Proteins are high molecular mass complex biomolecules of amino acid The important proteins required for our body are enzymes, hormones, antibodies, transport proteins, structural proteins, contractile proteins etc. Except for glycine, all o-amino acids have chiral carbon atom and most of them have L-configuration. The amino acids exists as dipolar ion called zwitter ion, in which a proton goes from the carboxyl group to the amino group. A large number of-amino acids are joined by peptide bonds forming polypeptides. The peptides having very large molecular mass (more than 10,000) are called proteins. The structure of proteins is described as primary structure giving sequence of linking of amino acids; secondary structure giving manner in which polypeptide chains are arranged and folded; tertiary structure giving folding, coiling or bonding polypeptide chains producing three dimensional structures and quaternary structure giving arrangement of sub- units in an aggregate protein molecule. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement. d) Assertion is wrong statement but reason is correct statement. i) Assertion .- All amino acids are optimally active. Reason : Amino acids contain asymmetric carbon atoms. ii) Assertion .- In o-helix structure, intramolecular H-bonding takes place whereas in β-pleated structure,intermolecular H-bonding takes place. Reason : An egg contains a soluble globular protein called albumin which is present in the white part. iii) Assertion .- Secondary structure of protein refers to regular folding patterns of continuous portions of the polypeptide chain Reason : Out of 20 amino acids, only 12 amino acids can be synthesized by human body. iv) Assertion .- The. helical structure of protein is stabilized by intramolecular hydrogen bond between —NH and carbonyl oxygen. Reason : Sanger’s reagent is used for the identification of N-terminal amino acid of peptide chain.

Read the passage given below and answer the following questions: When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. The denaturation causes change in secondary ann tertiary structures but primary structures remain intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk. Choose the most appropriate answer: i) Mark the wrong statement about denaturation of proteins. a) The primary structure of the protein does not change. b) Globular proteins are converted into fibrous proteins. c) Fibrous proteins are converted into globular proteins. d) The biological activity of the protein is destroyed. ii) Which structure{s) of proteins remains(s) intact during denaturation process? (a) Both secondary and tertiary structures (b) Primary structure only (c) Secondary structure only (d) Tertiary structure only iii) a-helix a n d β - pleated structures of proteins are classified as (a) primary structure (b) secondary structure (c) tertiary structure (d) quaternary structure. (iv) Secondary structure of protein refers to a) mainly denatured proteins and structure of prosthetic groups. b) three-dimensional structure, especially the bend between. amino acid residues that are distant from each other in the polypeptide chain. c) linear sequence of amino acid residues in the polypeptide chain. d) regular folding patterns of continuous portions of the polypeptide chain.

The sequence of bases along the DNA and RNA chain establishes its primary structure which controls the specific properties of the nucleic acid. An RNA molecule is usually a single chain of ribose-containing nucleotide. On the basis of X-ray analysis of DNA, J.D., Watson and F.H.C. CYST (shared noble prize in 1962) proposed a three dimensional secondary structure for DNA. DNA molecule is a long and highly complex, spirally twisted, double helix, ladder like structure. The two polynucleotide chains or strands are linked up by hydrogen bonding between the nitrogenous base molecules of their nucleotide monomers. Adenine (purine) always links with thymine (pyrimidine) with the help of two hydrogen bonds and guanine (purine) with cytosine (pyrimidine) with the help of three hydrogen bonds. Hence, the two strands extend in opposite directions, i.e., are antiparallel and complimentary. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement d) Assertion is wrong statement but reason is correct statement. i) Assertion - DNA molecules and RNA molecules are found in the nucleus of a cell. Reason : There are two types of nitrogenous bases, purines and pyrimidines. Adenine (A) and guanine (G)are substituted purines; cytokine (C), thymine (T) arid uracil (U) are substituted pyrimidines. ii) Assertion .- In both DNA and RNA, heterocyclic base and phosphate ester linkages are at C- 1’ and C-5’respectively of the sugar molecule. Reason : Nucleotides and nucleosides mainly differ from each other in presence of phosphate units. iii)Assertion .- The backbone of RNA molecule is a linear chain consisting of an alternating units of heterocylic base, D-ribose and a phosphate. Reason : The segment of RNA which acts as the instruction manual for the synthesis of protein is ribose. iv) Assertion.- In DNA, the complementary bases are, adenine and guanine; thymine and cytosine. Reason : The phenomenon of mutation is chemical change in DNA molecule.

