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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.
| 8651. |
(A) C_(2)H_(5)OH,C_(3)H_(7)OH and C_(4)H_(9)OH are members of homologous series. ( R ) These compounds do not have same functional group. |
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Answer» Both ASSERTION (A) and REASON ( R ) are TRUE. And reason ( R ) is the true explanation of the assertion (A). |
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| 8652. |
A bud of petunias becomes reddish purple after first shower of rain. What does it indicate ? |
| Answer» Solution :COLOUR of petunia BECOMES REDDISH purple in acidic medium. It indicates that first SHOWERS of RAIN contained acidic impurities | |
| 8653. |
A brown substance 'X' on heating in air forms a substance 'Y'. When hydrogen gas is passed over heated 'Y', it again changes back into 'X'. Name the substances X and Y. |
| Answer» SOLUTION :X: copper METAL , Y: Copper (II) oxide. | |
| 8654. |
A brown substance 'X' on heating in air forms a substance 'Y'. When hydrogen gas is passed over heated 'Y', it again changes back into 'X'. (i) Name the substance 'X' and 'Y'. (ii) Name the type of chemical reactions occuring during both the changes. (iii) Write the chemical equations of the reactions. |
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Answer» Solution :(i) `X-Cu,Y-CuO`. (II) When copper is heated in air, OXIDATION take PLACE. When hydrogen gas is passed over heated copper oxide, reduction takes place. (iii) `2Cu+O_(2)overset(DELTA)rarr2CuO` (Oxidation) `CuO+H_(2)overset(Delta)rarrCu+H_(2)O` (Reduction) |
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| 8655. |
A brass strip of dimensions 4 cm and 6 cm has to be coated with a silver layer of 2 min thickness. Calculate the amount of charge required for the deposition of silver and the amount of magnesium deposition in another electrolytic cell when same amount of charge is passed (density of Ag = 10.5 g/cc) |
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Answer» Solution :DENSITY of Ag = 10.5 g//cc Mass of silver REQUIRED for coating brass spoon `= 10.5 xx 4 xx 0.2 xx 6= 50.4 g` 108 g of Ag requires 1 FARADAY of charge. Therefore, 50.4 g of Ag require ------ ? `= (50.4)/(108) = 0.466` faradays 1 faraday of charge DEPOSITE 12 g of Mg. Therefore, 0.466 faraday charge deposite ? `= 0.466 xx 12 = 5.6 g` |
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| 8656. |
A blue - coloured salt becomes white on heating. Give reason for the above observation. What happens when we add water to the salt obtained after heating ? Also write its formula. |
| Answer» Solution :The blue salt is hydrated copper sulphate `(CuSO_(4).5H_(2)O)`, which LOSES its water of crystallization and becomes WHITE anhydrous copper sulphate `(CuSO_(4))` UPON heating it. On adding water, the anhydrous form again becomes hydrated and the blue colour is REGAINED. | |
| 8657. |
A black dot used as a full stop at the end of a sentence has a mass of about one attogram.Assuming that the dot is made up of carbon, calculate the approximate number of carbon atoms present in the dot. |
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| 8658. |
A binary compound contains 50% A (at. mass = 16) and 50% B (at. mass 32). The empirical formula of the compound is _____________. |
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| 8659. |
A + BC toAC+Bisa soubledisplacementreaction . |
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| 8660. |
A basic salt that is used in daily life is |
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Answer» SODIUM acetate |
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| 8662. |
A balanced chemical equation always obeys (a) Law of Conservatio of Mass |
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Answer» LAW of Conservatio of MASS |
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| 8663. |
A baker found that the cake prepared by him is hard and small in size. Which ingredient had he forgotten to add that would have caused the cake to rise and become light ? Give reason. |
| Answer» Solution :The BAKER had forgotten to ADD BAKING powder . Baking powder is a mixture of baking soda (sodium hydrogencarbonate) and a mild edible ACID such as tartaric acid. When this baking powder is added with WATER , then sodium hydrogencarbonate `(NaHCO_3)` reacts tartaric acid to evolve carbon dioxide gas. This `CO_2` gas causes the cake to rise and become soft and spongy. | |
