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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.
| 2351. |
Define sublimation. |
| Answer» Solution :Certain solid CHANGE DIRECTLY to gas without passing through the LIQUID is CALLED sublimation. | |
| 2352. |
(a) 2A_((g)) hArr 2B_((g))+C(g) (b) X_((g)) hArr Y_((g)) + Z_((g)) (c ) 2P_((g)) hArr Q_((g))+R_((g)) (d) 2R_((g)) hArr 3S_((g))+2T_((g)) The above reactions are carried out in four separate closed vessels of same volume and same temperature is maintained throughout. If all the reactions are inititained throughout. If all the reactions are initiated with with equal number of moles of the reactants and their degrees of dissociation are same, arrange them in the increasing order of pressure after the reactions attain equilibrium. |
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Answer» c a d B `(II) 2A(g) hArr 2B(g)+C(g)` `(III) X(g) hArr Y(g)+Z(g)` (iv)` 2R(g) hArr 3S(g)+2T(g)` |
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| 2353. |
Define solubility. |
| Answer» Solution :The amount of solute that can be DISSOLVED in a solvent, at a given temperature, to make a SATURATED solution is CALLED its solubility at that temperature. | |
| 2354. |
(a) 110 g of a solute are present in 550 g of solution. Calculate the concentration of solution. (b) Give any three points of difference between true solution, colloidal solution and suspension. |
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Answer» Solution :(a) MASS of solute = 110 g Mass of solution = 550 g Concentration of solution = `("Mass of solute")/("Mass of solution")XX 100 = ((110G))/((550g)) xx 100 = 20%` |
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| 2355. |
Define rate of a reaction. How do we represent it mathematically ? |
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| 2356. |
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of buron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight. |
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Answer» SOLUTION :The COMPOUND of OXYGEN and boron `to` boron + oxygen `0.24 g to 0.096 g + 0.144 g` Percentage of burun in the sample = `(0.096)/(0.24) xx 100 = 40` Percentage of oxygen in the sample = `(0.144)/(0.24) xx 100 = 60` Boron : 40%, Oxygen : 60% |
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| 2357. |
Definemelting point . |
| Answer» SOLUTION :The specific TEMPERATUREAT which a SOLID gets converted into liquid is CALLED the meltingpoint of the solid . | |
| 2358. |
Define inonisation potential |
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| 2359. |
0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.144g of oxygen and 0.096g of boron. Calculate the percentage composition of the compound by weight. (AS-1) |
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Answer» Solution :0.24 gm sample of COMPOUND of OXYGEN and boron 0.096 g of boron 0.144 gm of oxygen 0.250 gm `0.25 -0.24 = 0.1` `THEREFORE (0.1)/(0.25) xx 100 = (10)/(2cancel5) xx 10cancel0 = 40 %` `0 * 24 to 0 * 144` `(100 xx 0*144)/(0*24) = 60` % |
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| 2360. |
""_(92)^(234)U, " "_(92)^(239)U and " "_(92)^(238)U are .... Of uranium . |
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| 2361. |
Define electronegativity .What are its units? |
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| 2362. |
5.4 g of a trivalent metal is deposited by passing 5 amp current during the electrolysis of its molten chloride. If the atomic weight of the metal is 27, calculate the time taken for the deposition. |
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Answer» Solution :(i) calculation of number of moles of the METAL deposited (II) calculation of AMOUNT of charge required (iii) calculation of the time taken for the deposition (IV) 3.22 H |
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| 2363. |
Define electronegativity? |
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| 2364. |
5 moles of CO_(2) and 5 moles of H_(2)O do not have the same mass. Justify. |
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Answer» Solution :Molar mass of `CO_(2) = 12 + (2 xx 16)` = 12 + 32 = 44 g Molar mass of 5 moles of `CO_(2) = 5 xx 44` = 220g ...... (1) Molar mass of `H_(2)O = (2 xx 1) + 16` = 2 + 16 =18 g Molar mass of 5 moles of `H_(2)O = 5 xx 18` = 90 g .... (2) From (1) and (2), 5 moles of `CO_(2)` and 5 moles of `H_(2)O` do not have the same mass. |
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| 2365. |
Define distillation. What type of mixtures can be separated by distillation ? Draw a labelled diagram of the apparatus used for fractional distillation. |
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Answer» Solution :For the definition of distillation, The PROCESS of distillation, ALSO called simple distillation can be used : (i) to SEPARATE two miscible liquids which differ in their boiling point temperature by `25^(@)C` or more. (ii) to purify an IMPURE liquid containing soluble non-volatile impurities. For the labelled diagram of the apparatus used for fractional distillation. |
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| 2366. |
Define condensation ? |
| Answer» SOLUTION :The CHANGE of SOLID STATE from liquid state on cooling is known as condensation. | |
| 2367. |