An organic compound A having molecular formula \(C_6\)\(H_6\)O turn blue litmus solution into red but does not react with sodium bicarbonate, but when treated with bromine water then form a white ppt of compound B. when compound A react with chloroform in presence of aqueous caustic soda solution at 340K then form two compound C and D. When compound A treated with caustic soda then form compound E compound E when treated with methyl halide then form compound F. Read the above passage carefully and answer the following questions: (i) The name of compound is: (a) 2-methyl propene-2-ol (b) 2-methyl phenol (c) 2,4,6-tribromophenol (d) Butane 1-ol (ii) Which are isomers of each other (a) A and C (b) B and C (c) C and D (d) D and E (iii) The IUPAC name of compound F is (a) Anisole (b) Methoxybenzene (c) Salicylaldehyde (d) 2-methyl propene-2-ol (iv) When compound E treated with ethyl iodide then ...... form. (a) Ehoxybenzene (b) Ethoxy hexane (c) Propoxypropane (d) Benzaldehyde (v) On oxidation with sodium dichromate and conc \(H_2\)\(SO_4\) phenol gives (a) Benzaldehyde (b) p-Benzoquinone (c) o-Benzoquinone (d) m-benzoquinone

Read the passage given below and answer the following questions: Phenol contains -OH group directly attached to carbon atoms of an aromatic system \(C_6\)\(H_5\)OH in phenol the group is attached to \(sp^2\) hybridised carbon of aromatic ring. The carbon oxygen bond length is 1:36 pm in phenol is slightly less than that in in methanol this is due to first point partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring s point \(sp^3\) hybridised state of carbon to which oxygen is attached it can be prepared by various means or methods. Some important methods are alkali fusion of sulphonates, hydrolysis of diazonium salts decarboxylation of salicylic acid and from Grignard reagent, it is prepared from Dow's process and from cumene. Aerial oxidation of human produce cumene peroxide which on hydrolysis produce phenol and acetone. In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are true and the reason is the correct explanation of assertion. (b) Both assertion and reason are true but the reason is not the correct explanation of assertion. (c) Assertion is true but reason is false. (d) Assertion is false but reason is true. (i) Assertion: C-O bond length in phenol is less than C-O bond length in ethanol. Reason: In phenol carbon atom is \(sp^2\) hybridize while in ethanol carbon atom is \(sp^3\) hybridised. (ii) Assertion: p-nitrophenol is more acidic than phenol. Reason: Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance. (iii) Assertion: Phenol is prepared by the reaction of chlorobenzene with caustic soda at 623K and 300 atm pressure. Reason: The substitution of chorine atom from chlorobenzene is difficult due to resonance. (iv) Assertion: Methanol is less acidic than phenol. Reason: Due to resonance phenate ion become more stable. (v) Assertion: On nitration phenol forms 2-nitrophenol and 4- nitrophenol. Reason: The presence of –OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions.

