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ISC Class XII Board Exam 2025 : Chemistry

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ISC EXAMINATION PAPER 2025 CHEMISTRY Class 12th (Solved) Maximum Marks: 70 Time allowed: 3 hours Reading Time: Additional Fifteen minutes Instructions to Candidates 1. You are allowed an additional fifteen minutes for only reading the question paper. 2. You must NOT start writing during the reading time. 3. It is divided into four sections and has twenty one questions in all. 4. Answer all questions. 5. Section A has fourteen subparts. Each question carries 1 mark. 6. While attempting Multiple Choice Questions in Section A, you are required to write only ONE option as the answer. 7. Section B has ten questions. Each question carries 2 marks. 8. Section C has seven questions. Each question carries 3 marks. 9. Section D has three questions. Each question carries 5 marks. 10. Internal choices have been provided in one question each in Sections B, C and D. 11. The intended marks for questions are given in brackets [ ]. 12. All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer. 13. Balanced equations must be given wherever possible and diagrams where they are helpful. 14. When solving numerical problems, all essential workings must be shown. 15. In working out problems, use the following data: Gas constant R = 1.987cal deg 1 mol 1 = 8.314 JK 1 mol 1 = 0.082l dm3 atm K 1 mol 1 1 l atm = 1 dm3 atm = 101.3 J, 1 Faraday = 96500 coulombs Avogadro s number = 6.023 1023 SECTION A [14 Marks] Question 1 (A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets: [4 1] [+2, ethane, tetrahedral, square planar, zero dry cell, nickel cadmium cell, propane, Wolff Kishner, Stephen, completely filled, incompletely filled, paramagnetic, diamagnetic] (i) ___________ is an example of a primary cell but __________ is an example of a secondary cell (ii) The complex compound [Ni(CO)4] is _________ in shape and nickel is in __________ oxidation state in this complex compound. (iii) When acetaldehyde is treated with hydrazine and KOH in a high boiling solvent glycol, __________ is formed and the reaction is known as _________ reduction. (iv) The transition metal ions having _________ d-orbitals are colourless and _________ in nature. (B) Select and write the correct alternative from the choices given below. [7 1] (i) Which one of the following can produce the foul smelling compound methyl isocyanide in presence of alcoholic KOH? (a) Chloroform and aniline (b) Chloroform and methanol (c) Chloroform and dimethyl amine (d) Chloroform and methyl amine (ii) The osmotic pressure of a solution: (P) increases with an increase in number of moles of solute. (Q) decreases with an increase in number of moles of solute. (R) increases at a higher temperature. (S) is dependent on the nature of solute. Which one of the following is correct? (a) Only (P) and (Q) are correct (b) Only (P) and (R) are correct (c) Only (P) and (S) are correct (d) Only (Q) and (S) are correct (iii) Which one of the following alcohols is the strongest acid? (a) Phenol (b) Methanol (c) Ethanol (d) t-butyl alcohol (iv) Which one of the following does NOT form a silver mirror on heating with Tollen s reagent? (a) Glucose (b) Fructose (c) Sucrose (d) Lactose 2 Oswaal ISC Question Bank Chapterwise & Topicwise, CHEMISTRY, Class-XII (v) The coordination number and the oxidation state of the central metal D in the complex [D(en)2(H2O)2] Cl3 are: (a) 6 and 2, respectively (b) 4 and 2, respectively (c) 3 and 6, respectively (d) 6 and 3, respectively (vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option. Assertion: The process of halogenation of benzene takes place in the presence of anhydrous FeCl3. Reason: Anhydrous FeCl3 prepares nucleophile to attack the benzene ring. (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Both Assertion and Reason are false. (vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option. Assertion: The Zr Hf pair of elements has the same value of atomic radii