Maharashtra Board Class 12 Chemistry Solutions 2022 PDF

www.collegedunia.com Maharashtra Board Class 12 Chemistry Solutions 2022 SECTION A Question 1. Select and write the correct answer for the following multiple choice type of questions: (i) The co-ordination number of atoms in body centred cubic struct...

www.collegedunia.com Maharashtra Board Class 12 Chemistry Solutions 2022 SECTION A Question 1. Select and write the correct answer for the following multiple choice type of questions: (i) The co-ordination number of atoms in body centred cubic structure (bcc) is _______. (a) 4 (b) 6 (c) 8 (d) 12 Answer. (c) Solution. (c) 8 The coordination number of an atom in a body-centered cubic (BCC) structure is 8. This means that each atom in a BCC lattice is directly bonded to 8 other atoms. This can be visualized by considering the unit cell of a BCC lattice, which contains one atom in the center and eight atoms at the corners. The central atom is in contact with all eight corner atoms, so its coordination number is 8. (ii) In calculating osmotic pressure, the concentration of solute is expressed in _______. (a) molarity (b) molality (c) mole fraction  www.collegedunia.com (d) percentage mass Answer. (a) (iii) The enthalpy change for the chemical reaction H2 O(s) H2 O(l) is called enthalpy of _______. (a) vapourisation (b) fusion (c) combustion (d) sublimation Answer. (b) Solution. (b) fusion The enthalpy change for the chemical reaction H2 O(s) H2 O(l) is called the enthalpy of fusion. This is because the reaction involves a change in the physical state of water from solid (ice) to liquid (water). The enthalpy of fusion is the amount of energy required to melt one mole of a solid substance at its melting point. The other options are not correct because: Enthalpy of vaporization is the amount of energy required to vaporize one mole of a liquid substance at its boiling point. Enthalpy of combustion is the amount of energy released when one mole of a substance is completely burned in oxygen. Sublimation is the transition of a solid directly to a gas, and it does not involve a liquid phase. Therefore, the correct answer is (b) fusion.  www.collegedunia.com (iv) Which of the following transition element shows maximum oxidation state? (a) Sc (b) Fe (c) Mn (d) V Answer. (c) Solution. (c) Mn Manganese (Mn) exhibits the highest oxidation state among the given options, with an oxidation state of +7. This is due to its electronic configuration, [Ar]3d54s2, which allows it to lose up to seven electrons. This high oxidation state is observed in compounds like potassium permanganate (KMnO4), where manganese is in the +7 oxidation state. Scandium (Sc) typically exhibits an oxidation state of +3, iron (Fe) commonly exhibits oxidation states of +2 and +3, and vanadium (V) exhibits a range of oxidation states from +2 to +5. However, manganese's ability to reach an oxidation state of +7 sets it apart from the other options. Therefore, the correct answer is (c) Mn. (v) The correct formula for the complex compound, sodium hexacyanoferrate (III) is _______. (a) Na [Fe(CN)6 ] (b) Na 2 [Fe(CN)6 ] (c) Na 3 [Fe(CN)6 ] (d) Na 4 [Fe(CN)6 ] Answer. (c) Solution. The correct formula for sodium hexacyanoferrate(III) is (c) Na3[Fe(CN)6].  www.collegedunia.com This compound is an ionic coordination complex, where the central metal ion, iron (Fe3+), is surrounded by six cyanide ligands (CN−). The charge of the complex is −3, which is balanced by three sodium ions (Na+). The formula Na3[Fe(CN)6] represents this balance of charges. The other options are incorrect because: (a) Na[Fe(CN)6] would result in a negative charge of −6, which is not possible for a neutral compound. (b) Na2[Fe(CN)6] would result in a negative charge of −12, which is even further from the correct charge of −3. (d) Na4[Fe(CN)6] would result in a positive charge of +3, which is not possible for a neutral compound. Therefore, the correct answer is (c) Na3[Fe(CN)6]. (vi) Isopropylbenzene on air oxidation followed by decomposition by dilute acid gives _______. (a) C6H5OH (b) C6H5COOCH3 (c) C6H5COOH (d) C6H5CHO Answer. (a) Solution. The correct answer is (a) C6H5OH. Isopropylbenzene, also known as cumene, undergoes a two-step reaction when subjected to air oxidation followed by decomposition using dilute acid. Step 