Read the passage given below and answer the following questions: Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of -OH group of alcohols with halogens using phosphorus halides, thionyl chloride or halogen acids. Aryl halides are prepared by electrophilic substitution to arene. Fluorine and iodides are best prepared by halogen exchange method. These compounds find wide applications in industry as well as in day-to-day life. These compounds are generally used as solvents and as starting material for the synthesis of a large number of organic compounds. (i) The best method for the conversion of an alcohol into analkyl chloride is by treating the alcohol with (a) \(PCl_5\) (b) dry HCl in the presence of anhydrous \(ZnCl_2\) (c) \(SOCl_2\) in presence of pyridine (d) None of these (ii) The catalyst used in the preparation of an alkyl chloride bythe action of dry HCl on an alcohol is (a) anhydrous \(AlCl_3\) (b) \(FeCl_3\) (c) anhydrous \(ZnCl_2\) (d) Cu (iii) An alkyl halide reacts with metallic sodium in dry ether. The reaction is known as : (a) Frankland’sreaction (b) Sandmeyer’sreaction (c) Wurtz reaction (d) Kolbe’s reaction (iv) Fluorobenzene (\(C_6\)\(H_5\)F) can be synthesized in the laboratory (a) by direct fluorination of benzene with \(F_2\) gas (b) by reacting bromobenzene with NaF solution (c) by heating phenol with HF and KF (d) from aniline by diazotisation followed by heating thediazonium salt with \(HBF_4\) (v) When 2-bromobutane reacts with alcoholic KOH, thereaction is called (a) halogenation (b) chlorination (c) hydrogenation (d) dehydrohalogenation

Read the passage given below and answer the following questions: The replacement of hydrogen atom in a hydrocarbon, aliphatic or aromatic results in the formation of haloalkanes and haloarenes respectively. Haloalkanes contain halogen atom attached to \(sp^3\) hybridised carbon atom of an alkyl group whereas haloarenes contain halogen atom attached to \(sp^2\) hybridised carbon atom of an aryl group. Haloalkanes and haloarenes may be classified on the basis of number of halogen atoms in their structures as mono, di or poly halogen compounds and also on the basis of the state of hybridisation of carbon atom to which the halogen atom is bonded. (i) Which of the following halide is 2°? (a) Isopropyl chloride (b) Isobutyl chloride (c) n-propyl chloride (d) n-butyl chloride (ii) Which of the following is a Gem-dibromide is : (a) \(CH_3\)\(CH(Br)CH_2\)(Br) (b) \(CH_3\)\(CBr_2\)\(CH_3\) (c) \(CH_2\)\((Br)CH_2\)\(CH_2\) (d) \(CH_2\)\(BrCH_2\)Br (iii) IUPAC name of (\({\\(CH_3\\)}_3\))CCl is: (a) 3-Chlorobutane (b) 2-Chloro-2-methylpropane (c) t-butyl chloride (d) n-butyl chloride (iv) Which of the following is a primary halide? (a) Isopropyl iodide (b) Secondary butyl iodide (c) Tertiarybutylbromide (d) Neohexylchloride (v) Which one of the following is not an allylic halide? (a) 4-Bromopent-2-ene (b) 3-Bromo-2-methylbut-1-ene (c) 1-Bromobut-2-ene (d) 4-Bromobut-1-ene

Read the passage given below and answer the following questions: Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (SN2) and substitution nucleophilic unimolecular (SN1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards SN1 and SN2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. SN2 reaction mechanism is favoured mostly by primary alkyl halide or transition state and polarity of solvent, SN2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of SN1 reactions. (i) Which of the following is most reactive towards nucleophilic substitution reaction? (a) \(C_6\)\(H_5\)Cl (b) \(CH_2\)=CHCl (c) \(ClCH_2\)\(CH=CH_2\) (d) \(CH_3\)CH=CHCl (ii) Isopropyl chloride undergoes hydrolysis by (a) SN1 mechanism (b) SN2 mechanism (c) SN1 and SN2 mechanism (d) neither SN1 nor SN2 mechanism (iii) Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of (a) insolubility (b) instability (c) inductive effect (d) steric hindrance (iv) Which of the following is the correct order of decreasing SN2 reactivity? (a) \(RCH_2\)X > \(R_2\)CHX > \(R_3\)CX (b) \(R_3\)CX > \(R_2\)CHX > \(RCH_2\)X (c) \(R_2\)CHX > \(R_3\)CX > \(RCH_2\)X (d) \(RCH_2\)X > \(R_3\)CX > \(R_2\)CHX (v) An organic molecule necessarily shows optical activity if it- a) contains asymmetric carbon atoms b) is non-polar c) is non-superimposable on its mirror image d) is superimposable on its mirror image

