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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.
| 8451. |
A milkman adds a very small amount of baking soda to fresh milk. (a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ? |
| Answer» Solution :The pH of milk changes from pH 6 to slightly alkaline on addition of a very small AMOUNT of baking soda. This is because sodium HYDROGENCARBONATE is BASIC in NATURE. This prevents the milk from souring. | |
| 8452. |
A milkman adds a very small amount of baking soda to fresh milk. Why does this milk take a long time to set as curd ? |
| Answer» Solution :Lactic acid FORMED as a result of FERMENTATION is NEUTRALISED by SODIUM hydrogen-carbonate. This prolongs the time taken by milk to set as curd. | |
| 8453. |
A milkman adds a very small amount of baking soda to fresh milk. (a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ? (b) Why does this milk take a long time to set as curd ? |
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Answer» Solution :(a) MILK is made slightly alkaline so that it may not get sour EASILY due to the formation of lactic acid in it. (b) The alkaline milk takes a longer time to set into curd because the lactic acid formed has to first NEUTRALISE the alkali present in it . |
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| 8454. |
A metallic ore 'X' reacts with dilute HCl to liberate a gas which turns lime water milky. Another ore 'Y' gives off a gas with the smell of rotten eggs on treatment with the same acid. Which metallurgical processes are used for the extraction of the metals X and Y ? |
| Answer» SOLUTION :X is a carbonate ORE since it gives off carbon DIOXIDE on treatment with HCL. So, X is extracted by the process of calcination. Y gives off hydrogen SULPHIDE gas on treatment with dilute HCl. So, it is a sulphide ore which is extracted by roasting. | |
| 8455. |
A metal 'X' which is used in thermite process, when heated with oxygen gives an oxide 'Y' which is amphoteric in nature. Identify X and Y. Write down balanced chemical equations of the reactions of oxide Y with HCl and NaOH. |
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Answer» SOLUTION :`X - Al , Y - Al_(2)O_(3)` `Al_(2)O_(3) + 6HCl rarr 2AlCl_(3) + 3H_(2)O` `Al_(2)O_(3) + 2NAOH rarr 2NaAlO_(2) + H_(2)O` |
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| 8456. |
A metal 'X' forms a salt XSO_(4). The salt forms a clear solution in water which reacts with sodium hydroxide solution to form a blue precipitate Y. Metal X is used in making electric wires and alloys like brass. |
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Answer» Solution :(a) COPPER. (b) Copper hydroxide `[CU(OH)_(2)]`. (c) `CuSO_(4)+2NaOH to Cu(OH)_(2)+NaSO_(4),` Double - displacement reaction. |
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| 8457. |
A metal 'X' combines with a non-metal 'Y' by the transfer of electrons to form a compound Z. (i) State the type of bond in compound Z. (ii) What can you say about the melting point and boiling point of compound Z ? (iii) Will this compound dissolve in kerosne or petrol ? (iv) Will this compound be a good conductor of electricity ? |
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Answer» Solution :(i) Ionic BOND (ii) High (III) No (IV) YES. |
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| 8458. |
A metal 'X' combines with a non-metal ‘Y by the transfer of electrons to form a compound Z.(i) State the type of bond in compound Z.(ii) What can you say about the melting point and boiling point of compound Z? (iii) Will this compound dissolve in kerosene or petrol ? |
| Answer» Solution : (i) IONIC or electrovalent BOND. (ii) It has a HIGH melting point and high boiling point.(III) No, ionic compounds do not dissolve in kerosene or PETROL. | |
| 8462. |
A metal that exists as a liquid at a room temperature is obtained by heating its sulphide in the presence of air, Identify the metal and its ore and give the reaction involved. |
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Answer» Solution :Metals low in activity SERIES can be obtained by REDUCING their sulphides or oxides by heating. Mercury is the only metal that EXISTS as liquid at room temperature. It can be obtained by heating cinnabar `(HgS)` the SULPHIDE ORE of mercury. The reactions are as follows: `2HgS(s)+3O_2(g) to^(Heat) 2HgO(s)+2SO_2(g)` `2HgO(s) to^(Heat) 2Hg(l)+O_2(g)` |
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| 8463. |
A metal object is to be electroplated with silver. The electrolyte selected is sodium argentocyanide . Write the reaction taking place at the anode . |
| Answer» SOLUTION :`AG(s) to Ag^(-) to Ag(s)` | |
| 8464. |
A metal object is to be electroplated with silver. The electrolyte selected is sodium argentocyanide . Write the reaction taking place at the cathode . |