1: Air Oxidation In the presence of air and a catalyst, such as a transition metal ion, isopropylbenzene undergoes cumene hydroperoxide formation. This reaction involves the addition of oxygen to the tertiary carbon atom of  www.collegedunia.com isopropylbenzene, resulting in the formation of a hydroperoxide group (-OOH). Step 2: Decomposition by Dilute Acid Cumene hydroperoxide is unstable and readily decomposes in the presence of dilute acid. This decomposition reaction generates phenol (C6H5OH) and acetone (CH3COCH3) as the major products. Therefore, the correct answer is (a) C6H5OH. (vii) The name of metal nanoparticle which acts as highly effective bacterial disinfectant in water purification process is _______. (a) carbon black (b) silver (c) gold (d) copper Answer. (b) Solution. The correct answer is (b) silver. Silver nanoparticles have been widely studied and demonstrated to be highly effective bacterial disinfectants in water purification processes. Their antimicrobial activity is attributed to their ability to interact with bacterial cell membranes and disrupt their function, leading to cell death. Silver nanoparticles also possess broad-spectrum antimicrobial properties, making them effective against a wide range of bacteria. While other metal nanoparticles, such as carbon black, gold, and copper, have also shown some antimicrobial properties, silver nanoparticles have consistently demonstrated superior efficacy and are more widely used in water purification applications. Therefore, the correct answer is (b) silver.  www.collegedunia.com (viii) Acid anhydride on reaction with primary amine gives compound having a functional group _______. (a) amide (b) nitrile (c) secondary amine (d) imine Answer. (a) Solution. The correct answer is (a) amide. When an acid anhydride reacts with a primary amine, the resulting compound is an amide. Amides are a class of organic compounds that contain the carbonyl group (-C=O) bonded to a nitrogen atom (-NH2). The general reaction for the formation of an amide from an acid anhydride and a primary amine is as follows: RCO2O + R'NH2 → RCONHR' + RCOOH In this reaction, R and R' represent alkyl or aryl groups. The reaction proceeds through a nucleophilic attack of the amine on the carbonyl carbon of the anhydride, leading to the breaking of the anhydride bond and the formation of an amide bond. The other options are incorrect because: Nitriles (-C≡N) do not contain the amide functional group. Secondary amines (-NHR2) are not produced in the reaction between an acid anhydride and a primary amine. Imines (-C=N) are formed by the reaction between an aldehyde or ketone and an amine, not between an acid anhydride and an amine. Therefore, the correct answer is (a) amide.  www.collegedunia.com (ix) The standard potential of the cell in the following reaction is _______. Cd(s) + Cu2+(IM) → Cu2+(IM) + Cd(s) (E°Cd = –0.403V, E°Cu = 0.334 V) (a) – 0.737 V (b) 0.737 V (c) – 0.069 V (d) 0.069 V Answer. (b) Solution. (b) 0.737 V The standard potential of a cell is the potential difference between its electrodes when all the reactants and products are at standard concentrations (1 M) and the temperature is 25°C. It is denoted by E°cell and is calculated using the following equation: E°cell = E°cathode - E°anode In the given reaction, Cd is the anode and Cu is the cathode. Therefore, the standard cell potential is calculated as follows: E°cell = E°Cu - E°Cd = 0.334 V - (-0.403 V) = 0.737 V Therefore, the correct answer is (b) 0.737 V. (x) The value of [H3O+] in mol lit–1 of 0.001 M acetic acid solution (Ka = 1.8 *10–5 ) is _______. (a) 1.34*10–1 (b) 1.34*10–2 (c) 1.34*10–3 (d) 1.34*10–4  www.collegedunia.com Answer. (a) Question 2. Answer the following questions: (i) Write the product formed when alkyl halide reacts with silver nitrite. Answer. Nitroethane Solution. When an alkyl halide reacts with silver nitrite in an ethanolic medium, the product formed is a nitroalkane. The reaction proceeds through nucleophilic substitution, where the nitrogen atom of the silver nitrite attacks the carbon atom of the alkyl halide, resulting in the replacement of the halide ion with a nitro group (-NO2). The general reaction equation is as follows: R-X + AgNO2 → R-NO2 + AgX where R represents an alkyl group and X represents a halide ion (Cl, Br, I). For example, when chloroethane (CH3CH2Cl) reacts with silver nitrite, the product formed is nitroethane (CH3CH2NO2). (ii) Write the name of product formed, when acetone is treated with 2, 4-dinitrophenyl hydrazine. Answer. 