aluminium dissolve forming metal aluminates. This type of corrosion is due to the electrochemical reaction at the metal surface.

• Answer: Precautions include:
  • (i) Always add acid to water (never add water to concentrated acid) to prevent splashing.
  • (ii) Use safety goggles and gloves.
  • (iii) Stir continuously while adding acid.
  • (iv) Use proper glassware and handle carefully.
  • (v) Work in a well-ventilated area.
  • (vi) Prepare solution in a heat-resistant glass beaker as dilution generates heat.

• Answer:
  • Formula: N₁V₁ = N₂V₂
  • Given: N₁ = 10 N, N₂ = 2 N, V₂ = 250 ml
  • Calculation:
    V₁ = (N₂ × V₂) / N₁
    V₁ = (2 × 250) / 10
    V₁ = 50 ml
  • Conclusion: Therefore, take 50 ml of 10 N HCl and make the volume up to 250 ml by adding distilled water slowly with constant stirring.

 Practical No. 08: Determination of Equivalent Weight of Metal using Faraday’s Second Law

• Conclusion: Same quantity of electricity deposits weights proportional to equivalent weights, verifying Faraday’s second law.
• Recommendation: Clean electrodes carefully, maintain constant current, and accurately measure weights for reliable results.

• Answer: Cleaning removes impurities, oxide layers, or grease from electrode surfaces. This ensures proper adhesion of deposited metal and accurate weight measurement. Clean surfaces provide better electrical contact and uniform deposition. It prevents contamination of the electrolyte and ensures the reaction occurs properly at the electrode surface.

• Answer: During electrolysis, the anodes (positive electrodes) undergo oxidation and dissolve into the electrolyte as metal ions. The copper anode dissolves as Cu²⁺ ions, and the Zinc anode dissolves as Zn²⁺ ions. This dissolution causes a decrease in the weight of both anodes. The mass lost from the anode equals the mass deposited at the cathode.

• Answer: According to Faraday’s laws, the amount of substance deposited is directly proportional to the time for which current is passed when current is constant. W ∝ t. Therefore, longer electrolysis time results in greater deposition of metal on the cathode, provided current remains constant throughout the process.

• Answer: The increase in cathode weight indicates successful electrodeposition of metal on the cathode. It confirms that the ions from the electrolyte are being reduced and deposited on the electrode. This increase is used to calculate the electrochemical equivalent and equivalent weight of the metal, making it essential for verifying Faraday’s laws.

 Practical No. 07: Electrochemical Equivalent of Copper using Faraday’s First Law

• Conclusion: Weight of substance deposited is directly proportional to quantity of electricity passed.
• Recommendation: Clean cathode thoroughly, maintain constant current, and accurately record time for precise results.

• Answer: Chemical equivalence refers to the equivalent weight of a substance (atomic weight / valency). Electrochemical equivalent is the weight of substance deposited by 1 coulomb of electricity. They are related by Faraday’s laws: Electrochemical equivalent (Z) = Equivalent weight / 96500. Thus, electrochemical equivalent is directly proportional to chemical equivalent.

• Answer: According to Faraday’s first law, the weight of substance deposited (W) is directly proportional to the time (t) for which current is passed, provided current (I) is constant. W ∝ t, or W = Z × I × t, where Z is electrochemical equivalent. Thus, longer time results in greater deposition of substance.

• Answer: The copper anode is a consumable/sacrificial anode in this electrolysis process. As current passes, the copper anode dissolves into Cu²⁺ ions, which migrate to the cathode where they get deposited. The anode loses mass, and this type of anode is called a “soluble anode.”

 Practical No. 06: Determination of Voltage Generated from Chemical Reaction using Daniel Cell

• Conclusion: Voltage depends on electrolyte concentrations, with maximum voltage at maximum concentrations.
• Recommendation: Clean electrodes, use fresh salt bridge, and ensure tight connections for accurate voltage measurement.

• Answer: In a Daniel cell, Zinc electrode (Zn) acts as the Anode (negative terminal) where oxidation occurs. Copper electrode (Cu) acts as the Cathode (positive terminal) where reduction occurs. Electrons flow from zinc anode through the external circuit to copper cathode.

• Answer:
  • At Anode (Zinc): Zn → Zn²⁺ + 2e⁻ (Oxidation).
  • At Cathode (Copper): Cu²⁺ + 2e⁻ → Cu (Reduction).
  • Net reaction: Zn + Cu²⁺ → Zn²⁺ + Cu.
  • Observation: The zinc dissolves, and copper deposits on the copper electrode.

• Answer: A half cell is a single electrode immersed in a solution of its own ions. It consists of a metal electrode and an electrolyte solution containing the corresponding metal ions. When two half cells are connected, they form an electrochemical cell. Each half cell has its own electrode potential, and the combination of two half cells gives the total cell potential.

• Answer: The salt bridge maintains electrical neutrality by allowing the flow of ions between the two half cells. It prevents the accumulation of excess positive charges near the anode and negative charges near the cathode. It completes the electrical circuit and allows the continuous flow of electrons in the external circuit, maintaining the cell potential.

• Answer: Suitable electrolytes for a salt bridge include:
  • (i) KCl (Potassium chloride)
  • (ii) KNO₃ (Potassium nitrate)
  • (iii) Na₂SO₄ (Sodium sulphate)
  • (iv) NH₄NO₃ (Ammonium nitrate)
  • Note: The ions should not react with the electrode solutions and should have similar ionic mobilities.