| Answer» SOLUTION :`AG^(+) (AQ) +E^(-) to Ag(s)` | |
| 8465. |
A metal object is to be electroplated with silver .The electroplated with silver . The electrolyte selected is sodium argentocyanide . State one condition to ensure that the deposit is smooth , firm and long lasting . |
| Answer» SOLUTION :The ELECTRIC CURRENT should be PASSED SLOWLY . | |
| 8466. |
A metal object is to be electroplated with silver .The electroplated with silver . The electrolyte selected is sodium argentocyanide . Why is it preferred to use silver nitrate as an electrolyte ? |
| Answer» Solution :Because SODIUM argentocyanide is a weak electrolyte , its ionisation is SLOW as COMPARED to SILVER nitrate . This makes the electrolysis slow , resulting in a smooth PLATING . | |
| 8467. |
A metal object is to be electroplated with silver .The electroplated with silver . The electrolyte selected is sodium argentocyanide . What type of salt is sodium argentocyanide ? |
| Answer» SOLUTION :SODIUM argentocyanide is a COMPLEX SALT . | |
| 8468. |
A metal object is to be electroplated with silver . The electrolyte selected is |
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Answer» silver nitrate |
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| 8469. |
A metal .M. kept in air turns green and when it is heated, it turns black. Name the metal and the compound formed in both cases. |
| Answer» Solution :METAL M is copper. GREEN compound is due to formation of copper CARBONATE and BLACK coloured compound is due to formation of copper oxide. | |
| 8470. |
A metal M is found in nature as MCO_3 It is used in galvanising iron articles. Name the metal. |
| Answer» SOLUTION :The METAL is ZINC (ZN). | |
| 8471. |
A metal 'M' is found in nature as its carbonate. It is used in the galvanization of iron. Identify 'M' and name its ore. How will you convert this ore into free metal ? |
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Answer» Solution :Zinc Ore of zinc is Zinc blende (ZnS). Sulphide ore is heated below its melting point in the presence of air to convert it into METAL OXIDE. It is called Roasting. `2ZnS(s) + 3O_(2)(s) overset("Roasting")rarr 2ZnO(s)+2SO_(2)(g)` REDUCTION of oxide ore : It is the process of conversion of metal oxide ore into metal. It can be done by heating the OXIDES. `ZnO(s) + C(s) rarr ZN(s) + CO(g)` |
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| 8472. |
A metal M found in nature as sulphide ore (M_(2)S) is one of the good conductors of heat and electricity and is used in making electric wires .Identify the metal M. |
| Answer» Solution :(i) The AVAILABLE clue suggests that the metal M is COPPER (CU) and the SULPHIDE ore is cuprous sulphide `(Cu_(2)S)` | |
| 8473. |
A metal M forms an oxide having the formula M_2O_3. It belongs to 3rd period in the Modem Periodic table. Write the atomic number and valency of the metal. |
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Answer» SOLUTION :Atomic NUMBER = 13 Electronic configuration= (2, 8, 3) VALENCY =3 |
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| 8474. |
A metal M belongs to 13th group in the Modern Periodic Table, Write the valency of the metal. |
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Answer» Solution :Atomic NUMBER = 13 (Al) Electronic configuration = 2, 8, 3 Valency = 3 |
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| 8475. |
A metal is treated with dilute sulphuric acid. The gas evolved is collected by the method shown in the figure: (i) Name the gas (ii) Is the gas soluble or insoluble in water? (iii) Is the gas lighter or heavier than air? (iv) How will you test the gas (v) IF the metal used above in zinc then write the chemical equation for the evolution of gas (vi) Write one industrial use of the gas evolved. |
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Answer» Solution :(i) Hydrogen gas (ii) Is soluble in water (iii) Gas is LIGHTER than air. (iv) Test for `H_2` gas: Bring a burning matchsticknear the gas jar. It burns with a pop SOUND. (v) `ZN(s)+H_2SO_4(dil) to ZnSO_4(aq)+H_2(g) uparrow` (vi) Liquid hydrogen is USED as a FUEL in rockets. |
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| 8476. |
A metal is treated with dil. H_(2)SO_(4). The gas evolved is collected by the method shown in the figure. Answer the following : (i) Name the gas. (ii) Name the method of collection of the gas. (iii) Is the gas soluble of insoluble in water ? (iv) Is the gas lighter or heavier than air ? |
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Answer» Solution :(i) The GAS evolved is HYDROGEN. (ii) The METHOD of collection of the gas is the downward DISPLACEMENT of water. (iii) The gas is insoluble in water. That is why it can be collected over water. (iv) The gas is LIGHTER than air. |
