BT-101 Engineering Chemistry - Jun 2025 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aWrite short note on i) Sludge and Scale ii) Caustic Embrittlement iii) Boiler Corrosion
   How to answer

   • Define sludge and scale, their formation, differences, disadvantages, and prevention.
   • Explain caustic embrittlement mechanism (caustic soda concentration at hair cracks) and prevention (adding sodium sulphate, tannin, lignin).
   • Describe boiler corrosion causes (dissolved oxygen, dissolved carbon dioxide, dissolved salts) and their removal methods.
2. 7 Marks
   b$100\text{ ml}$ water sample required $20\text{ ml}$ of $\text{N}/50\ \text{H}_2\text{SO}_4$ for neutralisation to Phenolphthalein end point. After Methyl orange was added to this and further acid required was $2.5\text{ ml}$. Calculate the alkalinity type and extent in terms of $\text{CaCO}_3$ equivalent.
   How to answer

   • Calculate phenolphthalein alkalinity $P = \frac{20 \times (1/50) \times 50 \times 1000}{100} = 200\text{ ppm}$.
   • Calculate total alkalinity $M = \frac{(20 + 2.5) \times (1/50) \times 50 \times 1000}{100} = 225\text{ ppm}$.
   • Determine relationship: $P > \frac{M}{2}$, hence both $\text{OH}^-$ and $\text{CO}_3^{2-}$ alkalinity are present.
   • Calculate $\text{CO}_3^{2-}$ alkalinity $= 2(M - P) = 2(225 - 200) = 50\text{ ppm}$.
   • Calculate $\text{OH}^-$ alkalinity $= 2P - M = 2(200) - 225 = 175\text{ ppm}$.
3. 7 Marks
   aAn oil sample under test has a Saybolt Universal Viscosity same as that of standard Gulf oil (low viscosity standard) and Pennsylvanian oil (high viscosity standard) at $210^\circ\text{F}$. Their Saybolt Universal Viscosities at $100^\circ\text{F}$ are $61$, $758$, and $420$ seconds, respectively. Calculate viscosity index of the sample.
   How to answer

   • Identify given parameters: $L = 758\text{ s}$, $H = 61\text{ s}$, $U = 420\text{ s}$.
   • State the Viscosity Index formula: $\text{VI} = \frac{L - U}{L - H} \times 100$.
   • Substitute values: $\text{VI} = \frac{758 - 420}{758 - 61} \times 100 = \frac{338}{697} \times 100$.
   • Calculate final result: $\text{VI} \approx 48.5$.
4. 7 Marks
   bDescribe ion exchange process used for softening of water with neat diagram. A zeolite softener was $90\%$ exhausted by removing the hardness completely when $10,000\text{L}$ of hard water sample passed through it. The exhausted zeolite bed required $200\text{L}$ of $3\%\ \text{NaCl}$ solution for its complete regeneration. Calculate the hardness of water sample.
   How to answer

   • Describe ion exchange (demineralization) resin process: cation exchange resin ($\\text{RH}_2$) and anion exchange resin ($\text{R}'\text{OH}$), with chemical equations.
   • Draw schematic diagram of ion-exchange column setup with regeneration units.
   • Calculate mass of $\text{NaCl} = 200 \times 0.03 = 6\text{ kg} = 6000\text{ g}$.
   • Convert to $\text{CaCO}_3$ equivalent: $6000 \times \frac{50}{58.5} = 5128.2\text{ g}$.
   • Account for $90\%$ exhaustion: Total capacity $= 5128.2\text{ g}$, hardness removed $= 0.90 \times 5128.2 = 4615.38\text{ g}$ of $\text{CaCO}_3$ equivalent (or directly using bed exhaustion values).
   • Calculate water hardness: $\frac{4615.38 \times 10^3\text{ mg}}{10,000\text{ L}} \approx 461.5\text{ mg/L (ppm)}$.
5. 7 Marks
   aExplain the following properties of the following (any two) i) Aniline Point ii) Cloud and Pour Point iii) Flash and Fire Point
   How to answer

   • Aniline Point: Minimum equilibrium solution temperature for equal volumes of aniline and oil; indicates aromatic content.
   • Cloud and Pour Point: Cloud point is the temperature at which wax crystals begin to separate; pour point is the lowest temperature at which oil flows.
   • Flash and Fire Point: Flash point is the lowest temperature giving ignitable vapors; fire point is the temperature where oil burns continuously for at least 5 seconds.
6. 7 Marks
   bDistinguish between thermoplastic and thermosetting resins. Give preparation and application of Urea-Formaldehyde and PVC.
   How to answer

   • Tabulate key differences: structure (linear/branched vs cross-linked), action of heat (soften reversibly vs set permanently), recyclability, solubility.
   • Preparation of Urea-Formaldehyde resin with reaction scheme and state major applications.
   • Preparation of PVC from vinyl chloride monomer and state common applications (pipes, cables, insulation).
7. 7 Marks
   aDifferentiate chain growth and step growth polymerization. Explain Free radical and Ionic polymerization with the help of suitable example.
   How to answer

