InterviewSolution
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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.
| 2401. |
Convert the following temperatures to the Celsius scale: a. 293 K b. 470 K |
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Answer» Solution :a. `293K=(293-273)^(@)=20^(@)` B. `470K=(470-273)^(@)C=197^(@)C` |
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| 2402. |
12 g of common salt is dissolved in 150 g of solution. |
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Answer» SOLUTION :Mass of common salt (solute) = 12 G Mass of solution = 150 g Percentage of concentration of solution, `(%w//w)=("Mass of solute")/("Mass of solution")XX100` = `(12g)/(150G)xx100` = 8% |
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| 2403. |
Convert the folowing temperature to the kelvin scale. (a) 25^(@)C (b)373^(@)C. |
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Answer» Solution :(a)See SAMPLE Problem 1 on page 23 of this book. Temp. on Kelvin SCALE = Temp. on Celsius scale + 273 = 373 + 273 = 646 K |
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| 2405. |
Convert the following temperature to the Celsius scale (a) 293K (b) 470K |
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Answer» Solution :(a) 293 K is EQUAL to `293-273 = 20^(@)C` `293K=20^(@)C` (b) 470K is equal to `470-273=197^(@)C` `470K=197^(@)C` |
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| 2406. |
100 mLof waterat room temperatureof 25^(@)Cis taken in abeakeranda littleof solidS s isdissolved in itby stirringof obtaina solution X. Moreand moreof solidS isaddedto thesolutionwithconstantstirringwhilekeepingthetemperatureofsolutionconstantat 30^(@) C Aftersometimeit isobservedthatno moresoliddissolvesin waterand itthe same timesomesolidis alsoleft undissolvedat thebottomof thebeaker .thecontentsof beakerarefilteredthrough a filterpaperto obtainsolutionY inthe fromof afiltrate . (a)whatnameis giventosolutions likeX? ( b)whatnameisgivento solutionslikeY ? (c )whatwillyouobserveitthesolutiony30^(@)Cis cooleddownto 10^(@)Cby keepingthebeakerin crushedice ?why ? (d ) whattermis usedtodenotetheamount of solid dissolvedin 100 gramsof waterin asolutionlikeY ? |
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| 2407. |
Convert the following temperatures to celsius scale :(a) 300 K(b)573 K |
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Answer» SOLUTION :(a) SEE Sample Problem 2 on page 24 of this book. (b) Temp. on KELVIN scale = Temp. on Celsius scale + 273 573 = Temp. on Celsius scale + 273 And, Temp. on Celsius scale = 573 - 273`=300^(@)C` |
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| 2408. |
100 g of 25% (w/w) sodium hydroxide is prepared in a laboratory. If the density of water is 0.9 g/cc at room temperature, calculate the volume of water taken |
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Answer» SOLUTION :(i) determination of the mass of sodium hydroxide present in the solution (ii) calculation of the mass of water taken (III) calculation of molarity (IV) calculation of VOLUME of water taken from the DENSITY given |
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| 2409. |
Convert the following temperature to the Kelving scale: a. 25^(@)C b. 373^(@)C |
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Answer» SOLUTION :a. `25^(@)C=(25+273)K=298K` B. `373^(L@)C=(373+273)K=646K` |
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| 2410. |
Convert the following temperatures to Celsium scale: a. 300K b. 573 K |
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Answer» SOLUTION :a. `300K=(300-273)^(@)C=27^(@)C` (B) `573K=(573-273)^(@)C=300^(@)C` |
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| 2411. |
1% solutionof Iodoform is used as …………………. . |
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Answer» antipyretic |
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| 2412. |
1 mole of ethyne on complete combustion gives |
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Answer» 2 moles of carbon monoxide HALF -mole of WATER When one mole of ethyne is subjected to complete combustion, 2 moles of `CO_2` and 1 mole of waterare formed |
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| 2413. |
(1) Classify the following as chemicalphysical changes : (1) cutting of trees (2) melting of butter in a pan (3) rusting of almirah (4) boiling of water to form steam (5) passing of electric current through water and the water breaking down into hydrogen and oxygen gases (6) dissolving common salt in water (7) making a fruit salad with raw fruits (8) burning of paper and wood. |
