 
                 
                InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2701. | 52. Which of the following mixtures contain less numberof ions?(2) 2M NaCl; 2M KCI(3) 2M MnSO 2M K,SO4(4) 2M K SO4 2M (NH4),SO4 | 
| Answer» | |
| 2702. | Balance the following chemical equátions. (ASI)a) NaOH + H,SO4 → Na,so, + H2O2.c) H, + O2 → H2O | 
| Answer» Balanced equations(a) 2NaOH + H2SO4 ---> Na2SO4 + 2H2O(b) Hg(NO3)2 + 2KI ---> HgI2 + 2KNO3(c) 2H2 + O2 ---> 2H2O(d) 2KClO3 ---> 2KCl + 3O2(e) C3H8 + 5O2 ---> 3CO2 + 4H2O | |
| 2703. | Find the power of a concave lens of focal length 2 m | 
| Answer» P=1/fP=1/2 DP=0.5D | |
| 2704. | Balance the following eq" stepwise\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}(\mathrm{I})+\mathrm{H}_{2} \mathrm{O}(1) \mathrm{m}_{2} \mathrm{SO}_{4}(\mathrm{I}) | 
| Answer» Hit like if you find it useful! | |
| 2705. | line in Balmer series corresponds to an electronic transition between which Bohr'sin hydrogen(A) 53(D) 4 2(B) 5 2(C) 43 | 
| Answer» balmer series always ends at n =2 so, first line will be 3→2 , 2nd will be 4→2 and 3rd will be 5→2 option B | |
| 2706. | 7. The only liquid non-metals is8. Fizzy drinks are9. The most expensive metal isin nature. | 
| Answer» 7.bromine8.carbonated9.rhodium | |
| 2707. | how would you confirm that a colourless liquid is pure water?(without tasting or smelling it). | 
| Answer» A very easy solution is to take a sample ofcolourless liquidand put on stove if it starts boiling exactly at 100 ºC then it ispure water. Any other colourless liquidsuch as vinegar always have different boiling point. | |
| 2708. | The color of vapours formed on sublimation of iodine solid is(a) Purple(b) yellow(c) colorless(d) orange | 
| Answer» At room temperature, elemental iodine is a black-bluish solid. When heated, solid iodine sublimates and enters the gas phase. Iodine got its name from the Greek word 'ἰοειδής', meaning 'purple'. Indeed, iodine vapor has a beautifulpurplecolor | |
| 2709. | The enthalpy of vaporization of a liquid is 30 kJ mol and entropy of vaporization is 75 Jmol-K-1. The boiling point of the liquid at 1 atm is | 
| Answer» | |
| 2710. | 2) Answer the following questions.a. Name one liquid metal and one liquid nonb. Arrange the substances inces given below in the order of increasing hardness.Sodium, Diamond, CopperCheck | 
| Answer» mercury is a liquid metal. bromine is a liquid non metal.diamond is the hardest material available due to its structure. it is then followed by copper. sodium is the weakest among the three.Diamond>Copper>Sodium | |
| 2711. | की ी। HH“"‘}"‘J—l}—flH Hwduwoo Suimoyof ayp ouwoN (9 | 
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| 2712. | 61. Calculate the magnetic moment ofdivalent ion in aqueous solution, if itsatomic number is 25.(1) V7 BM(3) V5 (5+2) BM (4) V5 BM(2) V5 (5+1) BM | 
| Answer» thanks | |
| 2713. | ipole moment of HBr is 2.6x10cmand the interatomic spacing is 1.41A. The percentage ofionic character in HBr is:(a) 10.5(b) 11.5(c) 12.5(d) 13.5TF | 
| Answer» HBr is an ionic compound in which H has + 1 and Br has -1 ion e.g charge one each have equal to charge of one electron or charge of proton .dipole moment =charge x distance between them =1.6 x 10^-19 x 141 x 10^-12 Cm=225.6 x 10^-31 Cm=22.56 x 10^-30 Cmnow,% of ionic characters =dipole moment by experimental/dipole moment by theoretical x100=(2.6 x 10^-30/22.56 x 10^-30 ) x 100=260/22.56 %=11.52 % | |
| 2714. | 69. Amongst LiCI, RbCI, BeCl, and MgCla, the compounds with the greatest and least ionic character respectivelyare(b) RbCl and BeCl(d) MgCl, and BeCl(a)LiCI and RbCi(c)RbCI and MgCl | 