| 8664. |
A, B C and D are four metals. Both A and B react with concentrated alkali and also with steam. Reaction of A with steam is rapid in the beginning and becones slow later. Oxide of C on treatment with A liberates huge amount of heat and C produces borwn-coloured flakes on exposure to humid atmosphere. D forms two types of oxides below 1370 K and above 1370 K. Identify A, B, C and D and write all the balanced chemical equations. |
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Answer» Solution :Since A and B reacts with concentrated alkali, they are Al and Zn. Reaction of A with steam is rapid in the beginning and becomes slow later and hence A is Al. `2H_(2)O+underset((A))2AI+2NaOHtoNaAlO_(2)+2H_(2)uarr` `Zn""underset((B))+NaOHtoNa_(2)ZnO_(2)+H_(2)uarr` `2Al+3H_(2)OtoAl_(2)O_(3)+3H_(2)` `Zn+H_(2)OtoZnO+H_(2)` Since oxide of C produces huge AMOUNT of HEAT with A and FLAKES with humid atmosphere, it is iron. `Fe_(2)O_(3)+2Al""to2Fe+Al_(2)O_(3)+"Heat"` As D forms TWO TYPES of oxides above and below 1370 K, it is Cu. `underset((D))Cu+O_(2)overset(lt1370K)toCu_(2)O` `Cu+O_(2)overset(gt1370K)toCu_(2)O` |
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| 8665. |
A, B and C are elements of Dobereiner's triad. If the atomic mass of A is 7 and that of C is 23, what will be the atomic mass of B. |
| Answer» Solution :Atomic mass of `B=(7+23)/(2) = (30)/(2) =15`. | |
| 8666. |
A, B and are three elements which undergo chemical according to the following equations : {:(""A_(2)O_(3)2B to B_(2)O_(3)+2A),(3CSO_(4)+2B to B_(2)(SO_(4))_(3)+3C),(" "3CO+2A to A_(2)O_(3)+3C):} Answer the following questions with reactions : Which element is the least reactive ? |
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Answer» SOLUTION :C is the LEAST reactive element. C is DISPLACED both by A and B. |
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| 8667. |
A, B and are three elements which undergo chemical according to the following equations : {:(""A_(2)O_(3)2B to B_(2)O_(3)+2A),(3CSO_(4)+2B to B_(2)(SO_(4))_(3)+3C),(" "3CO+2A to A_(2)O_(3)+3C):} Answer the following questions with reactions : What is the type of reactions listed above ? |
| Answer» SOLUTION :The REACTIONS GIVEN above are all DISPLACEMENT reactions | |
| 8668. |
A, B and are three elements which undergo chemical according to the following equations : {:(""A_(2)O_(3)2B to B_(2)O_(3)+2A),(3CSO_(4)+2B to B_(2)(SO_(4))_(3)+3C),(" "3CO+2A to A_(2)O_(3)+3C):} Answer the following questions with reactions : Which element is the most reactive ? |
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Answer» Solution :B is the most reactive element. B displaces both A and C from their SOLUTIONS. |
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| 8669. |
(A) Animal fats should be chosen for cooking. ( R ) Animal fats are not harmful for health. |
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Answer» Both ASSERTION (A) and REASON ( R ) are true. And reason ( R ) is the true EXPLANATION of the assertion (A). |
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| 8670. |
a) An object is kept between centre of curvature and principal focus of a concave mirror. Write the nature of the image formed. b) Define focal length of a convex mirror. Write the relationship between focal length and radius of curvature of a convex mirror. |
Answer» Solution :(a)  Real and INVERTED (B) The distance between the pole and the principal focus of a convex MIRROR is called the focal length of a convex mirror. The relationship between focal length and radius of curvature of a convex mirror. The radius of curvature is FOUND to be equal to twice the focal length. R=2f This implies that the principal focus of a spherical mirror lies midway between the pole and centre of curvature. |
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| 8671. |
(a) All ores are minerals but all minerals are not ores. Justify. (b) An iron knife kept in blue copper sulphate solution turns the blue solution into light green. Explain. (c ) An athlete won a bronze medal in a race competition. After some days, he found that the medal had lost its lustre due to the formation of a greenish layer on it. Name the metals present in the medal. What is the reason for the formation of a greenish layer on its surface ? |