   • Compare chain growth and step growth polymerization with at least 4 clear distinctions.
   • Explain Free Radical mechanism: initiation, propagation, and termination using an example like ethylene/styrene.
   • Explain Ionic polymerization (cationic and anionic) with initiators and mechanisms using isobutylene/styrene.
8. 7 Marks
   bDefine Corrosion. Write types and prevention of corrosion. Explain theory of mechanism of corrosion. How much rust ($\text{Fe}_2\text{O}_3\cdot 3\text{H}_2\text{O}$) will be formed, when $100\text{ kg}$ of iron have completely rusted away? (Mol. Wt. of rust = $214\text{ g}$).
   How to answer

   • Define corrosion: destruction or deterioration of a metal by chemical or electrochemical reaction with its environment.
   • List types (dry/chemical and wet/electrochemical) and prevention methods (cathodic protection, coatings, inhibitors).
   • Explain electrochemical mechanism: anodic oxidation ($\text{Fe} \rightarrow \text{Fe}^{2+} + 2\text{e}^-$) and cathodic reduction ($\text{O}_2$ absorption or $\text{H}_2$ evolution).
   • Calculate rust formed: $2\text{ moles of Fe}$ ($2 \times 55.85 = 111.7\text{ g}$) yield $1\text{ mole of rust}$ ($214\text{ g}$).
   • Mass of rust $= 100\text{ kg} \times \frac{214}{111.7} \approx 191.58\text{ kg}$.
9. 7 Marks
   aDefine and explain the terms involved in the phase rule. Draw a neat labelled phase diagram of water system and explain the areas and curves in it. What is the significance of the triple point and meta stable curve in this system?
   How to answer

   • Define phase ($P$), component ($C$), and degrees of freedom ($F$) with Gibbs phase rule $F = C - P + 2$.
   • Draw pressure-temperature phase diagram of the water system with labeled curves (OA, OB, OC) and regions (ice, water, steam).
   • Explain curves (univariant, $F=1$) and areas (bivariant, $F=2$).
   • Explain significance of triple point ($F = 0$, $T = 0.0098^\circ\text{C}, P = 4.58\text{ mm Hg}$) and the metastable curve (OA', supercooled water).
10. 7 Marks
    bWhat are the important properties that an efficient lubricant should possess? $1.5\text{ gm}$ of a lubricant oil dissolved in alcohol require $24\text{ ml}$ of $0.0125\text{ N}$ alcoholic $\text{KOH}$ for neutralization. What is neutralization value of the lubricant? If an oil to be used in machinery should not have neutralization value greater than $8.0$, is it safe to use the test sample as a lubricant in the machinery?
    How to answer

    • List desirable lubricant properties: suitable viscosity and viscosity index, high flash/fire point, low pour point, low acid value/emulsification tendency.
    • Calculate neutralization value: $\text{NV} = \frac{24 \times 0.0125 \times 56}{1.5}$.
    • Perform calculation: $24 \times 0.0125 = 0.3$; $\text{NV} = \frac{0.3 \times 56}{1.5} = 11.2\text{ mg KOH/g}$.
    • Conclude safety: Since calculated $\text{NV} = 11.2 > 8.0$, the oil is NOT safe to use in the machinery.
11. 7 Marks
    aHow can a combination of UV and IR spectroscopy be used to determine the structure of an unknown organic compound? Illustrate with an example.
    How to answer

    • Role of UV spectroscopy: detects conjugation, chromophores, and auxochromes ($\pi \rightarrow \pi^*$, $n \rightarrow \pi^*$ transitions).
    • Role of IR spectroscopy: identifies specific functional groups via characteristic vibrational frequencies (e.g., carbonyl at ~1715 cm⁻¹, O-H at ~3300 cm⁻¹).
    • Illustrate with an example (e.g., distinguishing acetone vs allyl alcohol, or identifying an $\alpha,\beta$-unsaturated ketone).
12. 7 Marks
    bWhat is Electronic spectroscopy? An unknown substance absorbs $40\%$ of the light of wavelength $43.5\text{ nm}$ in a cell of $2\text{ cm}$ dimension when has a concentration of $20\text{ gL}^{-1}$. Under the similar conditions, what percentage of light will be absorbed by a solution containing $40\text{ gL}^{-1}$ of the same unknown substance?
    How to answer