2,4-dinitrophenylhydrazone Solution. When acetone (CH3COCH3) is treated with 2, 4-dinitrophenyl hydrazine (DNP, C6H4N4O4), the product formed is 2, 4-dinitrophenylhydrazone (DNP hydrazine). This reaction is a common method for identifying carbonyl compounds, such as ketones and aldehydes. The reaction proceeds through a condensation reaction, where the carbonyl group of the acetone attacks the amino group of the DNP hydrazine, resulting in the formation of a hydrazone bond (-N=N-C=). The DNP hydrazine also acts as a reducing agent, transferring two hydrogen  www.collegedunia.com atoms to the carbonyl carbon atom of the acetone, forming an alcohol group (-OH). The DNP hydrazone product is typically a yellow or orange precipitate, which can be used to identify and characterize the carbonyl compound. (iii) Write the name of biodegradable polyamide copolymer. Answer. Nylon-2 nylon-6 Solution. There are two main biodegradable polyamide copolymers: 1. Nylon 2-nylon 6: This copolymer is synthesized from glycine (H2N-CH2-COOH) and aminocaproic acid (H2N-(CH2)5-COOH). It is biodegradable under both aerobic and anaerobic conditions, making it a promising material for a variety of applications, such as packaging, medical implants, and textiles. 2. Polyhydroxybutyrate-co-polyhydroxyvalerate (PHBV-co-PHV): This copolymer is synthesized from the monomers 3-hydroxybutyrate (HB) and 3-hydroxyvalerate (HV). The ratio of HB to HV can be varied to control the properties of the copolymer, such as its biodegradability, flexibility, and toughness. PHBV-co-PHV is also biodegradable under both aerobic and anaerobic conditions and has a wide range of potential applications, including packaging, medical implants, and agricultural films. (iv) Identify the molecularity of following elementary reaction: NO(g) + O3(g) NO3(g) + O(g) Answer. 2 Solution. The molecularity of a reaction is the total number of reactant molecules that come together to form products in a single elementary step.  www.collegedunia.com In the given reaction, NO(g) + O3(g) → NO3(g) + O(g), two reactant molecules (NO and O3) collide to form two product molecules (NO3 and O). Therefore, the molecularity of this reaction is 2. (v) What is the action of selenium on magnesium metal? Answer. Magnesium selenide Solution. Selenium reacts with magnesium to form magnesium selenide, a brown-colored crystalline solid. The reaction proceeds through direct combination, where the selenium atoms share electrons with the magnesium atoms to form a solid compound. The general reaction equation is as follows: Mg + Se → MgSe Magnesium selenide is a semiconductor with a band gap of about 2 eV. It has potential applications in solar cells, light-emitting diodes (LEDs), and other electronic devices. (vi) Write the name of isomerism in the following complexes: [Cu(NH3 )4 ] [PtCl4 ] and [Pt(NH3 )4 ] [CuCl4] Answer. Coordination isomerism Solution. The type of isomerism exhibited by the complexes [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4] is coordination isomerism. Coordination isomerism occurs when the ligands in a complex are exchanged between the central metal ions. In this case, the ligands NH3 and Cl- are exchanged between the copper and platinum ions. The other types of isomerism are: Ionization isomerism: This type of isomerism occurs when two complexes have the same formula but different ions dissociate from  www.collegedunia.com them in solution. Linkage isomerism: This type of isomerism occurs when a ligand can be bonded to a central metal ion in two different ways. Structural isomerism: This type of isomerism occurs when two compounds have the same formula but different arrangements of atoms. Geometric isomerism: This type of isomerism occurs when two compounds have the same formula but different spatial arrangements of atoms. In the case of [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4], the ligands are not different, and the two complexes do not have different ions dissociating from them in solution. Therefore, the only type of isomerism that can