though Zr and Hf are placed in different periods in the periodic table. Reason: The lanthanoid contraction prevents the expected increase in atomic radii of Hf. (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion. (c) Assertion is true and Reason is false. (d) Both Assertion and Reason are false. (C) Read the passage carefully and answer the questions that follow. [3 1] The rate of a reaction depends on the concentration of reactants. The rate law for a hypothetical reaction aA + bB cC + dD is rate = k[A]x[B]y where x and y are calculated experimentally and known as order of reaction. In most cases, the mechanism of a reaction is not straight forward but broken down in simple elementary steps. These steps represent the progress of overall reaction at the molecular level. (i) What will be the order of reaction if the unit of k is mol 2 L2s 1? (ii) State any one difference between order of reaction and molecularity of reaction. (iii) For the reaction 2A B + C the rate law is rate = k[A]3/2 What is the order and molecularity of the reaction? SECTION B [20 Marks] Question 2 [2] Calculate the number of coulombs required to electroplate 4.75 g of aluminium when electrode reaction is Al3+ +3e Al (Given: Atomic weight of Al = 27 g mol 1, 1 Faraday = 96,500 coulombs) Question 3 [2] An organic compound [A] has molecular formula C7H6O2. When compound [A] is treated with SOCl2, it yields compound [B]. On heating with NH3, compound [B] forms compound [C]. Compound [C] forms compound [D] on reaction with Br2/KOH. Compound [D] responds to carbylamine test. Identify compounds [A], [B], [C] and [D]. Question 4 [2] In the reaction 2NO + O2 2NO2, the rate law is rate = k[NO] [O2]2. (i) How will the rate of reaction change if [NO] concentration is doubled and [O2] concentration is halved at the same time? (ii) Write the order of reaction if [NO] concentration is in large excess. Question 5 [2] When an organic compound [A] having molecular formula C4H9Br is treated with aqueous KOH, the rate of reaction depends on concentration of compound [A] only. But when compound [B], with the same molecular formula, reacts with aqueous KOH, the rate of reaction depends on the concentration of compound [B] as well as of KOH. Compound [B] is a structural isomer of compound [A]. Identify compounds [A] and [B]. Question 6 [2] Write a balanced chemical equation for each of the following: (i) Ethyl cyanide is reduced with LiAlH4. (ii) Aniline is treated with bromine water. Question 7 [2] While conducting an experiment on coordination compounds, Shirin observed a white precipitate when AgNO3 solution was added to the aqueous solution of the complex [Co(NH3)3 (en) SO4] Cl. How can the ionisation isomer of this complex be detected by a chemical test? Write the structure and the IUPAC name of the ionisation isomer. Question 8 [2] (i) Arrange the following in the increasing order of their Acidic strength: CH3CH2OH, CF3CH2OH, CCl3CH2OH (ii) Arrange the following in the increasing order of their Boiling points: CH3CH2OH, CH2OH, CH3CH2Cl | CH2OH SOLVED PAPER - 2025 Question 9 [2] (i) Identify the compounds [A], [B], [C] and [D] in the following reaction: H2 CH3COOH + PCl5 [A] [B] Pd/BaSO 4 Boiling xylene K 2 Cr2 O7 + H 2 SO4 [O] Ca(OH)2 , heat [D] [C] OR (ii) Write chemical equations to convert the following: (a) Benzaldehyde to benzene (b) Benzoic acid to benzaldehyde Question 10 [2] o Calculate the Ecell of the following if the cell potential (Ecell) is 0.59 V. (Given: Ni/Ni 2+ (0.1M) ||Cu2+ (0.01 M)/Cu) Question 11 [2] Write the chemical test to distinguish between each of the following pairs of compounds. (i) Ethanol and propan-1-al (ii) Phenol and benzoic acid SECTION C [21 Marks] Question 12 [3] Identify the compounds [A], [B] and [C] in each of the following reactions: LiAlH HNO 4 2 [B] 0 [C] (i) C2H5Br KCN [A] [ 4 H] 5 C Br / KOH 3 (i) Which of the two complex compounds, [A] or [B], is a low spin complex? (ii) Write the number of unpaired electrons in complex compounds [A] and [B]. (iii) Does complex [A] have