| Answer» option b) ans is B) | |
| 2715. | (a)IV lslThe dipole moment of HBr is 2.6x10-0 cmand the interatomic spacing is 1.41A. The percentage ofionic character in HBr is:0(a) 10.5(b) 11.5(c) 12.5(d) 13.5If t | 
| Answer» We know ,HBr is an ionic compound in which H has + 1 and Br has -1 ion e.g charge one each have equal to charge of one electron or charge of proton .dipole moment =charge x distance between them =1.6 x 10^-19 x 141 x 10^-12 Cm=225.6 x 10^-31 Cm=22.56 x 10^-30 Cmnow,% of ionic characters =dipole moment by experimental/dipole moment by theoretical x100=(2.6 x 10^-30/22.56 x 10^-30 ) x 100=260/22.56 %=11.52 % Thanks | |
| 2716. | 12. In gaseous state, ionic character is greatest ina. HBOb. HFc. HC1d. HI | 
| Answer» BETWEEN HF, HCl, HBr and HI. HF has the highest ionic character because:1. F has the highest electron affinity. 2. In HF, electronegativity difference is highest 3. F - ion has the highest Value of ionic radius. 4. Atomic Orbitals of H and F have almost similar energy. (b) is correct option | |
| 2717. | 1.3.3 THE GASEOUS STATE | 
| Answer» Gasparticles spread out to fill a container evenly, unlike solids and liquids.Gasis astateof matter that has no fixed shape and no fixed volume.Gaseshave lower density than otherstatesof matter, such as solids and liquids. ... The particles exert more force on the interior volume of the container. Gas is a state of matter that has no shape and no fixed volume. Gasis astateof matter that has no fixed shape and no fixed volume.Gaseshave lower density than otherstatesof matter, such as solids and liquids. ... The particles exert more force on the interior volume of the container. This force is called pressure the state which can evaporate from gas | |
| 2718. | The rate of a reaction increases four times when the temperature changes from 300 K to 320 K.Calculate the energy of activation of the reaction, assuming that it does not change with temperature.(R = 8.314 JK mol') | 
| Answer» 320k-273=47300k-273=27 | |
| 2719. | At 300 K the half-life of a sample of a gaseous compound initially at 1 atm is 100 sec. When thepressure is 0.5 atm the half-life is 50 sec. The order of reaction is:(a) o65.(b) 1(c) 2(d) 3 | 
| Answer» we can se that half life is directly proportional to the pressure ( amount of gas) => t½ ∞ [A] so, order is 0 option a | |
| 2720. | How to convert 288 pm into centimetre | 
| Answer» 288 x 10^-8 show the solution. I can tell you as I know chemistry | |
| 2721. | 8.2 L of an ideal gas weight 9.0 gm at 300 K and 1 atm pressure. The molecular mass of gas is-(A) 91.(B) 27(C) 54(D) 81 | 
| Answer» | |
| 2722. | (2) 250ml, round bottom flask containing air washeated from 27°C to 477°C calculate whatfraction of the air is left behind in the flask | 
| Answer» Let the volume of the round bottomed flask beV. Then, the volume of air inside the flask at 27° C isV. Now, V1=V T1= 27°C = 300 K V2=? T2= 477° C = 750 K According to Charles’s law, V1/T1= V2/T2 => V2=V1T2/T1 =V(750/300) = 2.5V Therefore, volume of air expelled out = 2.5V–V= 1.5V Hence, fraction of air expelled out = (1.5V)/2.5V) = 3/5 Fraction remaining in the flask = 2/5 | |
| 2723. | (A) Only one answer is correllen flask contains air at 27C. The temperature at which it should be heated so that 1/3rd of ameasured at final temperature escapes out is(a) 100-C4. An op(c) 1S7 C4l. The density in g/L of a mixture containing equal number of moles of methane and ethane at S.T.P.is(c) 1-01(b) 127C(d) 177C.(a)0-90(b)1.00(d) 1-03. | 
| Answer» | |
| 2724. | o).orthyofaSolu tpox)diffe'ŕ¸"Q nt. | 