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| 8477. |
A metal is found in liquid state. It is widely used in instrument for measuring blood pressure. In what form does it occur in nature ? How can we extract this metal from its ore ? |
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Answer» Solution :Mercury is the metal which is found in LIQUID state and is used in INSTRUMENTS for measuring blood PRESSURE. It occurs in nature as sulphide ore in the FORM of Cinnabar (HgS). Mercury is obtained from purified cinnabar by roasting in the presence of air. `HgS underset("Heat")OVERSET("Air")rarr Hg + SO_(2)` |
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| 8478. |
A metal E is stored under kerosene. When a small piece of it is left open in air, it catches fire. When the product formed is dissolved in water, it turns red litmus to blue : (i) Name the metal E. (ii) Write the chemical equation for the reaction when it is exposed to air and when the product is dissolved in water. (iii) Explain the process by which the metal E is obtained from its molten chloride |
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Answer» Solution :(i) The available INFORMATION SUGGESTS that the metal E is sodium (Na) (II) `underset(("E"))(4Na)(s) + underset(("Air"))(O_(2))(g) rarr2Na_(2)O(s)` `Na_(2)O(s) + H_(2)O(aq) rarr 2NAOH(aq)` The solution is basic and it TURNS red litmus blue. (iii) The metal is obtained by the process of electrolytic-reduction. |
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| 8479. |
A metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle . Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride. |
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Answer» Solution :The GAS the extinguishes a burning candle is carbon DIOXIDE. It is formed by the action of dilute hydrochloric acid on metal carbonate or metal hydrogencarbonate and produces effervescence . Since one of the COMPOUNDS formed is calcium chloride, this shows that the metal compound is calcium carbonate. It cannotbe calcium hydrogencarbonate because calcium hydrogencarbonate is found in solution . Thus , the metal compound A is calcium carbonate `(CaCO_3)` . Calcium carbonate reacts with dilute hydrochloric acid to form calcium chloride. carbon dioxide and water . `CaCO_(3)(s)+2HCl(aq) rarrCaCl_(2)(aq)+CO_(2)(G)+H_(2)O(l)` |
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| 8480. |
A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y, gives a compound Z, Z is used for disinfecting drinking water. Identify X, Y, G and Z. |
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Answer» Solution :The gas evolved at ANODE during electrolysis of brine is chlorine (G) When chlorine gas is passed through dry `CA(OH)_(2)` (Y) products bleaching powder (Z) used for disinfecting DRINKING water. `underset("Slaked lime")(Ca(OH)_(2)) + Cl_(2) rarr underset("Bleaching powder")(CaOCl_(2) + H_(2)O)` Since Y and Z are calcium salts, therefore X is also a calcium salt and is calcium carbonate. `CaCO_(3) + 2HCl rarr CaCl_(2) + CO_(2) + H_(2)O` `Ca(OH)_(2) + CO_(2) rarr CaCO_(3) + H_(2)O` |
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| 8481. |
A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y , it gives a compound Z, used for disinfecting drinking water. Identify X,Y, G and Z. |
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Answer» Solution :The gas evolved at ANODE during electrolysis of brine is chlorine (G). When chlorine gas is passed through dry `Ca(OH)_2` (Y) produces bleaching POWDER bleaching powder (Z) used for disinfecting drinking water. `underset("Slaked lime")(Ca(OH)_2)+Cl_(2)rarrunderset("Bleaching powder")(CaOCl_2)+H_(2)O` SINCE Y and Z are CALCIUM salts , therefore X is also a calcium salt and is calcium carbonate. `CaCO_3+2HCl rarr CaCl_(2) + CO_(2)+H_(2)O` `Ca(OH)_2+CO_(2) rarrCaCO_(3)+H_(2)O` . |
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| 8482. |
A metal belonging to group 13 which is widely used to make household untensils. |
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Answer» |
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| 8483. |
A metal A which is used in thermite process, when heated with oxygen gives an oxide B, whichis amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH. |
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Answer» Solution :The metal A is ALUMINIUM AL. The substance B is `Al_2O_3`which is AMPHOTERIC because it reacts with both acid and base. `Al_2O_3 + 6HCl to 2Al_2O_3 + 3H_2O` `Al_2O_3 + 2NaOH to 2NaAlO_2 + H_2O` |
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| 8484. |
A metal A, which is used in thermit process, when heated with oxygen gives an oxide B,which is amphoteric in nature. Identify A and B.Write down the reaction of oxide B with HCl and NaOH. |