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Answer» Solution :Physical changes : (1) cutting of trees Stand (2) melting of butter in a pan (4) BOILING of water to form steam (6) dissolving common salt in water (7) making a fruit SALAD with raw fruits Chemical changes : (3) rusting of almirah (5) PASSING of electric current through water and water breaking down into hydrogen and oxygen gases (8) burning of PAPER and wood. |
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| 2414. |
Convert the following temperatures to the Kelvin scale. 25^(@)C |
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Answer» SOLUTION :(a) `25^(@)C` is EQUAL to `25+273=298K` `25^(@)C=298K` (B) `373^(@)C` is equal to `373+273=646K` `373^(@)C=646K` |
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| 2417. |
Convert the following temperature to Celsius scale (a) 300K (b) 573K |
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Answer» SOLUTION :(a) `300K=(300-273)^(@)C=27^(@)C`. (B) `573K=(573-273)^(@)C=300^(@)C`. |
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| 2418. |
1 amu = (1)/(12) the mass of one C-12 atom . |
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| 2419. |
Convert the following number "molecules"//"atoms" into number of moles: 6.022 xx 10^(25) molecules of ammonia |
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Answer» SOLUTION :Number of moles = `("given number of particles")/("AVOGADRO number")` `=(6.022 XX 10^(25))/(6.022 xx 10^(23))` `= 1 xx 10^(2)` = 100 MOLE ammonia |
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| 2420. |
0.5 gram common salt is dissolved in 25 g of water. Calculate the concentration in term of mass by mass percentage of the solution. |
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Answer» Solution :Mass of SOLUTE (common SALT) = 0.5 G Mass of solvent (water) = 25 g Mass of solution = (0.5 + 25)g = 25.5 g Now, Mass percentage of solution = `("Mass of solute")/("Mass of solution")XX100` = `0.5/25.5xx100` = 1.96% |
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| 2421. |
Convert the following number "molecules"//"atoms" into number of moles: 3.6132 xx 10^(20) molecules of water |
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Answer» SOLUTION :Number of MOLES = `("given number of particles")/("Avogadro number")` `=(3.6132 xx 10^(20))/(6.022 xx 10^(23))` `= 0.6 xx 10^(-3)` = `6.0 xx 10^(-4)` mole water |
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| 2422. |
Convert the following number "molecules"//"atoms" into number of moles: 3.011 xx 10^(25) molecules of helium |
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Answer» Solution :NUMBER of moles = `("GIVEN number of PARTICLES")/("AVOGADRO number")` `=(3.011 xx 10^(25))/(6.022 xx 10^(23))` `=0.5 xx 10^(2)` = 50 mole helium |
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| 2423. |
Convert the following number "molecules"//"atoms" into number of moles: 2.4088 xx 10^(23) molecules of chlorine |
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Answer» Solution :Number of moles = `("GIVEN number of PARTICLES")/("Avogadro number")` `=(2.4088 XX 10^(23))/(6.022 xx 10^(23))` = 0.4 mole chlorine |
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| 2424. |
0.5 g of a substanceis dissolvedin 25 gof a solventcalculate the percentageamount of the substance in the solution |
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Answer» Solution :MASSOF the SUBSTANCE = 0.5 G Mass of the solvent = 25 g `THEREFORE` percentage of thesubstance `=(0.5)/(0.5+25)xx100=1.96` |
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| 2425. |
Convert the following number "molecules"//"atoms" into number of moles: 12.044 xx 10^(22) molecules of sulphur dioxide |
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Answer» SOLUTION :Number of moles = `("GIVEN number of PARTICLES")/("Avogadro number")` `=(12.044 xx 10^(22))/(6.022 xx 10^(23))` `= 2 xx 10^(-1)` = 0.2 mole sulphur dioxide |
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| 2426. |
Convert the following into mole: 43.5 g K_(2)SO_(4) [Molecular weight of K_(2)SO_(4) is 174 u.] |
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Answer» Solution : Number of MOLES of 43.5 G `K_(2)SO_(4)= ("MASS of "K_(2)SO_(4)" in g")/("molecular mass u")` `= (43.5 g)/(174 u)` = 0.25 MOLE |
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| 2428. |