| Answer» Molarity, M, is the number ofmoles of solutein everyliter of solution. Thus it can be used to convert between moles of solute and volume of solution. The advantage of this is that one can use reagents in the form of solutions and measure out a given number of moles by measuring out the equivalent volume of reagent.Molality, m, is the number ofmoles of solutedissolved in eachkg of solvent. The units of molality are moles/kg. The symbol m is read "molal". Thus 2.3 m is "2.3 molal". | |
| 2725. | Chapter 1 [From First Test Syllabus1. State the characteristics of particles of matterSyllabus] | 
| Answer» Few important characteristics of particles of matters are:1) They are of small size.2) They are in continuous motion because they possess high kinetic energy.3) They occupy space4) They attract each other due to high intermolecular forces. | |
| 2726. | Exercise 11.The shape and hybridisation of XeF is(A) Planar, sp(c) octahedral, sp'd(B) Planar, sp'd(D) Tetrahedral, sp | 
| Answer» It has 4 bonds and two lone pairs total 6 → so hybridisation is sp3d2 square planer option B | |
| 2727. | Page NoDateDandayKin | 
| Answer» this is the diagram you need to draw | |
| 2728. | 6. How can salicylic acid be prepared from phenol? (Give 2 separate reactions) | 
| Answer» | |
| 2729. | Distinguish between physÄąsorpaunt aiu u6. How can salicylic acid be prepared from phenol? (Give 2 separate reactions)ich i more acidic and why? | 
| Answer» Phenols can be converted to salicylic acid by kolbe's reaction | |
| 2730. | 6.How can salicylic acid be prepared from phenol? (Give 2 separate reactions)nro gidic and urhu? | 
| Answer» | |
| 2731. | Aspirin CoH,O, is prepared by heating salicylicacid, C,H,O, with acetic anhydride, C4H,O3 Theother product is acetic acid, C,H,O2242when 2.00 g of salicylic acid is heated with 4.00 gof acetic anhydride? If the actual yield of aspirin is2.1 g, what is the percentage yield? | 
| Answer» | |
| 2732. | An ideal gas "A" having volume of 1 L at 27 °C is kept in a container havingmovable piston and adiabatic walls in ambient condition. If 133 L atm ofenergy is supplied inside the system, find out the final temperature of theo system?(A) 399 K(B) 499 K(C) 599 K(D) 299 KV(E) 450 K | 
| Answer» | |
| 2733. | ee i e e | 
| Answer» Mineralsare naturally occurring inorganic solidswith acrystalline structure and a definite range of chemical formula.Oresare concentrations ofmineralsin rock that are high enough to be economically extracted for use. | |
| 2734. | Define sublimation, name two solids that sublime on heating | 
| Answer» Sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase substance changes from a solid to a gas without ever passing through a liquid phase. Dry ice, solid CO2, provides a commonexampleofsublimation. The conversion of any substance directly from solid to gaseous state (without passing through liquid state) is known as sublimation examples - camphor and naphthalene balls | |
| 2735. | molarity is calculated as | 
| Answer» 👍👍👍 | |
| 2736. | 0.The molarity of pure water is:(1) 100 M(3) 50 M(2) 55.5 M(4) 18M | 
| Answer» The density of water is approximately 1000 g/L and its molar mass = 8.02 g/mol (or 1/18.02=0.055mol/g).Therefore, the molar concentration of water is: (H2O) = 1000 g/L / (18.02 g/mol) =55.5mol/L.(molar concentration=density/molar mass) | |
| 2737. | A solution of NaOH contain 0.04 gm of NaOH perlitre. Its pH is(1) 10(3) 11(2) 9(4) 12 | 
| Answer» | |
| 2738. | If 100 mL of 1 M NaOH is diluted to 1 litre, the resulting solution has molarity100O 0.1 M0 1 MO 10 M | 
| Answer» | |
| 2739. | 2g of NaOH is dissolved in water to give 1litre solution. What is the pH of the solution ? | 
