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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.
| 2301. |
Differentiate between absorption and adsorption. |
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Answer» Solution :Absorption- It is the PROCESS by which atoms, molecules, or ions ENTER a bulk PHASE (LIQUIDS, gas, solid) Adsorption- It is the adhesion of atoms, ions or molecules from a gas, liquid or DISSOLVED solid to a surface. |
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| 2302. |
Different stages involved in the nitrogen cycle are given below. Arrange them in correct sequence starting from nitrates. (1) conversion of animal protein to excretory product (2) conversion of ammonia to nitrites (3) conversion of plant protein to animal protein (4) conversion of nitrites to nitrites (5) conversion of nitrates to plant protein (6) conversion of excretroy product to ammonia |
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Answer» 425163 (II) conversion of plant protein to ANIMAL protein (iii) conversion of animal protein to excretory PRODUCT (iv) conversion of excretory product of AMMONIA (V) conversion of ammonia to nitrates (vi) conversion of nitrites to nitrates. |
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| 2303. |
Different stages involved in the manufacture of hydrogen gas by Bosch process are given below. Arrange them in the correct sequence. (1) removal of unoxidised CO (2) preparation of water gas (3) removal ofCO_2 (4) removal of water vapour (5) removal of CO. |
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Answer» 35421 (ii) removal of CO (III) removal of `CO_2` (iv) removal of water VAPOUR (V) removal of unoxidised CO |
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| 2304. |
Different methods of formation of carbon is the main reason for its __________ |
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| 2305. |
Different gases in the discharge tube produce different colours under suitable conditions of pressure and voltage. Explain. |
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Answer» Solution :(i) ENERGY of ELECTRON Excintation and deexcitation (ii) FACTORS affecting the energy of electron (III) comparison of the energy emitted during the deexcitation of electron in different atoms. |
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| 2306. |
A diver in able to cut through water in a swimming pool. Which property of matter does this observationshow? |
| Answer» SOLUTION :The observation SHOWS that the PARTICLES of matter have intermolecular space. The intermolecularspace in LIQUIDS is fair enough to let the diver pass through it. | |
| 2307. |
Dicky, Micky and Vicky had three liquids, A, B and C respectively. They mixed these liquids and observed that they form a homogeneous mixture. They wer unable to separate the liquidsand asked their teacher to separate these for them. The teacher subjected the given mixture to fractional distillation. Liquid B was obtained in the receiver flask. On further distillation, A was left behind in the distillation flask. On the basisof the results, comment on the critical temperatures of A, B and C in their respective gaseous states. |
| Answer» SOLUTION :Fractional distillation is employed to separate miscible liquids which have a difference in their BOILING points. In this process of separation, the liquid with a low boiling point, i.e., a high vepour pressure, distils. THEREFORE, from the given data, it can be said the boiling point of B is less than that of C which is less than that of A. For liquids with high boiling points the intermolecular forces of attraction are also high, and hence, they have high critical temperatures. Therefore, the critical TEMPERATURE of A is greater than that of C which is greater than that of B. | |
| 2308. |
Diamond is lustrous because |
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Answer» It is COLOURLESS |
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| 2309. |
(a) Define polyatomic ions. Write an example. (b) Calculate the formula unit mass of CaCO_(3) (Atomic mass of C=12u, Ca=40u, O=16u) (c) Calculate the molecular mass of the following (i) HNO_(3) (ii)CH_(3)COOH Atomic mass of H=14N, N=14u, O=16u, C=12u |
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Answer» Solution :(a) Polytomicionsare the IONS represting a group of atoms whichcarry eitherpostivite or negative charge. For example, `(SO_(4))^(2-)` ions . (b)Formula units mass of `CaCO_(3) = ( 1xx"Atomicmass of Ca") + ( 1 xx "Atomic mass ofC")+ ( 3 xx "Atomic mass of O")` ` = 4 (1xx 40 u) + (1xx12 u) + (3 xx 16U) = 100u` (c) Molecularmass of `CH_(3)COOH = (2xx "Atomic mass of C")+( 4 xx "Atomic mass of H") +(2 xx "Atomic mass of O")` `=(2 xx 12 u) + (4 xx 1 u) + (2 xx 16 u) = 60u` |