Group 18 elements are called noble gases and not inert gases because compounds of Kr, Xe and Rn have been prepared. Their general electronic configuration is \(ns^2\)\(np^6\) except He(\(1s^2\) ). They have highest ionisation enthalpy and positive electron gain enthalpy due to stable electronic configuration. Helium is found in sun and stars. Noble gases have low boiling points due to weak van der Waals’ forces of attraction. Xenon forms \(XeF_2\), \(XeF_4\), \(XeF_6\), \(XeOF_4\), \(XeO_3\), \(XeO_2\)\(F_2\), their structures can be drawn on bases of VSEPR theory. Helium is mixed with oxygen by deep sea divers to avoid pain. Neon is used in coloured advertising lights. Argon is used in bulbs as inert gas. Kr and Xe are used in high efficiency lamps, head light of cars. Radon is radioactive formed by a-decay of Radium 226 88Ra Argon is most abundant (0.9%) noble gas in atmosphere. The following questions are multiple choice questions. Choose the most appropriate answer. 1) What are the elements in group 18 (the far right) of the periodic table called? a) Alkali metals b) Alkaline earth metals c) Halogens d) Noble gases 2) Out of (i) \(XeO_3\) (ii) \(XeOF_4\) and (iii) \(XeF_6\) , the molecules having the same number of lone pairs on Xe are - a) (i) and (ii) only b) (i) and (iii) only c) (ii) and (iii) only d) (i) , (ii) and (iii) 3) Which one has linear shape? a) \(XeF_2\) b) \(XeF_4\) c) \(XeF_6\) d) \(XeO_3\) 4) Which of the outer electronic configuration represent Argon? a) \(ns^2\)\(np^4\) b) \(ns^2\)\(np^3\) c) \(ns^2\)\(np^6\) d) \(ns^1\)\(np^6\) 5) Which of the following statement is false? a) Radon is obtained from the decay of radium b) Helium is an inert gas c) Xenon is the most reactive among the rare gases d) The most abundant rare gas found in the atmosphere is helium

Group 16 elements are called chalcogens i.e., ore forming elements (oxygen, sulphur, selenium etc.) because most of the ores are oxides and sulphides. Oxygen is gas where as other elements of group 16 are solids. Oxygen shows anomalous behaviour. Oxygen is diatomic where is sulphur exists as \(S_8\) which has crown shaped structure. It shows allotropy. Sulphur is present in onion and garlic that is why they have pungent smell. Sulphur is used for manufacture of sulphuric acid which is called ‘King of chemicals’, used in fertilizer, detergents, dyes and drugs. The following questions are multiple choice questions. Choose the most appropriate answer. 1) Group 16 elements are also known as a) Noble elements b) Halogens c) Pnictogens d) Chalcogens 2) Acidic character of hydrides of group 16 elements is in the order a) \(H_2\)O < \(H_2\)S < \(H_2\)Se < \(H_2\)Te b) \(H_2\)S < \(H_2\)Se < \(H_2\)Te < \(H_2\)O c) \(H_2\)O < \(H_2\)Se < \(H_2\)Te < \(H_2\)S d) \(H_2\)O < \(H_2\)S < \(H_2\)Te < \(H_2\)Se 3) Hybridisation of S in \(SF_4\) and geometry of \(SF_4\) are respectively a) \(sp^3\)d, trigonal pyramidal b) \(sp^3\)d, see saw c) \(sp^3\), tetrahedral d) \(dsp^2\), square planner 4) Which is not an acidic oxide? a) \(CO_2\) b) \(SO_2\) c) \(Na_2\)O d) \(Cl_2\)\(O_7\) 5) Which is not correct about allotropes of sulphur a) The stable form at room temperature is rhombic sulphur b) Monoclinic sulphur is stable above 369 K and transforms into rhombic sulphur below it c) At 369 K both the forms are stable d) Monoclinic sulphur is soluble in \(CS_2\) while rhombic sulphur not