strong field ligands or weak field ligands? Give a reason. Question 16 [3] (i) Answer the following questions. (a) By referring to electrochemical series, how can anode and cathode half cells be identified in a galvanic cell? (b) What is the role of salt bridge in a galvanic cell? (c) Write an advantage and a disadvantage of a fuel cell. OR (ii) Answer the following questions. (a) Specific conductance of a solution decreases upon dilution. Why? (b) The emf of a cell should be positive for a spontaneous reaction. Give a reason. (c) Name the type of cell in which reaction occurs only in one direction and cannot be reversed by an external energy source. Write any one disadvantage of this type of cell. Question 17 [3] (i) The structure of amino acid exists in the following two forms: CHCl + KOH 2 3 (ii) C2H5CONH2 [A] [B] Na/C H OH 2 5 [C] Question 13 [3] The scientist van t Hoff introduced a factor (i) to account for the extent of association or dissociation of solutes. It is mathematically expressed as: normal molecular mass i= experimental molecular mass In case of association, i < 1 and in case of dissociation i > 1. (i) In the calculation of molecular mass of K4[Fe(CN)6] by using a colligative property, what will be the value of van t Hoff factor if the solute is 25% dissociated? (ii) Find the value of van t Hoff factor for a dilute aqueous solution of benzoic acid in water when it is completely associated to form a dimer. Question 14 [3] Write chemical equations to illustrate the following name reactions: (i) Finkelstein reaction (ii) Williamson s synthesis (iii) Reimer Tiemann reaction Question 15 [3] According to Crystal-Field Theory, the electronic configuration of complex compound [A] is t42geg2 and that of complex compound [B] is t62ge0g. The above structure is an example of __________. If the side chain R is replaced by hydrogen, the amino acid is known as __________. (ii) Janice notices that her gums bleed while brushing and eating food. Name the water soluble vitamin which she should consume to prevent bleeding of gums. (iii) Which linkage holds two units of monosaccharides in a disaccharide? Question 18 [3] The data given below is for the reaction between [NO] and [Cl2] to form NOCl at 25 C S.No. Conc. of [NO] mol L 1 Conc. of [Cl2] mol L 1 Rate: mol L 1 s 1 1. 2.0 2.0 2.0 10 3 2. 2.0 6.0 6.0 10 3 3. 6.0 2.0 1.8 10 2 Answer the following questions. (i) What is the order of reaction with regard to NO and Cl2? (ii) Calculate the overall order of the reaction. (iii) Find the value of rate constant (k). 4 Oswaal ISC Question Bank Chapterwise & Topicwise, CHEMISTRY, Class-XII SECTION D Question 19 [15 Marks] [5] Phenol is an aromatic alcohol that is used to prepare many important compounds such as picric acid. Phenol is widely used in household and industrial settings as a cleaner and disinfectant. It is also used as a primary chemical to make plastics. Phenol is less soluble in water as compared to aliphatic alcohol. Some aliphatic alcohols are toxic and can be addictive. (i) How is the acid mentioned above prepared from phenol? Write the chemical reaction involved in this preparation. (ii) Phenol is less soluble in water as compared to aliphatic alcohol . Explain. (b) K2Cr2O7 + H2SO4 + H2S _____ + _____ + ______ + ______ Question 21 [5] (i) Osmotic pressure is the external pressure which should be applied to stop the flow of solvent into the solution when the two are separated by a semipermeable membrane. The osmotic pressure is a colligative property. Two solutions having the same osmotic pressure are called isotonic. If there are two solutions and one of them is of lower osmotic pressure, it is called hypotonic while the other is called hypertonic. (iii) Write the chemical equation for the preparation of phenol from chlorobenzene. (iv) An organic compound [A] having molecular formula C4H10O gives positive Lucas test within five minutes at room temperature. Compound [A] upon oxidation with K2Cr2O7/H2SO4 forms compound [B] which does not respond to Tollen s test. Identify