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Answer» Solution :A is aluminium (Al). It REACTS with OXYGEN to FORM aluminium OXIDE `Al_2O_3`. `4Al(s) +3O_2(g) to 2Al_2O_3(s)` So, B is `Al_2O_3`. `Al_2O_3+6HCl to2AlCL_3+3H_2O` `Al_2O_3+2NaOH to 2NaAlO_2+H_2O` |
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| 8485. |
a] Mention the relative position of metals, non-metals, and metalloids in group 13-16 of the periodic table. b] Classify the following elements as metals, non-metals, and metalloids. (i) Calcium (ii) Sulphur (iii) Selenium |
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Answer» Solution :a] In group 13, first element is a METALLOID while the rest of the elements are metals. In group 14, Ist element is a non-METAL, 2nd and 3rd elements are metalloids, and the rest of the elements are metals. In group 15, 1st and 2nd elements are non-metals, 3rd element is a metalloid, and the rest are metals. In group 16, first 3 elements are nonmetals, next two are metalloids, and the remaining one is a metal. b] CALCIUM is a metal, sulphur and selenium are non-metals. |
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| 8486. |
(a) Mention the four informations given by an equation. (b)State the law of conservation of mass as applicable in a chemical reaction. |
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Answer» Solution :(a) (i) Physical state of reactants and products. (II) CONDITIONS such as stemperature, pressure, HEAT etc. (iii) CATALYST involved (iv) Change in state. (b) TOTAL mass of the elements present in the products in a chemical reaction has to be equal to the totalmass neither be created nor destroyed in a chemical reaction . |
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| 8487. |
A measured temperature on Fahrenheit scale is 200^(@)F. What will this reading be one celsium scale? |
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Answer» `40^(@)C` `200=9/5(""^(@)C)+32` `""^(@)C=(200-32)xx5/9=93.33^(@)C` |
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| 8489. |
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used ? |
| Answer» Solution :Aqua regia which is a mixture of 3 parts of concentrated HYDROCHLORIC acid and 1 part of concentratednitric acid dissolves gold. The man put the gold BANGLES in this solution. The outer dirty layer of gold bangles dissolved in aqua regia bringing out the shining bangles.As the outer layer of bangles dissolved in aqua regia, the weight was REDUCED DRASTICALLY. | |
| 8490. |
A man went door to door posing 2s a goldsmith he promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man had a hasty retreat. Canyou play the detective to find out the nature of the solution he had used ? |
| Answer» Solution :The solution he had USED is Aqua regia. It is a mixture of concentrated hydrochloric ACID and concentrated NITRIC acid in the ratio 3:1. Aqua regia has the ABILITY to dissolve the metals like GOLD and platinum. Since the outer layer of the gold bangles gets dissolved in aqua regia so their weight was reduced drastically. | |
| 8491. |
A man went door possing as goldsmith. He promised to bring back the glitter of old and gold ornaments. As unsuspecting lady gave a set of gold bangles to him, which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was ipset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used? |
| Answer» SOLUTION :the goldsmith dipped the set of the GOLD bangles in aqua regia solution. Aqua regia is a freshly PREPARED mixture of 1 part of conc. `HNO_3` and 3 parts of conc. HCL by volume . Aqua regia dissolved a considerable amount of gold from the dipped gold bangles and hence reduced their weight drastically. The dishonest goldsmith can recover the dissolved gold from aqua regia by a suitable treatment. | |
| 8492. |
A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y. (a) Write the chemical formulae of X and Y. (b) Write a balanced chemical equation, when X is dissolved in water. |
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Answer» Solution :`2Mg+O_(2)rarr2MgO+"Light"` `3Mg+N_(2)rarrMg_(3)N_(2)` (i) `X" is "MGO,Y" is "Mg_(3)N_(2)` (ii) `MgO+H_(2)OrarrMg(OH)_(2)` |
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| 8493. |
A lustrous non-metal is |
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Answer» DIAMOND |
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| 8494. |
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarly in the atoms of these elements ? (b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ? |