Convert the following into mole: 22 g carbon [Molecular weight of CO_(2) is 44 u.] |
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Answer» Solution :Number of moles of 22G carbon DIOXIDE = `("mass of carbon dioxide in g")/("molecular mass u")` `= (22 g)/(44 u)` (molecular mass carbon dioxide = 44 u) = 0.5 mole |
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| 2429. |
...... present in the nucleus of an atom possess positlve electric charge. |
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| 2430. |
Convert the following into mole: 12 g oxygen [Molecular weight of O_(2) is 32 u.] |
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Answer» SOLUTION :Number of moles of 12g oxygen = `("MASS of oxygen in G")/("molecular mass U")` `=(12 g)/(32 u)` (molecular mass of oxygen 32 ) = 0.375 MOLE |
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| 2431. |
.............. of water is very slow on a humid day . |
| Answer» SOLUTION :EVAPORATION | |
| 2432. |
Convert the following into mole: 20 g water [Molecular weight of H_(2)O is 18 u.] |
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Answer» Solution :NUMBER of MOLES of 20 G WATER = `("mass of water in g")/("MOLECULAR mass u")` `= (20 g)/(18 u)` (molecular mass of water 18 u) = 1.11 mole |
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| 2433. |
................. ofwater from water bodies followed by .............. causes rainfall. |
| Answer» SOLUTION :EVAPORATION , CONDENSATION | |
| 2434. |
Convert the following into mole: 102.5 g Ca(NO_(3))_(2) [Molecular weight of Ca(NO_(3))_(2) is 164 u.] |
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Answer» Solution :Number of moles of 102.5 g `Ca(NO_(3))_(2)= ("MASS of "Ca(NO_(3))_(2)" in g")/("molecular mass U")` `=(102.5 g)/(164 u)` = 0.625 mole |
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| 2437. |
Convert into mole : (a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of carbon dioxide. |
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Answer» Solution :(a) Molar mass of oxygen `(O_(2)) = 2 xx 16` = 32 g GIVEN mass of oxygen = 12 g Number of moles = `("given mass")/("molar mass")` `=(12)/(32) = 0.375` mole Mole of oxygen gas = 0.375 mole. (b) Molar mass of water `(H_(2)O) = (2 xx 1) + 16` = 2 + 16 = 18 g Given mass of water = 20 g Number of moles = `("given mass")/("molar mass")` `=(20)/(18) ~~ 1.11` mole Mole of water = 1.11 mole. (c) Molar mass of carbon dioxide `(CO_(2)) = 12 + (2 xx 16)` = 12 + 32 = 44 g Given mass of carbon dioxide = 22 g Number of moles = `("given mass")/("molar mass")` `=(22)/(44) = 0.5` mole Mole of carbon dioxide = 0.5 mole. |
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| 2438. |
______ method is used to separate coloured components from the dye. |
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| 2443. |
Convert into mole . 12 g of oxygen gas |
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Answer» Solution :`12 XX 2 xx 6.0.23 xx 10^(23)` `24 xx 6.023 xx 10^(23)` |
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| 2444. |
....... Is used in the treatment of cancer. |
| Answer» Answer :C | |
| 2445. |
Convert 575 K to Celsius scale. |
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Answer» SOLUTION :`575K=273.15+t` `ort=575-273.15=301.85^(@)C.` |
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| 2446. |
............... is the symbol for platinum (platina) proposed by Dalton. |
Answer» SOLUTION :
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| 2448. |
________ is the technique used for separation of those solutes that dissolve in the same solvent. |
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| 2449. |
Considers dobereinter traid with elementsX , Y and Z The sum of atomic weights of extreme elements X and Z is 46and the different of atomic weights Z and Y is two times the atomic number of oxygen .Identify X Y and Z. |
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Answer» Solution :Given X+Z=`46rarr(1)` `Z-Y=16 RARR(2)` `As (X+Y)/(2)=Y` (Dobereiner's law of TRIADS) `therefore Y=(46)/(22)=23rarr(3)` From EQUATION (2) and (3) Z-23 =16 `rarr` Z=39 From equation (1)X+Z=46 `rarr`X+39=49`rarr`X=7 So X,Y and Z are Li Na and K respectively |
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| 2450. |
Considering MgCl_(2) as ionic compound as ionic compound and CH_(4) a covalent compound give any two difference between these two compounds |
Answer» SOLUTION :
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