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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.
| 1. |
Assertion: Laboratory reagents are usually made up to a specific molarity rather than a given molality. Reason: The volume of a liquid is more easily measured than its mass.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 2. |
Which of the following statements is correct about the reaction given below:- `4Fe(s)+3O_(2)(g)rarr2Fe_(2)O_(3)(g)`A. Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product, therefore, it follows law of conservation of mass.B. Total mass of reactant = total mass of product, therefore, law of multiple proportions is followed.C. Amount of `Fe_2O_3` can be increased by taking any one of the reactants (iron or oxygen ) in axcess.D. Amount of `Fe_2O_3` produced will decreased if the amount of any one of the reactants (iron or oxygen ) is taken in excess. |
| Answer» According to law fo conservation of mass, total mass of elements in reactants is equal to the total mass of elements in products. | |
| 3. |
Assertion: On heating, a solid usually change to a liquid and the liquid on further heating change to the gaseous state. Reason : Arrangement of constituent particles is different in solid, liquid and gaseous state.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» On heating, the interparticle distance and freedom of movment of particles increased and interparticle force decrease. Hence, on heating solids changes to liquids, where the particles are comparatively less close to each other and hence can move around. Liquids on futher heating changes to gaseous state, in which the particles are far apart as compared to solids or liquids and their movement in easy and fast. | |
| 4. |
Mass of one atom of an element is `2.66xx10^(-23)g`. This mass is equal toA. 16 amuB. `2.39xx10^(-9)J`C. Both are correctD. None is correct |
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Answer» Correct Answer - C 1 a.m.u. `= (1)/(2)xx(12)/(6.02xx10^(23))g` `(1)/(6.02xx10^(23))g` `2.66xx10^(-23)xx((6.02xx10^(23)amu)/(1g)) = 16` a.m.u. `E = mc^(2)` `E = (2.66xx10^(-26)kg)(2.998xx10^(8)ms^(-1))^(2)` `= 23.91xx10^(-10)J = 2.39xx10^(-9)J` |
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| 5. |
Assertion : 1 amu equals to `1.66xx10^(-24)` g. Reason : `1.66xx10^(-24)g` equals to `1/12` th of mass of a `C^(12)` atom.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A 12 gm of C-12 contain `6.023xx10^(23)` atom `:. 12/6.023xx10^(-23)=1.66xx10^(-24)`. |
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| 6. |
Assertion: The average mass of one Mg atom is `24.305 amu`, which is not actual mass of one Mg atom. Reason: Three isotopes, `24 Mg, 25 Mg and 26 Mg`, of Mg are found in nature.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 7. |
Oxidation number of nickel in `Ni(CI)_(4)` |
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Answer» Correct Answer - A If any central atom combined with carbonyl group than central metal atom shows always zero oxidation state. |
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| 8. |
Hot concentrated sulpuric acis is a moderatly strong oxidizing agent. Which of the following reaction does not shwo oxidizing behaviour?A. `CaF_(2)+H_(2)SO_(4) rarr CaSO_(4)+2HF`B. `Cu+2H_(2)SO_(4) rarr CuSO_(4)+SO_(2)+2H_(2)O`C. `2S+2H_(2)SO_(4) rarr 2SO_(2) +2H_(2)O`D. `C+2H_(2)SO_(4) rarr CO_(2)+2SO_(2)+2H_(2)O` |
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Answer» Correct Answer - A 1st reaction is not a redox reaction as the oxidation number of elements remains unchanged |
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| 9. |
Bleaching powder and bleach solution are produced on a large scale and used in several hous-hold products. The effectiveness of bleach solution id often measured by iodometry. `25 mL` of household bleach solution was mixed with `30 mL` of `0.50 M KI` and `10 mL` of `4N` acetic acid. In the titration of the liberated iodine, `48 mL` of `0.25 N Na_(2)S_(2)O_(3)` was used to reach the end point. The molarity of the household bleach solution is :A. 0.48 MB. 0.96 MC. 0.24 MD. 0.024 M |