    • Define electronic spectroscopy: absorption of UV-visible radiation causing electronic transitions between energy levels.
    • Calculate absorbance $A_1$: Since $40\%$ light absorbed, $T_1 = 0.60$; $A_1 = -\log_{10}(0.60) = 0.2218$.
    • Calculate $A_2$ at doubled concentration ($c_2 = 2 c_1$): $A_2 = 2 \times A_1 = 0.4437$.
    • Find transmittance $T_2 = 10^{-0.4437} = (0.60)^2 = 0.36$.
    • Calculate percentage of light absorbed: $(1 - 0.36) \times 100 = 64\%$.
13. 7 Marks
    aWhat is Lime Soda process? Calculate the amount of lime and soda required for softening of $20,000\text{ liters}$ of water whose analysis is: $\text{Ca(HCO}_3)_2 = 40.0\text{ ppm}$; $\text{Mg(HCO}_3)_2 = 36.5\text{ ppm}$; $\text{CaSO}_4 = 34.0\text{ ppm}$; $\text{MgSO}_4 = 30.0\text{ ppm}$; $\text{CaCl}_2 = 27.75\text{ ppm}$.
    How to answer

    • Explain Lime Soda process: precipitation of soluble calcium and magnesium salts as insoluble $\text{CaCO}_3$ and $\text{Mg(OH)}_2$ using lime ($\text{Ca(OH)}_2$) and soda ($\text{Na}_2\text{CO}_3$).
    • Convert each constituent to $\text{CaCO}_3$ equivalents:
    • $\text{Ca(HCO}_3)_2 = 40.0 \times \frac{100}{162} = 24.69\text{ ppm}$; $\text{Mg(HCO}_3)_2 = 36.5 \times \frac{100}{146} = 25.0\text{ ppm}$; $\text{CaSO}_4 = 34.0 \times \frac{100}{136} = 25.0\text{ ppm}$; $\text{MgSO}_4 = 30.0 \times \frac{100}{120} = 25.0\text{ ppm}$; $\text{CaCl}_2 = 27.75 \times \frac{100}{111} = 25.0\text{ ppm}$.
    • Calculate Lime requirement $= \frac{74}{100} [24.69 + 2(25.0) + 25.0] \times \frac{20,000}{10^6} = \frac{74}{100} [99.69] \times 0.02 = 1.475\text{ kg}$.
    • Calculate Soda requirement $= \frac{106}{100} [25.0 + 25.0 + 25.0] \times \frac{20,000}{10^6} = \frac{106}{100} [75.0] \times 0.02 = 1.59\text{ kg}$.
14. 7 Marks
    bWrite brief note on (Any two): i) Electron affinity ii) Effective Nuclear Charge iii) Aufbau rule iv) Atomic and Ionic size
    How to answer

    • Electron affinity: Energy released when an electron is added to an isolated neutral gaseous atom, periodic trends.
    • Effective Nuclear Charge ($Z_{eff}$): Net positive charge experienced by valence electrons ($Z_{eff} = Z - S$), Slater's rules.
    • Aufbau principle: Electrons occupy orbitals in order of increasing energy ($(n+l)$ rule).
    • Atomic and Ionic size: Definition of radii, variations along periods (decreases) and groups (increases), cation < atom < anion.
15. 7 Marks
    aWrite brief note on (Any two) i) Electro negativity ii) Polarizability iii) Oxidation states
    How to answer

    • Electronegativity: Tendency of an atom in a molecule to attract shared pair of electrons toward itself, Pauling scale, periodic trend.
    • Polarizability: Ease with which the electron cloud of an atom/ion is distorted by an electric field/cation, Fajan's rules.
    • Oxidation states: Apparent charge an atom carries in a compound based on electronegativity, variable oxidation states of d-block elements.
16. 7 Marks
    bCalculate temporary, permanent and total hardness of water sample containing $\text{Mg(HCO}_3)_2 = 7.3\text{ mg/L}$, $\text{Ca(HCO}_3)_2 = 16.4\text{ mg/L}$, $\text{MgCl}_2 = 9.5\text{ mg/L}$ and $\text{CaSO}_4 = 13.6\text{ mg/L}$.
    How to answer

    • Calculate $\text{CaCO}_3$ equivalent for each salt:
    • $\text{Mg(HCO}_3)_2 = 7.3 \times \frac{100}{146} = 5.0\text{ mg/L}$.
    • $\text{Ca(HCO}_3)_2 = 16.4 \times \frac{100}{164} = 10.0\text{ mg/L}$ (using molecular weight $\approx 164$ or $16.2 \rightarrow 10$).
    • $\text{MgCl}_2 = 9.5 \times \frac{100}{95} = 10.0\text{ mg/L}$.
    • $\text{CaSO}_4 = 13.6 \times \frac{100}{136} = 10.0\text{ mg/L}$.
    • Temporary hardness $= 5.0 + 10.0 = 15.0\text{ mg/L}$ (or $15.12\text{ mg/L}$ if exact mw 162 used for $\text{Ca(HCO}_3)_2$).
    • Permanent hardness $= 10.0 + 10.0 = 20.0\text{ mg/L}$.
    • Total hardness $= 15.0 + 20.0 = 35.0\text{ mg/L}$ as $\text{CaCO}_3$ equivalent.