compounds [A] and [B]. (v) In the above reaction, compound [B] gets reduced with Zn/Hg and HCl and forms compound (C). Identify compound [C] and write the balanced reaction for the conversion of compound [B] to compound [C]. Question 20 [5] (i) Give a reason for each of the following: (a) Ti3+ salts are coloured whereas Ti4+ salts are colourless. [Given: Atomic number of Ti = 221 (b) Transition elements form alloys. (c) The pink coloured KMnO4 solution turns colourless when reacted with Mohr s salt (Fe2+) in acidic medium. (ii) Complete and balance the following reactions: (a) KMnO4 + H2SO4 + H2C2O4 _____ + _____ + ______ + ______ Answer the questions given below. (a) What will happen if red blood corpuscles are placed in a 5% NaCl solution which is a hypertonic solution? (b) Show that osmotic pressure ( ) is a colligative property. (c) Calculate the amount of pressure required to stop osmosis of a solution when 40 g of Na2SO4 is added to 1 L of water at 298 K. (Given: Na = 23, O = 16, S = 32 and R = 0.0821 L atm K 1 mol 1) (d) Briefly discuss the process of reverse osmosis followed to desalinate sea water and convert it into drinking water. OR (ii) (a) An aqueous solution is made by dissolving 10 g of glucose (C6H12O6) in 90 g of water at 300 K. If the vapour pressure of pure water at 300 K is 32.8 mm Hg, what would be the vapour pressure of the solution? (Given: C = 12, H = 1 and O = 16) (b) A solution containing 12.5 g of a non-electrolyte solute in 175 g of water gave boiling point of 100.70 C. Calculate the molecular mass of the solute. (Given: Kb for water = 0.52 K kg mol 1) (c) Why are soda water bottles sealed under high pressure? SOLVED PAPER - 2025 5 ANSWERS SECTION A 1. (A) Fill in the blanks: (i) Dry cell, Nickel-cadmium cell (ii) Tetrahedral, zero (iii) Ethane, Wolff- Kishner (iv) Completely filled, diamagnetic (B) (i) Option (d) is correct. 2. Explanation: Chloroform reacts with methyl amine in the presence of alcoholic KOH to form methyl isocyanide which is a foul/bad smelling liquid. (ii) Option (b) is correct. Explanation: p = cRT = n/V RT, Osmotic pressure increases with the number of moles of solute and temperature. (iii) Option (a) is correct. Explanation: As its conjugate base is the most stable. SECTION B For 1 mole of Al deposited, 3 moles of electrons are required. The number of moles of Al = w/At.wt = 4.75/27= 0.1759 moles Charge carried by 0.1759 3 moles = 0.1759 3 96500 = 50923 .05 Coulombs 3. A = Benzoic acid = B = Benzoyl chloride = C = Benzamide = D = Aniline = 4. (i) When we double the concentration of NO and halve the concentration of O2: Phenol conjugate base (iv) Option (c) is correct. Explanation: Sucrose does not have free aldehyde group that s why it does not gives positive tollen s reagent test. (v) Option (d) is correct. Explanation: Both (en) and (H2O) are neutral ligands, and (en) is a bidentate ligand. So, coordination number of metal D is 6. Three Cl- ion present in ionic sphere so oxidation number of metal D is +3. (vi) Option (c) is correct. Explanation: Benzene undergoes electrophilic attack, and anhydrous FeCl3 prepares the electrophile Cl+ rather than a nucleophile to attack the benzene ring. (vii) Option (a) is correct. Explanation: Lanthanide contraction prevents the increment of atomic radii of 5d series so assertion is correct and reason is the correct explanation of assertion. (C) (i) Third order reaction (ii) Order Molecularity Order of a reaction may Molecularity of a reaction be a whole number or a is always a whole number fractional number. and never fractional. 