be exhibited is coordination isomerism. (vii) Write the name of the alloy used in Fischer Tropsch process in the synthesis of gasoline. Answer. co-Th alloy Solution. The most commonly used alloy in the Fischer-Tropsch process for the synthesis of gasoline is a cobalt-thorium (Co-Th) alloy. This alloy is known for its high activity and selectivity towards the production of gasoline-range hydrocarbons. The thorium acts as a promoter, enhancing the activity of the cobalt and improving the selectivity of the reaction. Other alloys that have been used in the Fischer-Tropsch process include iron-cobalt (Fe-Co), iron-ruthenium (Fe-Ru), and cobalt-ruthenium (Co-Ru) alloys. However, the Co-Th alloy is generally considered to be the most effective catalyst for the production of gasoline. (viii) Henry’s law constant for CH3Br(g) is 0.159 mol dm–3 bar–1 at 25°C.  www.collegedunia.com What is solubility of CH3Br(g) in water at same temperature and partial pressure of 0.164 bar? Answer. 0.026 Solution. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. This relationship can be expressed mathematically as: S = kH * p where: S is the solubility of the gas in mol/dm3 kH is Henry's law constant in mol dm-3 bar-1 p is the partial pressure of the gas in bar In this case, kH = 0.159 mol dm-3 bar-1, p = 0.164 bar, and we want to find S. Plugging in these values, we get: S = 0.159 mol dm-3 bar-1 * 0.164 bar = 0.0261 mol dm-3 Therefore, the solubility of CH3Br(g) in water at 25°C and a partial pressure of 0.164 bar is 0.0261 mol dm-3.  www.collegedunia.com SECTION B Attempt any EIGHT of the following questions: Question 3. Explain pseudo-first order reaction with suitable example. Answer. Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction. Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate. CH3COOCH3(aq) + H2O(1) ⟶H+(aq) CH3COOH(aq) + CH3OH(aq) Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH3COOCH3] x [H2O] Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics. This is because solvent water being in a large excess, its concentration remains constant. Hence, [H2O] = constant = k” Rate = k [CH3COOCH3] x [H2O] = k [CH3COOCH3] x k” = k’ x k” x [CH3COOCH3] If k’ x k” = k, then Rate = k [CH3COOCH3], This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction. Question 4. Write the consequences of Schottky defect with reasons. Answer. 1) Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.  www.collegedunia.com 2) As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same. 3)This defect arises in ionic crystals like NaCl, AgBr, KCl, etc. Question 5. What is the action of following on ethyl bromide: (i) Na in dry ether (ii) Mg in dry ether Answer. 1. C2H5Br + 2 Na + BrC2H5 → C4H10 + 2 NaBr 2. C2H5Br + mg----------C2H5MgBr Question 6. Explain formation of peptide linkage in protein with an example. Answer. A peptide bond is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water (H2O). This is a dehydration synthesis reaction (also known as a condensation reaction), and usually occurs between amino acids Question 7. Derive an expression to calculate molar mass of non volatile solute by osmotic pressure measurement. Answer. (1) Consider V dm3 of a solution in which n1 moles of a solvent contains n2 moles of a nonvolatile solute at absolute temperature T. (2) The osmotic pressure, n of a solution is given by, π = nRT/V R is gas constant having value 0.08206 dm3 atm K-1 mol-1 (OR L atm K-1 mol-1). Since concentration, C of a solution is in mol dm-3 or molarity is, C = n/V mol dm-3 or M ∴ π = CRT (If concentration C is expressed in mol m-3 and R = 8.314 J K-1mol-1, then π will be in SI units, pascals or Nm-2.)  www.collegedunia.com Question 8. Explain monodentate and ambidentate ligands with example. Answer. Monodentate ligand-Only one donor site is present. Examples includes ammonia and chloride ions. Ambidentate ligand- Ligands can coordinate to a central metal through two different sites. Examples include (1) nitro group (N as donor atom) and nitrito group (O as donor atom) (2) Thiocyanate (S atom as donor atom) and isothiocyanate (N atom as donor atom). Question 9. Explain the trends in the following atomic properties of group 16 elements: (i) Atomic radii (ii) Ionisation enthalpy (iii) Electronegativity (iv) Electron gain enthalpy Answer. Atomic and ionic radii :In groups 16, the atomic and ionic radii increase down the group, due to increase in the number of quantum shells. Across a period atomic or ionic radii decrease due to increase in effective nuclear charge. Ionisation enthalpy : In groups 16, the ionisation enthalpy decreases down the group, due to increase in atomic size. Electronegativity : In groups 16,the elec-tronegativity decreases down the group. Electron gain enthalpy : In groups 16 the electron gain enthalpy becomes less negative down the group.  