5 mL of water was taken in a test tube and china dish separately. These samples were then kept under different conditions as below : (a) Both the samples are kept under a fan. (b) Both the samples are kept inside a cup board. State in which case evaporation will be faster ? Give reason to support your answer. (c) How will the rate of evaporation change if above activity is carried out on a rainy day ? Justify your answer. |
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Answer» Solution :(a) Evaporation will be faster under a fan. Since fan helps in the fast movement of the air, water will GET more opportunity to evaporate under a fan than inside a CUP board. In this case, the outside air will not COME in CONTACT with water. (b) On a rainy DAY, there is humidity in air. As a result, evaporation of water will slow down. |
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| 2368. |
Define boiling pont of water. What happens to the boiling point of water when the atmosphere pressure increases? |
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Answer» SOLUTION :The temperature at which pure WATER boils at ATMOSPHERE pressure is called the boiling point of water. The boiling point of water increases with the INCREASE of atmosphere pressure. |
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| 2369. |
5, Calculate the number of neutrons, protons and electrons (i) atomic number 3 and mass number 7 (ii) atomic number 92 and mass number 238 |
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Answer» Solution :(i)Number of NEUTRONS -4 Number of protons -3 Number of electrons-3 (ii) Number of neutrons - 146 Number of protons - 92 Number of electrons - 92 |
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| 2370. |
40 mL of alcohol is mixed with 160 mL of water. |
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Answer» SOLUTION :VOLUME of alcohol (solute) = 40 mL Volume of water (solvent) = 160 mL `THEREFORE` Volume of solute + Volume of solvent = 40mL + 160 mL = 200 mL Percentage of concentration of solution, `(%v//v)=("Volume of solute")/("Volume of solution")XX100` = `40/200xx100=20%` |
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| 2371. |
Define atom. |
| Answer» SOLUTION :The SMALLEST PARTICLE of an element which takes part in the chemical REACTION is KNOWN as an atom. | |
| 2372. |
30 moles of PCl_(5) are placed in a 2 L reaction vessel and heated. K_(c ) for the reaction is 0.5. What is the composition of the mixture at equilibrium ? PCl_(5) hArr PCl_(3)+Cl_(2) |
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Answer» Solution :`(i) K_(c )=([HI]^(2))/([H_(2)][I_(2)])` `K_(c )=(["Products"]^("COEFFICIENTS"))/(["Reactants"]^("coefficients"))` (ii) calculation of concentration of the reactants and products at EQUILIBRIUM (iii) 5 MOLES of `PCl_(3)`, 5 moles of `Cl_(2)` and 25 moles of `PCl_(5)` |
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| 2373. |
Define aquaregia. |
| Answer» Solution :METALS like gold and silver are not reactive with either HCl or `HNO_3.` But the mixture of these two acids can dissolve gold. This mixture is called Aquaregia. It is a mixture of hydrochloric acid and NITRIC acid prepared optimally in a molar ratio of 3:1. It is a yellow- orange fuming LIQUID. It is a highly corrosive liquid, ABLE to attack gold and other resistant substances. | |
| 2374. |
Define a reversible reaction. Give an example . |
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| 2375. |
30 g of glucose and 20 g common saltis dissolved in 500 mL of water to makea solution. Calculate the concentration of (a) glucose and (b) common salt in termsof mass by mass percentage of the solution. (Density of water = 1 g/mL) |
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Answer» Solution :DENSITY of water = 1 g/mL `THEREFORE` Mass of water (solvent) = 500 g Mass of solute (glucose) = 30 g Mass of solute (COMMON salt) = 20 g Mass of solution = (30 + 20 + 500)g = 550 g (a) Mass percentage of solution for glucose = `("Mass of solute (glucose)")/("Mass of solution")xx100` = `30/550xx100` = 5.45% (B) Mass percentage of solution for common salt = `("Mass of solute (common salt)")/("Mass of solution")xx100` = `20/550xx100` = `3.64%` |
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| 2376. |
Define a mixture,Givetwopointsofevidenceto showthatsugar solutionis amixture |
| Answer» Solution :A MIXTURE is formed from ELEMENTS and/or COMPOUNDS mixed in any PROPORTION . It may be homogeneous or heterogeneous . | |
| 2377. |
Define (a) Compressibility (b) Rigidity (c ) Fluidity. |
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Answer» Solution :(a) COMPRESSIBILITY. It is the property as a result of which the PARTICLES of any matter come closer on applying pressure. (b) Rigidity. It is the capacity of the particles of a matter to RESIST a change in shape and SIZE on applying stress. (c ) Fluidity. It is the property as a result of which particles of a matter have tendency to flow. |
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| 2378. |
2SO_(2)+O_(2) to 2SO_(3)+ " Heat". The favourable conditions for the above reaction are |
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Answer» LOW PRESSURE, HIGH temperature |
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| 2379. |
Amongst the following statements, select the set having statements which was porposed by Dalton. (1) All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. (2) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P (3) Chemical reaction involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction. (4) Matter consists of indivisible atoms |