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Answer» Solution :(a) In the earth's crust, metals are present in the FORM of minerals and there are more than one mineral for a PARTICULAR metal. However, metal may not be extracted from all of them. The mineral from which a metal can be profitably and conveniently extracted is known as ore. This clearly means that all ores are minerals but all minerals are not ores. For example, the different minerals of iron are : `{:("Haematite",:Fe_(2)O_(3),", Limonite",:Fe_(2)O_(3).3H_(2)O.),("Siderite",:FeCO_(3),", Iron PYRITES",:FeS_(2).):}` Iron is extracted from haematite `(Fe_(2)O_(3))`. Haematite mineral is the ore of iron while other minerals are not the ores. (b) Iron LIES above copper in the activity series. This means that iron or iron knife will displace copper from copper sulphate solution. As a result of the reaction, ferrous sulphate will be formed and the solution will be light green in colour. `{:(Fe(s),+,CuSO_(4)(aq),rarr,FeSO_(4)(aq),+,Cu(s)),("Iron",,"Copper sulphate",,"Ferrous sulphate",,"Copper"),(,,"(blue)",,"(light green)",,):}` (C ) The bronze medal is an alloy and the constituting metals are copper and tin. The loss of lustre by the medal is due to the formation of a coating of green layer. This layer is of basic copper carbonate. |
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| 8672. |
(A) Acidified K_(2)Cr_(2)O_(7) is used in oxidation of ethanol. ( R ) Acidic K_(2)Cr_(2)O_(7) is a reducing agent. |
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Answer» Both ASSERTION (A) and REASON ( R ) are true. And reason ( R ) is the true EXPLANATION of the assertion (A). |
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| 8673. |
(a) A security mirror used in a big showroom has radius of curvature 5 m. If a customer is standing at a distance of 20 m from the cash counter, find the position, nature and size of the image formed in the security mirror. (b) Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with a mirror. State the nature of this mirror and reason for its use in the instrument used by dentist. |
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Answer» Solution :The security mirror is a convex mirror whose radius of curvature R = +5 m or FOCAL length `f=(R )/(2)= +2.5m`, distance of object (customer) from the cash counter (i.e., the image) u = 20m and as per sign convention u is -ve i.e., u = -20 m. As per mirror formula `(1)/(v)+(1)/(u)=(1)/(f)`, we have `(1)/(v)=(1)/(f)-(1)/(u)=(1)/((+5))-(1)/((-20))=(1)/(5)+(1)/(20)=(1)/(4)` `RARR v=4 m and m=(-v)/(u)=(-4)/((-20))=+0.2` Thus, image is formed at a distance of 4 m BEHIND the security mirror. The image is virtual, erect and diminished one (of magnification 0.2). (B) The mirror used by dentist is a concave mirror. Dentist used this mirror so as to form large images of the teeth of patients so that he can examine them easily and in more details. |
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| 8674. |
(A) A mixture of oxygen and ethyne is burnt for welding. ( R ) Burning of the mixture releases a large amount of heat. |
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Answer» Both assertion (A) and reason ( R ) are true. And reason ( R ) is the true explanation of the assertion (A). |
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| 8675. |
a) A, B and C are 3 elements which undergo chemical reaction according to following equations a)A_2O_3 +2B to B_2O_3 + 2A b)3CSO_4 + 2B to B_2(SO_4)_3 + 3C c)3CO + 2Ato A_2 SO_3 + 3C Answer the following: i) Which element is most reactive? ii) Which element is least reactive? b) Why is ZnO called amphoteric oxide? Name another amphoteric oxide. |
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Answer» SOLUTION :a) i) Most reactive element is B as it has replaced both A and C from their compounds. ii) Element C is least reactive as it has been replaced both by A and B b) Zinc oxide (ZnO) is called amphoteric oxide as it behaves both as ACIDIC oxide and basic oxide. `underest("basic oxide")(ZnO) + underset("acid")(2HCl) to underset("zinc chloride")(ZnCl_2) + H_2O` `underset("acidic oxide")(ZnO)+ underset("base")(2NaOH) to underset("sodium zincate")(Na_2ZnO_2) + H_2O` Alluminium oxide `(Al_2O_3)`is another amphoteric oxide. |
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| 8676. |
A 600 mL vessel containing oxygen at 800 mm anda 400 mL vesselcontaining nitrogen at 600 mm, at the same temperature, are joined to each other . The final pressure of the mixture is |
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Answer» 1400MM |
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| 8677. |
A 25% alcohol solution means ........ |