www.collegedunia.com Question 10. Write preparation of phenol from aniline. Answer. Aniline is diazotized by treatment with nitrous acid (NaNO2 and HCl) under ice-cold conditions to form benzene diazonium chloride. The step is followed by hydrolysis with dilute sulphuric acid to form phenol. C6H5 − NH2 + HNO2 + HCl → C6H5 − N2+ Cl− + 2H2O C6H5−N2+Cl− + H2O → C6H5−OH + N2↑ + HCl Question 11. Write chemical reactions to prepare ethanamine from: (i) acetonitrile (ii) nitroethane Answer. Ethanamine from acetonitrile : Question 12. Identify A and B from the following reaction:  www.collegedunia.com Question 13. One mole of an ideal gas is expanded isothermally and reversibly from 10 L to 15 L at 300 K. Calculate the work done in the process. Answer. Given: Number of moles of gas (n) = 1 mol Gas constant (R) = 8.314 J/mol·K Temperature (T) = 300 K Initial volume (Vi) = 10 L Final volume (Vf) = 15 L Work done in an isothermal process can be calculated using the formula: W = -nRTln(Vf/Vi) where: W is the work done (J) n is the number of moles of gas R is the gas constant (J/mol·K) T is the temperature (K) Vi is the initial volume (L) Vf is the final volume (L) Substituting the given values into the formula, we get: W = -1 * 8.314 * 300 * ln(15/10) = -1011.31 J Therefore, the work done in the process is -1011.31 J. Since the work is negative, we know that the gas is doing work on the surroundings.  www.collegedunia.com Question 14. How many moles of electrons are required for reduction of 2 moles of Zn2+ to Zn? How many Faradays of electricity will be required? Answer. The balanced equation for the reduction of Zn2+ to Zn is Zn2+ + 2e- -------------> Zn The equation shows that 1 mole of Zn2+ is reduced to Zn by 2 moles of electrons for reduction of 2 mole of Zn2+, 4 mole of electron will be required. SECTION C Attempt any EIGHT of the following questions: Question 15. Write chemical composition of haematite. Write the names and electronic configurations of first two elements of group 17. Answer. Fe2O3 group 17 elements 9F – 1s2 2s2 2p5 17Cl – 1s2 2s2 2p6 3s2 3p5 Question 16. Write classification of polymers on the basis of structure. Answer. Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers. (1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.  www.collegedunia.com (2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches. (3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle. Question 17. Define green chemistry. Write two disadvantages of nanotechnology. Answer. (i) Green chemistry : Green chemistry is the use of chemistry for pollution prevention and it designs the use of chemical products and processes that reduce or eliminate the use or generation of hazardous  www.collegedunia.com substances. Disadvantages of nanotechnology 1. Nanoparticle can cause lung damage 2. Nano pollution is very danger for living organism. Question 18. Write commercial method for preparation of glucose. Write structure of adipic acid. Answer. Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose. Question 19. Write chemical reactions of following reagents on methoxyethane: (i) hot HI (ii) PCl 5 (iii) dilute H2 SO4 Question 20. Explain cationic, anionic and neutral sphere complexes with example. Answer. (1) Cationic sphere complexes : A positively charged coordination sphere or a coordination compound having a positively charged coordination sphere is called cationic sphere complex. For example : [Zn(NH3)4]2+ and [Co(NH3)5Cl] SO4 are cationic complexes. The  www.collegedunia.com latter has coordination sphere [Co(NH3)5Cl]2+, the anion SO42+ makes it electrically neutral. (2) Anionic sphere complexes : A negatively charged coordination sphere or a coordination compound having negatively charged coordination sphere is called anionic