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| 2380. |
C_((s))+CO_(2(g)) hArr 2CO_((g)) -" Heat". The favourable conditions for the above reaction are |
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Answer» LOW PRESSURE, high temperature Since `n_(p) GT n_(R ) implies ` low pressure favours forward reaction. Since the reaction is ENDOTHERMIC, high temperature favours forward reaction. |
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| 2381. |
2A + 4B to 3C + 5D In the given reaction, the rates of reaction with respect to A, B, C and D are (a) r_(A) "" (b) r_(B) (c ) r_(C ) "" (d) r_(D) Arrange them in the ascending order of their magnitude. |
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Answer» a c b d `(i) r_(A) "" (ii) r_(C )` (iii) ` r_(B) "" (iv) r_(D)` |
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| 2383. |
Covalent bond is formed by transfer of electrons (b) sharing of electrons(c) sharing a pair of electrons |
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| 2385. |
Correct the wrong statements. a. lonic compounds dissolve in non polar solvents b. Covalent compounds conduct electricity in molten or solution state |
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Answer» Solution :a. lonic compounds are soluble in polar SOLVENTS like water. They are insoluble in non-polar solvents like benzene `(C_6H_6)` carbon tetra CHLORIDE `(CCI_4)` . B. lonic compounds CONDUCT electricity in molten or solution state.Covalent compounds are non-conductors of electricity. |
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| 2386. |
2.0 of a sample of sodium chloride contains 0.785 g of sodium and 1.775 g of chlorine . In another sample , 2.925 g of sodium chloride was found to contain 1.15 g of sodium and 1.775 g of chlorine . Show that these data are in agreement with the law of constant proportion . |
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| 2387. |
Correct the statements : The formula of magnesium nitride is MgNO_(2). |
| Answer» Solution :The formula of MAGNESIUM nitride is `Mg_(3)N_(2)`. | |
| 2388. |
20 cm^(3)of an alcohol is dissoved in 80 cm^(3) of watercalculate the percentageof alcoholin the solution |
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Answer» SOLUTION :Volumeof alcohol = 20 `CM^(3)` Volumeof water = 80 `cm^(3)` `therefore` percentageof alcohol =`(20)/(20+80)xx100=20` |
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| 2391. |
Correct the statements : Chlorine is evolved when chlorine water is exposed to sunlight. |
| Answer» SOLUTION :OXYGEN is EVOLVED `(Cl_(2)+2H_(2)Oto2HCl+O_(2))`. | |
| 2393. |
Correct the statements : Boyle's law relates volume with temperature. |
| Answer» SOLUTION :Charles.s LAW RELATES VOLUME with TEMPERATURE. | |
| 2394. |
{:(1.,"Dalton",(a),"Hydrogen atom model"),(2.,"Thomson",(b),"Planetary model"),(3.,"Rutherford",(c ),"First atomic theory"),(4.,"Neils Bohr",(d),"Plum pundding model"),(,,(e ),"Discovery of neutrons"):} |
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| 2395. |
Correct the mistakes: (a) Washing soda is used for making cakes and bread soft, spongy. (b) Calcium sulphate hemihydrate is used in textile industry. |
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Answer» Solution :(a) Baking soda is USED for making CAKES and BREAD soft, spongy. (b) CALCIUM Oxychloride hemihydrate is used in textile industry. |
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| 2396. |
16.8 g of sodium hydrogen carbonate are added to 12.0 g of acetic acid .The residue left weighed 20.0 g what is the mass of CO_(2) escaped in the reaction ? |
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Answer» Solution :The chemical reaction TAKING place is : Sodium hydrogen carbonate + acetic ACID rarr sodium acetate + water + CARBON dioxide (16.8 g)(12.0 g)(20.0 g) Mass of `CO_(2)` RELEASED = (16.8 g + 12.0 g) -(20.0 g) =8.8 g |
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| 2398. |
120 g of salt is present in 600 mL of solution. Calculate the concentration in terms of mass by volume percentage of the solution. |
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Answer» Volume of solution = 600 mL Mass by volume percentage of solution `("Mass of SOLUTE")/("Mass of solution")xx100` = `120/600xx100=20%` |
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| 2399. |
Convert the following temperatures to the celsius scale. (a) 293k (b) 470k. |
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Answer» Solution :(a)Temp. on KELVIN SCALE = Temp. on Celsius scale + 273 293 = Temp. on Celsius scale + 273 So, Temp. on Celsius scale = 293 - 273`= 20^(@)C`. (b)Temp. on Kelvin scale = Temp. on Celsius scale + 273 470 = Temp. on Celsius scale + 273 So, Temp. on Celsius scale = 470 - 273`= 197^(@)C`. |
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| 2400. |
120 g of solution contains 16 g of urea. Calculate the concentration in terms of mass by mass percentage of the solution. |
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Answer» Solution :MASS of SOLUTE (urea) = 16 g Mass of solution = 120 g Mass percentage of solution = `("Mass of solute")/("Mass of solution")xx100` = `16/120xx100` = 13.33% |
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