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| 2311. |
Dheya takes some ice and water in a beaker and heat it on a low flame. She notes the temperature from time to time. Which of the following graphs shows the correct result. |
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Answer» a |
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| 2312. |
(a ) Define atomic number and mass number. (b) Which one of them is a more fundamental attribute of element ? State the reason. |
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Answer» Solution :(a )Atomic number: The number of protons PRESENT in the nucleus of an atom is called the atomic number. Mass number: The mass number is the sum of TOTAL number of protons and neutrons present in the nucleus of an atom. (b) (b) Refer to the ANSWER to subquestion 6 (a) in Questions and ANSWERS SECTION of 9. 10. |
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| 2313. |
Determine the molecular mass of water . |
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Answer» Solution :The molecular FORMULA of water is `H_(2)O` Atomic MASS of H = 1 Atomic mass of O = 16. `THEREFORE` molecular mass of water = (2 `xx` atomic mass of H ) + (1 `xx` atomic mass of O) `= 2 xx 1 + 1xx 16 = 18`. So , molecular mass of water = 18 AMU . |
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| 2314. |
A current of 9.65 amperes is passed through three different electrolytes, i.e., NaNO_(3), KCl and ZnSO_(4) for 30 min separately. Calculate the mass ratio of the metals deposited at the respective electrodes. Also find out the weights of various metals deposited at the respective electrodes. |
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Answer» Solution :Quantity of electricity passed through the electrolyte = current `XX` time of flow `= 9.65 xx 30 xx 60` coulombs `m_(Na) : m_(K) : m_(Zn) = E_(Na) : E_(K) : E_(Zn)` `{:(E_(Na) = 23/1 = 23,, (Na^(+) + E^(-) to Na)),(E_(K) = 39/1 = 39,, (K^(+) + e^(-) to K)),(E_(Zn) = 65/2 = 32.5,, (Zn^(+2) + e^(-) to Zn)):}` `m_(Na) : m_(K) : m_(Zn) = 23 : 39 : 32.5` Weight of sodium deposited `= (23 xx 9.65 xx 30 xx 60)/(96500) = 4.14 G` Weight of POTASSIUM deposited `= (39 xx 9.65 xx 30 xx 60)/(96500) = 7.02 g` Weight of zinc deposited `= (32.5 xx 9.65 xx 30 xx 60)/(96500) = 5.85 g` |
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| 2315. |
Determine the mass of the following : (i) 6.022 xx 10^(23) number of O_(2) molecules (ii) 6.022 xx 10^(23) number of O atoms . |
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Answer» Solution :(i) The NUMBER of moles = `("number of MOLECULAR")/("Avogadro constant") = (6.022 xx 10^(23))/(6.022 xx 10^(23)) = 1`. Now , mass = mole `xx` molecular mass = `1 xx 32 = 32` G (ii) The number of moles `= ("number of atoms")/("Avogadro constant") = (6.022 xx 10^(23))/(6.022 xx 10^(23)) = 1` . Now , mass = mole `xx` atomic mass = `1 xx 16 = 16` g |
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| 2316. |
Why is crystallisation better than evaporation for the separation of mixtures ? |
| Answer» SOLUTION :The SEPARATION can be DONE by the use of separating FUNNEL. | |
| 2317. |
Determine the mass of the following : (i) 0.7 mole of O_(2) gas (ii) 0.7 mol of O atoms . |
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Answer» Solution :(i) Molecular mass of `O_(2) ` GAS = 32 `therefore ` mass = MOLE `XX` molecular mass = `0.7 xx 32 = 22.4` G (ii) Atomic mass of O atom = 16 `therefore` mass = mole `xx` atomic mass = `0.7 xx 16 = 11.2` g . |
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| 2318. |
A crystal of copper sulphate can impart colour of large amout of water. Which characteristic of particles of matter is described by the above observation? |
| Answer» SOLUTION :The particles of matter consist of MILLIONS of TINY particles. | |
| 2319. |
A copper rod is placen in AgNO_3 solution and FeSO_4 solution. What changes do you obseve ? What type of reactions takes place? Justify your observation. |
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Answer» Solution :(i) REQUISITE for displacement of one metal by the other (ii) relative POSITIONS of AG, Cu and FE in activity series |
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| 2320. |
(a) conversion of solid to vapours is called sublimation. Name the term used to denote the conversion of vapours to solid. (b) Conversion of solid state to liquid state is called fusion, what is meant by latent heat of fusion ? |
| Answer» SOLUTION :(a) CONVERSION of vapurs to SOLID is KNOWN as FREEZING. | |
| 2321. |
A compound X on hudrolysis produces a gas Y which on hydrogenation peoduces a gas Z which can decolourise bromine water forming a compound with molecular weight 188.a.m.uIdentify X,Y and Z. |