3 (iii) Order = , Molecularity= 2 2 Initial rate: r0 = k[NO] [O2]2 1 New rate: rn = k (2[NO]) [O2 ] 2 2 2 rn 1 = k (2[NO]) [O2 ] / k[NO] [O2]2 r0 2 =2 1 1 = 4 2 The new reaction rate will be 1 of initial reaction 2 rate. (ii) Rate = k[NO][O2]2, is a third order reaction. When NO is in excess, its concentration does not change significantly during the reaction. Therefore, it can be considered constant. Rate law = k [O2]2 , k = k[NO] = constant Now the order of the reaction = 2 5. A = tertiary butyl bromide, as the molecule undergoes SN1 mechanism. Oswaal ISC Question Bank Chapterwise & Topicwise, CHEMISTRY, Class-XII 6 H , Pd/BaSO /xylene 2 4 C6H5CHO 10. The cell reaction: B = n-butyl bromide, is primary halide, as the molecule undergoes SN2 mechanism Ni + Cu2+ Ni2+ + Cu 0.0591 log[products] Ecell = Eocell [reactants] n 0.59 = Eocell 0.0591/2 log [Ni2+]/[Cu2+] = Eocell 0.0295 log 0.1/0.01 = Eocell 0.0295 log 10 Eocell = 0.59 + 0.0295 = 0.6195 V (i) Propanal gives deposition of silver mirror with Tollen s reagent, while ethanol does not respond to this test. This is a characteristic test for aldehydes. (ii) Benzoic acid gives brisk effervescence with the evolution of CO2, on reaction with sodium bicarbonate while phenol does not respond to this test. This is a characteristic test for carboxylic acids. 11. Compounds A and B are structural isomers. 6. 4 C2H5CH2NH2 (i) C2H5CN + 4[H] LiAlH SECTION C 12. (i) A = C2H5CN B = C2H5CH2NH2 C = C2H5CH2OH (ii) A = C2H5NH2 (ii) 7. The ionisation isomer is: [Co(NH3)3(en)Cl]SO4. IUPAC nomenclature: Triamminechlorido (ethylenediamine)cobalt(III)sulphate. This on ionisation generates SO42 ions, which can be tested by adding Barium chloride solution to give a thick white precipitate of BaSO4. 8. (i) CH3CH2OH < CCl3CH2OH < CF3CH2OH. Fluorine being the most electronegative atom exerts negative inductive effect and stabilises the conjugate base most and thus is most acidic. Alkyl groups exert positive inductive effect and decreases acid strength. (ii) CH3CH2Cl < CH3CH2OH< HOCH2CH2OH The presence of two alcoholic groups in CH2OHCH2OH results in stronger intermolecular hydrogen bonds, which are weaker in ethanol comparatively. Ethyl chloride does not form these bonds and has the lowest boiling point. 9.(i) A = CH3COCl = Acetyl chloride B = CH3CHO = Acetaldehyde C = CH3COOH = Acetic acid B = C2H5NC C = C2H5NHCH3 13. (i) K4[Fe(CN)6] 4K+ + [Fe(CN)6]4 The molecule generates 5 ions/molecule i 1 , = n 1 = degree of dissociation, n = no. of ions generated on dissociation i 1 i 1 = = 5 1 4 0.25 4 = i 1 1 = i 1, i = 2 (ii) For association 2 C6H5COOH (C6H5COOH)2 The van t Hoff factor i = nobs/ntheo, nobs = number of solute particles actually present in the solution ntheo = number of solute particles present in the solution without considering association or dissociation 1 i = 2 1 D = CH3COCH3 = Acetone OR (ii) (a) Benzaldehyde to benzene K Cr O /H+ 2 2 7 C6H5CHO C H COOH 6 5 NaOH C6H5COONa CaO/NaOH C6H6 (b) Benzoic acid to benzaldehyde +PCl C6H5COCl C6H5COOH 5 1 2 i < 1, indicates association. i = 14. (i) Finkelstein reaction acetone CH3CH2I + NaCl CH3CH2Cl + NaI (ii) Williamson s synthesis RCH2ONa + R X RCH2 O R + NaX SOLVED PAPER - 2025 (iii) Reimer-Tiemann reaction 15. (i) B, t62g e0g, as it has all electrons paired in lower energy level so it is low spin complex. (ii) A, t42g e2g has four unpaired electrons, two in t2g and two in eg. B, t62g e0g has no unpaired electrons. (iii) A has weak field ligand as the energy difference between t2g and eg is small, and it gives high spin complex. 16. (i) (a) At the top of the electrochemical series are good oxidising agents, i.e., those having a positive value of standard reduction potential. They act as cathode as they can be easily reduced. Those appearing on the bottom of the electrochemical series are good reducing agents, i.e., they have a negative value of standard reduction potential. They act as anode as they can be easily oxidised. 