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Answer» 25 ml ALCOHOL in 100 ml of WATER |
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| 8678. |
A 2 L flask contains 22 g of carbondioxide and 1 g of helium at 20^(@)C . Calculate the partial pressure exerted by CO_(2) and He if the total pressure is 3 atm. |
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Answer» SOLUTION :Number of moles of `CO_(2)=("WEIGHT of "CO_(2))/("Molecular Weight of "CO_(2))=22/44 = 0.5`. Number of moles of He = `1/4 = 0.25` Partial pressure = Mole fraction `xx` total pressure `pco_(2) =(0.5)/(0.5+0.25) xx3=2` `p_(He) =(0.25)/(0.75) xx 3 = 1` ATM. |
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| 8679. |
A 1.25g sample of octane (C_(8)H_(18)) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 to 300.78 K. If heat capacity of the calorimeter is 8.93 kJ K^(-1). Find the heat transferred to calorimeter. |
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| 8680. |
A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na_(2)CO_(3) to precipitate calcium as calcium carbonate. This CaCO_(3)is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 g. Calculate % by mass of NaCl in original solution. |
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| 8681. |
A 1 L reaction vessel contained1 mole each of solid NH_(4)HS, NH_(30 and H_(2)S at a temperature of 150^(@)C . When the decompositionof NH_(4)HS was carried out , equilibrium is established K_(p) value at that temperature is 100. Calculate the equilibrium partial pressures at which 60 per centdissociation of NH_(4)HS takes place at a lower temperature where K_(p) value is equal to 200 "atm"^(2) . |
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Answer» SOLUTION :`NH_(4)HS_((s)) hArr NH_(3(g)) + H_(2)S_((g))` `1-x""1+x""1+x` `K_(p)=P_(NH_(3)) xx P_(H_(2)S)` Total no. of moles at EQUILIBRIUM `=(1+x)+(1+x)=2x+2` Partial pressure=MOLE fraction `xx`P Mole fraction of `NH_(3)=(1+x)/(2x+2)` `therefore ` Partial pressure of `NH_(3)=(1+x)/(2x+2)xxP` Mole fraction of `H_(2)S=(1+x)/(2x+2)` `K_(p)=((1+x)/(2x+x)xxP)((1+x)/(2x+2)xxP)` `K_(p)=((1+x)^(2))/((2x+2)^(2))=100` `((1+x)^(2)P^(2))/(4(1+x)^(2))=100 implies P^(2)=100xx4` `P=sqrt(400)=200 ` atm `P_(H_(2)S)=((1+x)20)/(2(x+1))=10`atm `P_(NH_(3))=10` atm per cent DISSOCIATION of `NH_(3) and H_(2)S` corresponds to moles each of `NH_(3) and H_(2)S` `K_(p)=P_(NH_(3))xxP_(H_(2)S)` `P_(NH_(3))=(1.6)/(3.2)xxp,P_(H_(2)S)=(1.6)/(3.2)xxP` `K_(P)=((1.6)xxP^(2))/((3.2)^(2))=400` `P^(2)=(200xx(3.2)^(2))/((1.6)^(2))=400` `P=sqrt(400)=20` atm |
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| 8683. |
98% pure copper and 2% impurities is called ……………… . |
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Answer» Matte |
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| 8684. |
Name the functional group present in (a) CH_3CHO ""(b) C_2H_5COOH |
| Answer» SOLUTION :i] ALDEHYDE II] CARBOXYLIC ACID. | |
| 8685. |
8 g of mixture of NaNO_(3) and Na_(2)CO_(3) are dissolved and made into 500 mL of solution. 50 mL of this solution neutralises completely 25 mL of N/5 HNO_(3). Calculate the precentage composition of the mixture. |
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Answer» Solution :`Na_(2)CO_(3) + 2HNO_(3) rarr 2NaNO_(3) + H_(2)CO_(3)` Only sodium carbonate reacts with acid `(HNO_(3)) N_(1) V_(1) = N_(2) V_(2)` Normality of `Na_(2)CO_(3) " is " (0.2 xx 25)/(50) = 0.1 N` `:. 0.1 = ("wt of " Na_(2)CO_(3))/(53) xx (1000)/(500)` `:.wt " of " Na_(2)CO_(3) = (0.1 xx 53 xx 500)/(1000) = 2.65g` `:. wt " of " NaNO_(3) " is " = 8 - 2.65 = 5.35g` % of `NaNO_(3) = (5.35)/(8) xx 100 = 66.8%` % of `Na_(2)CO_(3) = (2.65)/(8) xx 100 = 33.2%` |
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| 8686. |
7^( N^( 15)) and 8^(O^(16)) are a pair of |
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Answer» isotopes |
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| 8687. |
6g of hydrogen is burnt in the presence of excess oxygen. The mass of water formed is: |
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Answer» Solution :`underset("2 MOLES")(2H_(2))+O_(2) to underset("2 moles")(2H_(2)O)` 6gm= 3 moles From EQUATION 2moles `H_(2)` = 2moles `H_(2)O` `:.` 3 moles `H_(2)` = 3 moles `H_(2)O` Now mass of `H_(2)O` = mass `xx` Molecular wt. `= 3 xx 18= 54 g` |
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| 8688. |