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Answer» SOLUTION :IDENTIFICATION of Z NATURE ofZ identification OFY,based on its hydrogenation reaction |
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| 2322. |
Determine the mass by percentage concentration of a 100 g salt solution which contains 20 g salt (AS_(1)) |
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| 2323. |
A compound cannot be broken into simpler substances chemically |
| Answer» SOLUTION :Correct statement: A compound can be BROKEN into simpler substances CHEMICALLY. | |
| 2324. |
Desertbe the tests used to determine acids and bases. |
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Answer» Solution :Take 10 ml of solution in a TEST TUBE and test with a litmus paper or Indicators LIKE phenolphthalein and methyl orange. (a) Test with a litmus paper: An noid turns blue litmus paper into red. A base turns red limus paper into blue. (b) Test with an indicator Phenolphthalein: In neid medium, plenolphthalein is colourlons, In basic medium, phenolphthalein in pink in COLOUR. (c) Test with an indientor Methyl orange: In acid modium, mothyl orange is pink in colour In basic medium, methyl orange is YELLOW in colour. |
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| 2325. |
(a) Compare metals and non-metals based on their physical properties (any two). (b) What are metalloids ? Give two examples. (c ) Identify metals from the following : boron, sodium, mercury, carbon. |
| Answer» SOLUTION :(a) SODIUM and MERCURY are METALS. | |
| 2326. |
(a) Colloidal solutions show Tyndall effect but true solutions do not. Discuss. (b) Explain how does soap help in cleaning dirty clothes ? |
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Answer» Solution :(a) In a colloidal solution, the particle size is such (1 nm to 100 nm), that these particles scatter the light rays as they fall on them. Because of scattering, the path of the light as well as the particles become visible. But in a true solution, the particle size is so small (less than 1nm) that these particles are not in a position to scatter the light. Therefore, true solution does not show any TYNDALL effect. (b) In dirty clothes, the dust particles are sticking on the oil drops present. Simple water cannot remove these oil drops from the clothes because water and oil as such do not form a stable emulsion. Soap plays the role of emulsifier and helps in forming a stable emulsion between the two. In other WORDS, it helps in mixing oil and water. This means that soap helps in removing these oil drops ALONG with the dirt sticking to them. The dirty clothes get washed by soap solution. |
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| 2327. |
Describe the trend of the metallic and non metallic character of the elements of a period |
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| 2328. |
Describe the trend of the metallic and non metallic character of the elements in a group |
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| 2329. |
fillin thefollowingblanks : (a)Milkis a ……………………. Solutionbutvinegar is a…….. Solution . (b)Acolloid is a………………mixtureanditscomponents can beseparatedbythetechniqueknown as ……………. |
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| 2330. |
A child wanted to separate the mixture of dyes constituting a sample of ink. Hemarked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in figure. The filter paper was removed when the water moved near the top of the filter paper. (a) What would you expect to see, if the ink contains three different coloured components? (b) Name the technique used by the child. (c) Suggest one more application of this technique. |
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Answer» Solution :(i) Three different bands will be seen on the FILTER papers. (ii) The technique is KNOWN as PAPER chromatography. (iii) It can be USED to separate the coloured PIGMENTS from chlorophyll. |
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| 2331. |
Describe the main features of Bohr'smodel of an atom. Draw a neat and labelled diagram of energy levels. |
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Answer» Solution :Niels Bohr was the first to show the arrangement of electrons in an atom. Bohr.s atomic model : (1) The atom consists of a small nucleus at the CENTRE. (2) The whole mass of the atom is concentrated in the nucleus. (3)The electrons of the atom revolve round the nucleus in circular paths known as ORBITS or shells. The shells are DESIGNATED as K, L, M, N, etc. or numbered n= 1, 2, 3, 4, etc. (4) The orbits are known as energy levels. (5) Each orbit is associated with a fixed amount of energy. (6) Electrons move in these orbits without losing or GAINING any energy. 