sphere complex. For example, [Ni(CN)4]2+ and K3 [Fe(CN)6] have anionic coordination sphere; [Fe(CN)6]3- and three K+ ions make the latter electrically neutral. (3) Neutral sphere complexes : A neutral coordination complex does not possess cationic or anionic sphere. [Pt(NH3)2Cl2] or [Ni(CO)4] are neither cation nor anion but are neutral sphere complexes. Question 21. Calculate spin only magnetic moment of divalent cation of transition metal with atomic number 25. Salts of Ti4+ are colourless. Give reason. Answer. For element with atomic number 25. electronic configuration of its divalent cation will be : [Ar] 3d5. The electronic configuration of Ti is [Ar]3d 2 4s 2 The electronic configuration of Ti 4+ is [Ar]3d0 4s0. Rao Junior College of Science Since no unpaired electron is present in Ti4+ So , it is colourless  www.collegedunia.com Question 22. What is lanthanoid contraction? Write preparation of acetic acid from (i) dry ice (ii) acetyl chloride Answer. Lanthanide contraction is the gradual decrease in the atomic and ionic size of lanthanoids with an increase in atomic number. Causes of lanthanide contraction: With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell. The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell. With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction. Question 23. Write the classification of aliphatic ketones with example. What is the action of sodium hypoiodite on acetone? Answer. Aliphatic ketones : The compounds in which group is attached to two alkyl groups are called aliphatic ketones. Ketones are classified into two types : 1. Simple or symmetrical ketones and 2. mixed or unsymmetrical ketones. 1. Simple or symmetrical ketone : The ketone in which the carbonyl carbon is attached to two identical alkyl groups is called a simple or symmetrical ketone.  www.collegedunia.com 2. Mixed or unsymmetrical ketone : The ketone in which the carbonyl carbon is attached to two different alkyl groups is called a mixed or unsymmetrical ketone. Question 24. Define half life of first order reaction. Obtain the expression for half life and rate constant of the first order reaction. Answer. The half-life of a first-order reaction is the time it takes for the concentration of a reactant to decrease to half of its initial concentration. It is a characteristic property of the reaction and is independent of the initial concentration of the reactant. The half-life is denoted by the symbol t₁/₂. The expression for the half-life of a first-order reaction is: t₁/₂ = 0.693 / k  www.collegedunia.com where k is the rate constant of the reaction. The rate constant is a measure of the speed of the reaction and is defined as the proportionality constant between the rate of the reaction and the concentration of the reactant. The units of the rate constant depend on the order of the reaction. For a first-order reaction, the units of the rate constant are s⁻¹. The expression for the rate of a first-order reaction is: rate = k[A] where [A] is the concentration of the reactant in mol/L. The rate of a reaction is the measure of how quickly the reaction is occurring and is defined as the change in concentration of a reactant or product per unit time. The units of the rate of a reaction are mol/L·s⁻¹. Question 25. Calculate the standard enthalpy of formation of CH3 OH(1) from the following data (i) CH3OH(1) + 3/2 O2(g) → CO2(g) + 2H2O(1) ⃤ H° = - 726 kJ mol-1 (ii) C(s) + O2(g) → CO2(g) ⃤ cH° = 393 kJ mol-1 (iii) H2(g) + 1/2 O2(g) → H2O(1) ⃤ fH° = -286 kJ mol-1 Question 26. Calculate the pH of buffer solution composed of 0.01 M weak base BOH and 0.02 M of its salt BA. [Kb = 1.8 *10-5 for weak base] Answer. The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution composed of a weak base and its conjugate acid. The equation is: pH = pKb + log([A-]/[HA]) where: pH is the pH of the buffer solution  www.collegedunia.com pKb is the base dissociation constant of the weak base [A-] is the concentration of the conjugate base (salt) [HA] is the concentration of the weak acid (base) In this problem, we are given that the concentration of the weak base (BOH) is 0.01 M and the concentration of its salt (BA) is 0.02 M. We are also given that the base dissociation constant (Kb) of the weak base is 1.8 * 10^-5. Substituting these values into