""_(6)^(12C and ""_(6)^(14)Care two elements. Do both these elements get different positions in modern periodic table? Explain your answer. Identify the period and group to which they belong in the modern periodic table with suitable reason. |
| Answer» SOLUTION :No, `""_(6)^(12)C` and `""_(6)^(14)C`will be placed in same slot (POSITION). The MODERN periodic table is based on atomic number, not on atomic mass. Since, both C-12 and C - 14 have some atomic number, hence they are placed in same slot. | |
| 8689. |
6.02xx10^(23) molecules of urea are present in 100 ml of its solution. The concentration of urea solution is - |
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Answer» 0.02 M |
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| 8690. |
6 xx 10^(–3) mole K_(2)Cr_(2)O_(7) reacts completely with 9 xx 10^(–3) mole X^(n+) to give XO_(3)^(-)and Cr^(3+). Find the value of X. |
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Answer» SOLUTION :`K_(2)Cr_(2)O_(7)+X^(n+)RARROVERSET(+5)(X)O_(3)^(-)+CR^(3+)` `6xx10^(-3)xx6=(5n)xx9xx10^(-3)rarrn=1` |
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| 8691. |
5.6 litre of oxygen at S.T.P |
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Answer» Solution :Given volume of `O_(2)` at `"NUMBER of MOLES"=("S.T.P.")/("Molar volume at S.T.P")` `=46//23` `"Number of moles of OXYGEN"=(5.6)/(22.4)="0.25 mole of oxygen"` |
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| 8692. |
50g common salt is dissolved in 150g of water. Find out the concentration of solution in terms of weight percent. |
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Answer» Solution :Weightpercent `=("Weightof THESOLUTE")/("Weightof solute+Weightof SOLVENT ")XX 100` `= (50 )/( 50+150 ) xx 100 = 25 % ` |
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| 8693. |
50 mL of a gaseous hydrocarbon is mixed with excess of oxygen and burnt and cooled to thelaboratory temperature. Thereduction in volume wasfound to be150 mL. The gas is then passed intocaustic potash, when there is a further reduction inthe volume of 150 mL. Provide all thevolumesare measured at the same conditions of temperature and pressure, find out themolecular formula of thehydrocarbon. |
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Answer» Solution :Chemical REACTION equation to combustion reaction is `{:(C_(x)H_(y)+O_(2) to CO_(2)+H_(2)O),("50ML"" 150mL""150mL"):}` We know volume of gas has same number of moles, at same temperature andpressure `{:("50 moles"" 150 moles""150 moles"),(C_(x)H_(y)+(x+y/4)O_(2) toxCO_(2) + y/2 H_(2)O),("50 MOLE""150 mole""150 mole"):}` Given that reduction in volume, when thegases mixture is cooled to the laboratory temperature = 150 ML `:. y/2 (50) = 150 rArr y = 6`. and `x (50) = 150 rArr x = 3` `:.` Hydrocarbon is `C_(3)H_(6)`. |
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| 8694. |
50 g of saturated solution of NaCl at 30^@ C is evaporated to dryness and 13.2 g of dry NaCl was obtained. Find the solubility of NaCl at 30^@ C in water. |
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Answer» SOLUTION :Massof water in solution `= 50-13.2=36.8 g` Solubility of NACL `=( "MASS of NaCl ")/("Mass of water")xx 100` `= ( 13.2 )/( 36.8 )xx 100 =36 g ` Solubilityof NaCl=36 g ( APPX.) |
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| 8695. |
5% sugar solution means ................ |
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Answer» 5 G of sugar in 95 g of water |
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| 8697. |
5 mL of acetic acid is dissolved in 20mL of water. The volume of the solution becomes. |
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Answer» 25 mL |
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| 8698. |
5 mL each of concentrated HCl and concentrated HNO3 are taken in test tubes labelled as A and B while a mixture of concentrated HCI (15 mL) and concentrated HNO_3(5 mL) is taken in test tube labelled as C. A small piece of metal is placed in each test tube. No change is observed in test tubes A and B, but metal got dissolved in test tube C. What would be the metal ? |
| Answer» Solution :Metal WOULD be silver, GOLD or PLATINUM | |
| 8699. |
4g of NaOH is dissolved in water and the volume is made upto 1 L. (1 mole of NaOH = 40 g) How will you make 1 M solution of NaOH using the same amount (4g) of NaOH? |
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| 8700. |
4.523 + 2.34 + 1.23 is equal to |
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Answer» 8.093 |
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