Molecular Nitrogen \(N_2\) comprises about 78% by volume of Earth’s atmosphere. It occurs as Sodium nitrate, \(NaNO_3\)(chile saltpeter) & Potassium nitrate, \(KNO_3\)(Indian altpeter) in earth’s crust. Since nitrate are very soluble in water so these are not wide spread in the earth’s crust. Nitrogen is also an important constituent of amino acids, protein & nucleic acids in plants & animals. Nitrogen shows anomalous behavior from rest of the elements due to following reasons; Smaller size, high ionization enthalpy, high electronegativity & absence of d-orbital. It has unique ability to form p∏-p∏ multiple bonds with itself & with small size atoms like C & O as they have small size & high electronegativity. Heavier elements of this group do not form p∏-p∏ bonds as their atomic orbitals are so large & diffuse that they can’t have effective overlapping. Thus Nitrogen exists as diatomic molecules \(N_2\) with a triple bond. Consequently, its bond enthalpy (941.4 KJ \(mol^{-1}\)) is very high. P, As & Sb form only single bonds as P-P, As-As & Sb-Sb. Due to much bond enthalpy N is much less reactive than P. Single N-N bond is weaker than single P-P bond due to high interelectronic repulsion of the non bonding electrons, owing to small bond length. As a result, the catenation tendency is weaker in nitrogen. Hence nitrogen exists as gas while phosphorus exists as solid. Nitrogen can’t form d∏- d∏ bond due to absence of d- orbitals so it can’t expand its covalency beyond four as heavier members can. The following questions are multiple choice questions. choose the most appropriate answer. 1) Among group 15 elements which exists as gas at room temperature a) Arsenic b) Bismuth c) Nitrogen d) Phosphorous 2) The stability of +5 oxidation state decreases and that of +3 state increases down the group in group 15 elements due to a) inert pair effect b) decrease in ionisation enthalpy c) increase in size d) shielding effect 3) Nitrogen is restricted to a maximum covalency of 4 because of a) absence of d-orbitals b) presence of d-orbitals c) absence of s and p-orbitals d) none of the above 4) Extra pure \(N_2\) can be obtained by heating a) \(NH_3\) with CuO b) \(NH_4NO_3\) c) \(\left(NH_4\right)_2Cr_2O_7\) d) \(Ba\left(N_3\right)_2\) 5) Catenation tendency is weaker in nitrogen, because of a) single N–N bond is weaker b) single N–N bond is stronger c) ability to form pi bonds by N atoms d) none of the above

Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Answer the following MCQs by choosing the most appropriate options: (i) When a non volatile solid is added to pure water it will- (a) boil above \(100^\circ\)C and freeze above \(0^\circ\)C. (b) b) boil below \(100^\circ\)C and freeze above \(0^\circ\)C. (c) boil above \(100^\circ\)C and freeze below \(0^\circ\)C. (d) boil below \(100^\circ\)C and freeze below \(0^\circ\)C. (ii) Colligative properties are (a) dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity. (b) dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity. (c) dependent on the identity of the solvent and solute and thus on the concentration of the solute. (d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity. (iii) Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1 M, 0.5 M and 0.2 M respectively. Freezing point will be highest for the fruit juice (a) A (b) B (c) C (d) all have same freezing point (iv) Identify which of the following is a colligative property? (a) Freezing point (b) Boiling point (c) Osmotic pressure (d) All of the above