7 Dividing 2 by 1: Comparing 1 and 3: 2 10 3 = k[2]x [2]1 1.8 10 2 = k[6]x [2]1 Dividing 3 by 1: Rate = k[NO]2[Cl2]1 Reaction is second order w.r.t NO, and first order w.r.t Cl2 3 = (3)y, y = 1 9 = [3]x, x = 2 (ii) Overall order of reaction = 2 + 1 = 3 (iii) Value of rate constant: 2 10 3 = k[2]2 [2]1 2 10 3 4 2 = 0.25 10 3 L mol 2 s 1 SECTION D k = 19. (i) Phenol to picric acid: (b) A salt bridge is a device used in an electrochemical cell for connecting its oxidation and reduction half-cells, where a inert electrolyte is used. It helps maintain electrical neutrality within the internal circuit. (c) Advantages of fuel cells: They have higher efficiency, lower emissions and are renewable. Disadvantages: Fuel cells are expensive to manufacture and install and are made up of delicate components that can be damaged by extreme temperatures. OR (ii) (a) The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions per unit volume decreases when the solution is diluted. (ii) Aliphatic alcohols form intermolecular hydrogen bonds with water and dissolves. Phenol does not effectively form such bonds and is therefore less soluble. (iii) Chlorobenzene to phenol: By Dow s process: As a result, the conductivity of a solution decreases with dilution. (b) If EMF of cell is positive, then the G of the overall reaction is less than zero, which results in a spontaneous reaction. G = nFEcell. (c) In a primary cell, the chemical reaction occurs in one direction only and cannot be reversed by an external energy source. Once the chemicals are depleted, the cell cannot be recharged, making it a single-use battery. 17. (i) Zwitter ion, Glycine. (ii) Vitamin C (iii) Glycosidic linkage 18. (i) Let rate = k [NO]x[Cl2]y Comparing (1) and (2) 2 10 3 = k[2]x [2]y 6 10 3 = k[2]x [6]y (iv) A = Butan-2-ol, CH3CH(OH)CH2CH3, secondary alcohol B = Butan-2-one, CH3COCH2CH3, a ketone (v) n-butane is formed. CH3COCH2CH3 + 4[H] Zn-Hg/HCl CH3CH2CH2CH3 + H2O 20. (i) (a) Ti: [Ar]3d24s2 Ti3+: [Ar]3d14s0: has one unpaired electron in d orbital to favour d-d transition that absorbs light in visible range and complementary colour is seen. Ti4+: [Ar]3d04s0: has no unpaired electrons and thus no d-d transition. So, it is colourless. (b) Transition metals have very similar atomic sizes. Metals can easily replace the one another in the lattice to form solid solution or alloy. They are miscible with one another in the molten state. 8 Oswaal ISC Question Bank Chapterwise & Topicwise, CHEMISTRY, Class-XII (c) Mohr salt acts as a reducing agent and potassium permanganate acts as an oxidising agent forming a redox reaction. In this reaction, Fe[+2] from Mohr s salt gets oxidised to Fe[+3] and pink coloured Mn [+7] in potassium permanganate, gets reduced to colourless Mn[+2] state. (ii) (a) 2KMnO4 + 3H2SO4 + 5H2C2O4 K2SO4 + 2MnSO4 + 8H2O + 10CO2 (b) K2Cr2O7 + 3H2S + 4H2SO4 K2SO4 + Cr2(SO4)3 + 7H2O + 3S (d) The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. The pure solvent flows out of the solution through the semi-permeable membrane. This phenomenon is called reverse osmosis. OR (ii) (a) Molar mass of glucose = 6 12 + 1 12 + 6 16 = 180 10 No. of moles of glucose = = 0.0555 180 90 =5 18 21. (i) (a) If blood cells are placed in a hypertonic solution of sodium chloride, water will flow out of the cells, causing them to shrink. No. of moles of water = (b) Osmotic pressure is a colligative property as it depends on the concentration of the solution and the number of solute molecules in solution, not on their identity. P0 Ps = X2 = mole fraction of solute P0 n2 32.8 P0 = n 32.8 2 + n1 32.8 P0 0.0555 = 32.8 0.0555 + 5 = iCRT = i n/V RT, C= concentration of solution n = number of moles of solute = w2/M2 w2 = weight of solute M2 = molecular weight of solute 32.8 P0 = 0.0109 32.8 32.8 P0 = 0.36 P0 = 32.8 0.36 = 32.44 mm (c) = iCRT= i n/V RT, (b) Tb = kb molality n = w2/M2 for Na2SO4 M2 = 2 23 + 32 + 4 16 = 142 g/mol W2 = 40 g 40 n= = 0.281 moles 142 100.70 100 = 0.52 W2 1000/M2 W1 M2 = molar mass of solute W2 = weight of solute W1 = weight of solvent 0.70 = 0.52 12.5 1000/M2 175 V= 1L T= 298 K M2 = 0.52 12.5 1000/0.70 175 For Na2SO4, i = 3 as it ionises into two Na+ and one SO42 M2 = 53.06 g Substituting: = 3 0.281 0.0821 298/1 = 20.62 atm (c) To increase the solubility of CO2 in soft drinks, the soda water bottles are sealed under high pressure. When the bottle is opened at room temperature under normal atmosphere conditions, the pressure inside the bottle decreases to atmospheric pressure and excess CO2 fizzes out.

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