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| 2333. |
Describe smelting of iron in a blast furnace. Write all the reactions which take in different zones all the reactions which take place in different zones of the blast furnace. |
| Answer» SOLUTION :N/A | |
| 2334. |
(a) Can we separate a mixture of water and alcohol by the use of a separating funnel ? If not, suggest a suitable method. (b) You are provided with two liquids, one is a mixture of two miscible liquids while the other is a pure compound. Suggest two ways to distinguish them from each other. |
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Answer» Solution :(a) No, it is not possible because water and alcohol form a miscible liquid mixture. The separation can be done with the help of FRACTIONAL distillation. Alcohol (ethyl alcohol) with lesser boiling point distils leaving behind water in the distillation flask. (b) The distinction can be made as FOLLOWS : (i) In case the constituents of a given liquid can be separated by methods like simple distillation or fractional distillation, it is a liquid mixture. If it is not possible,then the liquid is a COMPOUND. (II) Find the boiling points of both the liquids. In case it is sharp, the given liquid is a compound. In case it is not, then the given liquid is a mixture. |
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| 2335. |
(a) Calculate the relative molecular mass of water (H_(2)O). (b) Calculate the molecular mass of HNO_(3). |
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Answer» Solution :(a) Atomic mass of hydrogen = 1 u Atomic mass of oxygen = 16 u So, the MOLECULAR mass of water, which CONTAINS two atoms of hydrogen and ONE atom of oxygen = 2 (atomic mass of H) + 1 (atomic mass of O) = 2(1) + 1(16) = 18 u (b) The molecular mass of `HNO_(3)` = 1 (atomic mass of H) + 1 (atomic mass of N) + 3 (atomic mass of O) = 2(1) + 1(14) + 3(16) = 1 + 14 + 48 = 63 u |
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| 2336. |
Derive units for K_(C ) for the following equilibria. (i) CaCO_(3(s)) implies CaO_((s))+CO_(2(g)) (ii) PCl_(3(g))+Cl_(2(g))implies PCl_(5(g)) (iii) Na_(2(g))+O_(2(g)) implies 2NO_((g)) |
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Answer» SOLUTION :`(i) CaCO_(3(s)) implies CaO_((s))+CO_(2(g))` (II) `PCl_(3(g))+Cl_(2(g))implies PCl_(5(g))` (III) `Na_(2(g))+O_(2(g)) implies 2NO_((g))` 
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| 2337. |
(a) Butter is generally wrapped in wet cloth during summer if no refrigerator is available. Explain. (b) The Latent heat of vaporisation of steam is more than that of the boiling water. Explain. |
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Answer» Solution :(a) In summer, the WEATHER is a QUITE hot. As a result, WATER present in wet cloth READILY evaporates. Since cooling is caused during evaporation, the temperature of butter gets lowered. This checks the rancidity of butter or it does not give any foul odour. (b) When boiling water changes into steam, it absorbs a certain amount of heat energy. This shows that the latent heat of VAPORISATION of steam is more than that of boiling water. |
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| 2339. |
Density |
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Answer» Solution :The mass PER UNIT VOLUME of a substance is called DENSITY. Density `=("Mass")/("Volume")` |
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| 2340. |
A bivalent metal forms a salt with the oxyacid of a solid non-metal in which the non-metallic element has +5 oxidation state. The salt formed cannot be used as a fertiliser though it contains essential nutrients and is available as mineral in nature. However, the salt when treated with conc. H_3PO_4 gives an important and describe fertiliser when compared with conc H_2SO_4. Identify the various substances involved and give equations. |
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Answer» Solution :(i) IDENTIFICATION of the oxyacid of solid non-metal which can exhibit the given oxidation state. (ii) identification of salt formed (iii) solubility of the salt formed (iv) proudces obtained when the salt reacts with `H_3PO_4 and H_2SO_4` SEPARATELY (V) solubility of the PRODUCTS formed (VI) comparison of the extent of acidity imparted to the soil by both the proudcts. |