the Henderson-Hasselbalch equation, we get: pH = 4.7447 + log(0.02/0.01) = 9.7781 Therefore, the pH of the buffer solution is 9.7781. SECTION D Attempt any THREE of the following questions: Question 27. Define the following terms: (i) Isotonic solution (ii) Osmosis Gold crystallises into face-centred cubic cells. The edge length of unit cell is 4.08 *10-8 cm. Calculate the density of gold. [Molar mass of gold = 197gmol–1 ] Answer. Isotonic solution A solution that has the same solute concentration as blood plasma is called an isotonic solution. In other words, an isotonic solution has the same osmotic pressure as blood plasma. This means that the concentration of solutes, such as sodium chloride, is the same in the solution as it is in blood plasma. Isotonic solutions are used in medicine to prevent the dehydration of  www.collegedunia.com cells. For example, saline solution, which is a solution of sodium chloride in water, is an isotonic solution. Saline solution is often used to intravenously rehydrate patients who have lost fluids due to illness or injury. Osmosis Osmosis is the movement of water molecules across a semipermeable membrane from an area of lower solute concentration to an area of higher solute concentration. This movement of water molecules is driven by the difference in osmotic pressure between the two solutions. The osmotic pressure of a solution is a measure of the force that must be applied to the solution to prevent the inflow of water across a semipermeable membrane. The osmotic pressure of a solution is directly proportional to the concentration of solutes in the solution. Osmosis is a natural process that occurs in many biological systems. For example, osmosis is responsible for the uptake of water by plant cells and the filtration of blood by the kidneys. Calculation of the density of gold The density of a substance is defined as the mass of the substance per unit volume. The formula for density is: density = mass / volume In this problem, we are given that the molar mass of gold is 197 g/mol and the edge length of a unit cell of gold is 4.08 * 10^-8 cm. We can use this information to calculate the volume of a unit cell of gold. The volume of a unit cell of a face-centered cubic crystal is given by the formula:  www.collegedunia.com volume = a^3 where a is the edge length of the unit cell. Substituting the given values into this formula, we get: volume = (4.08 * 10^-8 cm)^3 = 6.9746 * 10^-23 cm^3 Now we can calculate the mass of a unit cell of gold. The mass of a unit cell of gold is equal to the molar mass of gold multiplied by the number of atoms in the unit cell. The number of atoms in a unit cell of a face-centered cubic crystal is 4. Therefore, the mass of a unit cell of gold is: mass = 197 g/mol * 4 atoms/unit cell = 788 g/mol Finally, we can calculate the density of gold. The density of gold is equal to the mass of a unit cell of gold divided by the volume of a unit cell of gold. Therefore, the density of gold is: density = 788 g/mol / 6.9746 * 10^-23 cm^3 = 1.12 * 10^21 g/cm^3 Therefore, the density of gold is 1.12 * 10^21 g/cm^3. Question 28. Write the mathematical equation for the first law of thermodynamics for (i) isothermal process (ii) adiabatic process Derive the relationship between pH and pOH. Answer. First Law of Thermodynamics The first law of thermodynamics is a fundamental principle of physics that states that the total energy of an isolated system remains constant. This principle can be expressed mathematically as:  www.collegedunia.com ΔU = Q + W where: ΔU is the change in internal energy of the system Q is the heat transferred into or out of the system W is the work done on or by the system Isothermal Process An isothermal process is a thermodynamic process in which the temperature of the system remains constant. For an isothermal process, the first law of thermodynamics can be written as: ΔU = W This means that the change in internal energy of the system is equal to the work done on or by the system. Adiabatic Process An adiabatic process is a thermodynamic process in which there is no heat transfer between the system and its surroundings. For an adiabatic process, the first law of thermodynamics can be written as: ΔU = -W This means that the change in internal energy of the system is equal to the negative of the work done on or by the system. Relationship between pH and pOH The pH of a solution is a measure of the acidity or alkalinity of the solution. The pH scale ranges from 0 to 14, with 7 being neutral. A solution with a pH less than 7 is acidic, while a solution with a pH greater than 7 is alkaline.  