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| 2740. | If 100 mL of 1 M NaOH is diluted to 1 litre, the resulting solution has molarityA1100O 0.1 MO10 M | 
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| 2741. | HCN 6.2 x 10andahu b22. One litre of 0.05 M HCI was completely neutralized by NaOH. Calculate the pH of resultingsolutionDuas Rond Nou Delhi 110005 Ph.011-47623456 | 
| Answer» | |
| 2742. | 16. Iron powder and suphur powder were mixed together and divided into two partsAand B. When part A was hThe part B was,substance C, then gas D was evolved and when dilute hydrochloric acid was added topart B then gas E was evolved.eated strongly over a burner, then a substance C was formed.however, not heated at all. When dilute hydrochloric acid was added to(a) What type of substance is B?(b) What type of substance is C?(c) Name the gas () D, and (i) E?(d) State one characteristic property of gas D. | 
| Answer» | |
| 2743. | “की न.ca (), ] 8. | है (5. | 
| Answer» Tetra-ammine dichloro cobaltTetraquamanganese(II) ion | |
| 2744. | 22 A compound exists in the gaseous state both as a monomer(A) and dimer (A2). The molecular weight ofthe monomer is 48. In an experiment, 96 g of the compound was confined in a vessel of volume33.6 litres and heated to 273° C. Calculate the pressure developed, if the compound exists as a dimer tothe extent of 50 per cent by weight, under these conditions. [Use R-0.082 atm-litre/mole-K | 
| Answer» | |
| 2745. | Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of1.84 g cm-3. Vvolume of the concentrâted acid is required to make So litre of 0.500 M H2SO, solution? | 
| Answer» 98% by weight means 98 gm H2SO4 in 100 gm solution or 98 g H2SO4 in 2 g water.Number of moles of H2SO4 = 98/98 = 1Density = 1.8 g/LVolume of solution = Mass of solution/ Density = 100/1.8 = 55.55 ml = 0.055 L Applying,M1V1 = M2V218.18 x V1 = 0.1 x1V1 = 0.0055 L = 5.5 ml Hence , volume of acid required = 5.5 ml | |
| 2746. | mention one type of ore that can be concentrated by froth floatation process | 
| Answer» Froth flotation method is used for the concentration ofsulphide ores. The method is based on the preferential wetting properties with the frothing agent and water. In the given options,Galena(PbS) is the onlysulphide ore. | |
| 2747. | Choose the correct answer.Sound can travel through(a) gases only(b) solids only(c) liquids only(d) solids, liquids and gases, | 
| Answer» (d) solids,liquids and gases correct answer answer is d solids, liquid and gases | |
| 2748. | calculate the pH of a solution prepared by mixing 5 ml pf 0.1 M NH3 with 10 ml of 0.02 M HCl | 
| Answer» 9.26 is the pH of a solution | |
| 2749. | 4. The molarity of a solution made by mixing 50 ml of conc. H,SO, (18 M) with 50ml of water,is:(A) 36 M(B) 18 M(C)9 M(D) 6M | 
| Answer» moles of H2SO4 = 18*(50)/1000 = 0.9 moles and volume = 50+50 = 100ml = 0.1 L so, M = 0.9/0.1 = 9 option C | |
| 2750. | 3.Withthehelpofasuitableexampleexplaininbrieftheprocessofhydrogenationmentioningtheconditions for the reaction and also state any one physical property of substances which changesdue to hydrogenation. | 
| Answer» Hrdrogenation is a process to obtain saturated hydrocarbons from unsaturated hydrocarbons. See a common example :-CH2=CH2 + H2 (Pd) ---> CH3CH3 The unsaturated hydrocarbon ethene is reduced to ethane. The conditions require pressure (different for different reactions) and catalyst such as Ni, Pd, Pt The Physical properties might change as follows :-(i) Unsaturated hydrocarbons are mainly insoluble in water but saturated are soluble. Therefore the insoluble becomes soluble .(ii) The boiling point and freezing point changes as accordance to the compounds. | |