The properties of dilute or ideal solutions which depend only upon the concentration of the solute in the solution and no other characteristics are known as colligative properties. There are in all four such properties i.e. relative lowering in vapour pressure, osmotic pressure, elevation in boiling point temperature and depression in freezing point temperature. All of them help in calculating the observed molar mass of the solute which is inversely proportional to the colligative property involved. Out of these, osmotic pressure may be regarded as the best for the determination of molecular mass of the solute. According to Van’t Hoff theory of dilute solution, π = CRT, where ‘π’ is the osmotic pressure while ‘C’ is the molar concentration of the solution. (i) When liquids A and B are mixed, hydrogen bonding occurs. The solutions will show: a) Positive deviation from Raoult’s law b) Negative deviation from Raoult’s law c) No deviation from Raoult’s law d) Slightly increase in volume (ii) The azeotropic mixture of water and HCl boils at \(108.5^\circ\)C when the mixture is distilled. It is possible to obtain: a) Pure HCl b) Pure water c) Pure water as well as pure HCl d) Neither HCl nor water in their pure states. (iii) On freezing an aqueous solution of sugar, the solid which starts separating out is: a. Sugar b. Ice c. Solution with the same composition d. Solution with different composition (iv) The value of osmotic pressure does not depend upon: a) Concentration of the solution b) Temperature of the solution c) Number of the particles of the solute present d) Structure of the solute particles (v) Effect of adding a non-volatile solute to a solvent is : a) to lower the vapour pressure b) to increase the freezing point c) to decrease the boiling point d) to decrease the osmotic pressure

The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry’s law “the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution”. Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry’s law can be modified as “the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution”. Answer the following MCQs by choosing the most appropriate options: (i) Henry’s law constant for the solubility of methane in benzene at 298 K is 4.27 × \(10^5\) mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is (a) 4.27 × \(10^{-5}\) (b) 1.78 × \(10^{-3}\) (c) 4.27 × \(10^{-3}\) (d) 1.78 × \(10^{-3}\) (ii) The partial pressure of ethane over a saturated solution containing 6.56 × \(10^{-2}\) g of ethane is 1 bar. If the solution contains 5.00 × \(10^{-2}\) g of ethane then what will be the partial pressure (in bar) of the gas? (a) 0.762 (b) 1.312 (c) 3.81 (d) 5.0 (iii) KH (K bar) values for Ar(g), \(CO_2\)(g), HCHO(g) and \(CH_4\)(g) are 40.39, 1.67, 1.83 × \(10^{-5}\) and 0.413 respectively. Arrange these gases in the order of their increasing solubility. (a) HCHO < \(CH_4\) < \(CO_2\) < Ar (b) HCHO < \(CO_2\) < \(CH_4\) < Ar (c) Ar < \(CO_2\) < \(CH_4\) < HCHO (d) Ar < \(CH_4\) < \(CO_2\) < HCHO (iv) When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 150 K bar. If the gas exerts a partial pressure of 2 bar, the number of millimoles of the gas dissolved in 1 L of water is (a) 0.55 (b) 0.87 (c) 0.37 (d) 0.66

The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg.(vapour pressure of water at \(20^\circ\)C is 17.5 mm of Hg). Answer the following MCQs by choosing the most appropriate options: (i) Relative lowering of vapour pressure for the given solution is- (a) 0.00348 (b) 0.061 (c) 0.122 (d) 1.75 (ii) The vapour pressure (mm of Hg) of Solution will be (a) 17.5 (b) 0.61 (c) 17.439 (d) 0.00348 (iii) Mole fraction of sugar in the solution is (a) 0.00348 (b) 0.9965 (c) 0.061 (d) 1.75 (iv) If weight of sugar taken is 5 g in 108 g of water then molar mass of sugar will be (a) 358 (b) 120 (c) 240 (d) 400 (v) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is (a) 17.2 (b) 17.4 (c) 17.120 (d) 17.02