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Answer» Solution :a] Yes, the atoms of lithium, sodium, and potassium have one ELECTRON in their outermost shell. Hence, they show similar chemical properties. B] Both helium and neon are noble gases. The common THING between their atoms is that both have their outermost shell completely filled. Helium has 2 electrons in its outermost shell (K-shell) while neon has 8 electrons in its outermost shell (L-shell). Since their valence shell is completely filled, they are CHEMICALLY inactive. |
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| 8495. |
(a) Lithium, sodium and potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements ? (b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ? |
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Answer» Solution :(a) Lithium, sodium and POTASSIUM have the same number of electrons (i.e., 1) in the outermost shell. (b) HELIUM and ARGON have completed outermost shells, 2 electrons in case of helium (completed DUPLET) and 8 electrons in case of argon (completed octet). |
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| 8496. |
Mentionthe Properties of Metals ? |
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Answer» Solution :PHYSICAL properties of metals are : i) Metals are usually hard. ii) They are Sonorous, III) They are lustrus iv) Metals exhibit malleability and ductility v) They exhibit high tensil strength and have high densities: Roasting :It is a process in which sulphide ore is HEATED in the presence of oxygen to convert into oxide `2Zns-3O to 2ZnO + 2SO_2` Calcination : It is a process in which carbonate ore in heated in the ABSENCE of air of form OXIDES. `ZnCO_(3(s)) overset("heat")to ZnO_((s)) + CO_2` By reduction process Zn can be extracted from its ore. Reduction: `2ZnO+ C to ZnO + CO_2` |
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| 8497. |
A list of common metals arranged in order of decreasing reactivity is known as ____ |
| Answer» SOLUTION :ACTIVITY SERIES | |
| 8498. |
(a) List four characteristics of homologous series. (b) Draw the electron dot structure of carbon dioxide. |
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Answer» Solution :(a) Four characteristics of HOMOLOGOUS series are : (i) Same functional group. (II) Similar CHEMICAL properties. (III) Regular gradation in physical properties. (iv) Successive member differ by `- CH_(2)` (b) 
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| 8499. |
(a) List any three observations which determine that a chemical reaction has taken place . Also list three informations that cannot be obtained abouta chemical reaction, merely by its chemical equation. (b) Balance the following chemical equations. (i) Fe+H_(2)OrarrFe_(3)O_(4)+H_(2) (ii) CO_(2)+H_(2)O rarr C_(6)H_(12)O_(6)+O_(2) |
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Answer» Solution :(a) Three observations which determine that achemical reactionhas taken place are : (i) Change in STATE and colour . (ii) EVOLUTION of gas . (iii) Change in temperature. Three informations that cannot be obtained about a chemical reaction, merely by its chemical reuation are : (i) The state of a matter of the reactants and products . (ii) Amount of heat evolved or absorbed in a reaction . (iii) The change of colour after the chemical reaction . (b) Balanced chemical equations : (i) `3Fe+4H_(2)OrarrFe_(3)O_(4)+4H_(2)` (ii) `6CO_(2)+6H_(2)OrarrC_(6)H_(12)O_(6)+6O_(2)` |
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| 8500. |
a) List any three observation that determine that a chemical reaction has taken place. Also list three in formation that cannot be obtained about a chemical reaction merely by its chemical equation. b) Balance the following chemical equations i) Fe+H_(2)O to Fe_(3)O_(4) +H_(2) ii) CO_(2)+H_(2)O to C_(6)H_(12)O_(6) +O_(2) OR a) Explain the following in terms of gain or lose of oxygen with two examples each. i) Oxidation ii) reduction. b) Balance the following chemical equations. 1)HNO_(3)+Ca(OH)_(2) to Ca(NO_(3))_(2)+H_(2)O ii) NaOH + H_(2)SO_(4) to Na_(2)SO_(4) + H_(2)O |
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Answer» Solution :The three observations are i) Change of colour ii) Change of temperature ii) Evolution of gas The three information are i) Atmospheric conditions ii) catalyst inolved iii) Physical state of reactants and PRODUCTS Balancing equations i) `3Fe+4H_(2)O to Fe_(3)O_(4) +4H_(2)` ii) `6CO_(2) + 6H_(2)O to C_(6)H_(12)O_(6) + 6O_(2)` OR a) i) Oxidation : It is defined as PROCESS which involves gain of oxygen for example `2Mg(s) + O_(2)(g) to 2MgO(s)` Magnesium oxygen Magnesiumoxide. ii) Reduction: It is defind as the process which involves loss of oxygen for example `CuO_((s)) + H_(2)(g) to Cu(s) + H_(2)O(g)` Copper Hydrogen Copper Water b)i) `2HNO_(3aq)+ Ca(OH)_(2aq) to Ca(NO_(3))_(2aq) + H_(2)O` ii) `2NaOH_(aq) + H_(2)SO_((4)(aq)) to Na_(2) SO_((4)(aq)) + 2H_(2)O(I)` |
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