www.collegedunia.com The pOH of a solution is a measure of the hydroxide ion concentration of the solution. The pOH scale ranges from 0 to 14, with 7 being neutral. A solution with a pOH less than 7 is basic, while a solution with a pOH greater than 7 is acidic. The pH and pOH of a solution are related by the following equation: pH + pOH = 14 This equation is known as the water dissociation constant, or the Kw equation. It states that the product of the pH and the pOH of a solution is always equal to 14. Question 29. Define reference electrode. Write functions of salt bridge. Draw neat, labelled diagram of standard hydrogen electrode (SHE). Question 30. Explain metal deficiency defect with example. Write chemical equation for preparation of sulphur dioxide from sulphur. Write uses of sulphur. Answer. Metal deficiency defect A metal deficiency defect is a type of point defect in a crystal in which one or more metal atoms are missing from the lattice. This type of defect can occur in any type of crystal, but it is most common in ionic crystals. Metal deficiency defects can have a significant impact on the properties of a material. For example, they can make a material more brittle and less conductive. In some cases, they can also lead to the formation of color centers. An example of a metal deficiency defect is iron(II) oxide (FeO). In this crystal, some of the Fe2+ ions are missing from the lattice. This creates a  www.collegedunia.com vacancy at the site of the missing ion. The vacancy is then filled by an electron, which creates a color center. The color center gives iron(II) oxide its black color. Chemical equation for preparation of sulfur dioxide from sulfur The chemical equation for the preparation of sulfur dioxide from sulfur is: S + O2 → SO2 This reaction is a combustion reaction, and it is exothermic. The reaction takes place in the presence of a catalyst, such as vanadium pentoxide (V2O5). Uses of sulfur Sulfur has a wide variety of uses. Some of the most common uses of sulfur include: Production of sulfuric acid: Sulfuric acid is one of the most important industrial chemicals. It is used in a wide variety of applications, including the production of fertilizers, detergents, and plastics. Vulcanization of rubber: Sulfur is used to vulcanize rubber. This process strengthens rubber and makes it more resistant to wear and tear. Production of pigments: Sulfur is used to produce a variety of pigments, including white, black, and yellow pigments. Production of pharmaceuticals: Sulfur is used to produce a variety of pharmaceuticals, including antibiotics and painkillers. Production of agricultural chemicals: Sulfur is used to produce a variety of agricultural chemicals, including insecticides and fungicides.  www.collegedunia.com Question 31. Write chemical reactions for the following conversions: (i) Ethyl bromide to ethyl methyl ether. (ii) Ethyl bromide to ethene, (iii) Bromobenzene to toluene. (iv) Chlorobenzene to biphenyl. Answer. (i) Ethyl bromide to ethyl methyl ether: CH3CH2Br + NaOH → CH3CH2OCH3 + NaBr This reaction is known as the Williamson ether synthesis. It is an example of an SN2 reaction, in which the bromide ion is displaced by the ethoxide ion. (ii) Ethyl bromide to ethene: CH3CH2Br + KOH (alc) → CH2=CH2 + KBr + H2O This reaction is known as the Zaitsev reaction. It is an example of an E2 elimination reaction, in which the bromide ion is eliminated along with the hydrogen atom from the β-carbon atom. (iii) Bromobenzene to toluene: C6H5Br + Mg → C6H5MgBr C6H5MgBr + H2O → C6H5H + MgBrOH This reaction is known as the Grignard reaction. It is a two-step reaction in which the magnesium atom is inserted into the carbon-bromine bond, followed by hydrolysis to form toluene. (iv) Chlorobenzene to biphenyl: 2 C6H5Cl + 2 Na → C6H5-C6H5 + 2 NaCl  www.collegedunia.com This reaction is known as the Wurtz reaction. It is an example of a coupling reaction, in which two aryl halides are coupled to form an aryl-aryl bond.