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| 2341. |
Degration of organic wastes in the absence of air gives |
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Answer» methane ,HYDROGEN neitrogen and `CO_2` |
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| 2342. |
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real) A-2,2 B- 2, 8, 5 C- 2, 7 D- 2, 8, 2 'C' belongs to which period and group? |
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| 2343. |
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real) A-2,2 B- 2, 8, 5 C- 2, 7 D- 2, 8, 2 Find the elements belongs to same group. |
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| 2344. |
Define valency by taking examples of silicon and oxygen . |
| Answer» Solution :The valency of an element is the combining capacity of that element. The valency of an element is determined by the number of valence ELECTRONS present in the atom of that element.Valency of SILICON. It has electronic configuration : 2, 8, 4. THUS, the valency of silicon is 4 as these electrons, can be SHARED with others to complete OCTET. Valency of Oxygen : It has electronic configuration : 2, 6. Thus, the valency of oxygen is 2 as it will gain 2 electrons to complete its octet. | |
| 2345. |
Define valency by taking examples of silicon and oxygen. |
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Answer» Solution :The combining capacity of an atoma is called its VALENCY. The number of electrons present in the OUTERMOST shell of an atom gives its valency. If the number of electrons in the outermost shell is close to OCTET, then valency of the atom is 8 minus the number of ELCTRONS in it. Atomic number of Electronic configuration silicon = 142 , 8 , 4 Atomic number of Electronic configuration oxygen= 82, 6 Silicon has 4 VALENCE electrons. `therefore` the valency of silicon is 4. Oxygen has 6 valence electrons It is nearer to 8 . `therefore` he valency of oxygen = 8 - 6 = 2. Thus, slicon requires 4 electrons and oxygen requires 2 electrons to complete the octet. |
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| 2346. |
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real) A-2,2 B- 2, 8, 5 C- 2, 7 D- 2, 8, 2 Find the elements belongs to same period? |
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| 2347. |
(a) A solution contains 20 g of common salt in 160 g of water. Calculate the concentration in terms of mass percentage of the solution. (b) Write dispersion medium and dispersed phase of fog (aerosol). |
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Answer» Solution :MASS of common salt (solute) = 20 G Mass of water (solvent) = 160 g Mass of solution = Mass of solute + Mass of solvent = (20 + 160)g = 180 g CONCENTRATION = `("Mass of solute")/("Mass of solution")xx100` = `20/180xx100` `~~11.1%` (b) In FOG, the dispersion medium is gas and the dispersed phase is LIQUID. |
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| 2348. |
Define the atomic mass unit. |
| Answer» Solution :One atomic mass UNIT is equal to exactly one TWELFTH the mass of one ATOM of carbon-12. | |
| 2349. |
Define the atomic mass unit ? |
| Answer» Solution :It is a RATIO of MASS of an element to the mass of `(1^(th))/(12)` the mass of carbon -12 ATOM | |
| 2350. |
A 3 L flask contains 0.5 moles of sulphur dioxide at 127^(@)C temperature. Some amount of oxygen is introduced into the flask along with a catalyst. When the pressure reached 8.2 atm, 0.3 moles of SO_(3) is formed and there is no further change in the pressure of the reaction mixture. Calculate K_(c ) value for the above reaction. |
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Answer» Solution :V= 3L P= 8.2 ATM R=0.0821 L atm `K^(-1) " mole"^(-1)` T=127+123=400 K TOTAL number of moles at equilibrium is `n=(PV)/(RT)=(8.2xx3)/(0.0821xx400)=0.75` `2SO_(2)+O_(2) hArr 2SO_(3)` Number of moles of `SO_(2)` at equilibrium is `0.5-0.3 n_(O_(3))0.3=0.2` `therefore` Number of moles of `O_(2)` at equilibrium `n_(O_(2))=n-(n_(SO_(3))+n_(SO_(2)))=0.75-0.5=0.25` `K_(c )=((0.3)^(2))/((0.2)^(2)xx(0.25))=9` |
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