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Answer» Correct Answer - C Milli mole of Hypo `=0.25xx48` `=2xx` milli moles of `Cl_(2)` milli mole of `Cl_(2)=(0.25xx48)/2=6` milli mole of `CaOCl_(2)` = milli mole of `Cl_(2)=` milli mole of `CaOCl_(2)` So, molarity `=6/25 M=0.24 M`. |
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| 10. |
Bleaching powder and bleach solution are produced on a large scale and used in several hous-hold products. The effectiveness of bleach solution id often measured by iodometry. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:A. `Cl_(2)O`B. `Cl_(2)O_(7)`C. `ClO_(2)`D. `Cl_(2)O_(6)` |
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Answer» Correct Answer - A `CaOCl_(2)=Ca(OCl)Cl` `OCl^(-) rarr` Hypochlorite ion which is anion of `HOCl` Anhydride of `HOCl=Cl_(2)O` |
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| 11. |
In the following reaction `3Br_(2)+6CO_(3)^(2-)+3H_(2)O to 5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-)`A. Bromine is oxidised and carbonate is reducedB. Bromine is reduced and water is oxidisedC. Bromine is neither reduced nor oxidisedD. Bromine is both reduced and oxidised |
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Answer» Correct Answer - D `3 overset(0)(B)r_(2)+6CO_(3)^(2-)+3H_(2)O rarr 5 overset(-1)(Br^(-))+overset(+5)(BrO_(3)^(-))+6HCO_(3)`. In this reaction bromine is oxidised as well as reduced. |
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| 12. |
Assuming that petrol is octane `(C_(8)H_(18))` and has density `0.8 g//ml, 1.425 litre` of petrol on complete combustion will consumeA. (a)`50` mole of `O_(2)`B. (b)`125` mole of `O_(2)`C. (c )`100` mole of `O_(2)`D. (d)`200` mole of `O_(2)` |
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Answer» Correct Answer - b Mass of octane `=1.425xx103xx0.8 g` Moles of octane `=(1425xx.8)/114=10` moles `C_(8)H_(18)+25/2 O_(2) rarr 8CO_(2)+9H_(2)O` From the equation it can be seen For 1 mole octane oxygen required `25/2` moles `:.` For 10 moles octane required `25/2xx10=125` moles |
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| 13. |
Borax in water gives `B_(4)O_(7)^(2-)+7H_(2)O rarr 4H_(3)BO_(3)+2OH^(-)` How many grams of Borax `(Na_(2)B_(4)O_(7). 10H_(2)O)` are required to prepare 50 ml of 0.2 M solution.A. (a)0.32 gB. (b)3.82 gC. (c )0.28 gD. (d)None of these |
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Answer» Correct Answer - b The n-factor or valency factor of borax in the above reaction is 2 Meq. Of borax in 50 ml solution `=50xx0.2xx2=20` Let weight of borax required =W `W/382xx2xx1000=20` `:. W_((Na_(2)B_(4)O_(7). 10H_(2)O))=3.82` |
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| 14. |
When `10 ml` of propane (gas) is combusted completely, volume of `CO_(2)(g)` obtained in similar condition isA. (a)`10 ml`B. (b)`20 ml`C. (c )`30 ml`D. (d)`40 ml` |
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Answer» Correct Answer - C `C_(3)H_(8)+5O_(2)rarr3CO_(2)+4H_(2)O` Value of `CO_(2)=3xx10 ml` `=30 ml` |
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| 15. |
The density of a solution prepared by dissolving 120 g of urea (mol. Mass=60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution isA. (a)0.50 MB. (b)1.78 MC. (c )1.02 MD. (d)2.05 M |
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Answer» Correct Answer - D Total mass of `solution=1000 g water+120 g urea=1120 g` Density of solution`=1.15 g//mL` Thus, volume of solution=mass/density`=(1120g)/(1.15 g//mL)` `=973.91 mL=0.974 L` Moles of solute`=120/60=2 mol` `"Molarity"=("moles of solute")/("volume (L) of solution")` `=(2 "mol")/(0.974 L)=2.05 "mol" L^(-1)` |
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| 16. |
16 of oxygen has same number of molecules as inA. 16 g of COB. 28 g of `N_(2)`C. `14 g "of" N_(2)`D. `1.0g "of" H_(2)` |
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Answer» The number of molecuels can be calculated as follows Number of molecuels `=("Mass")/(" Molar mass")xx ` Avogadro number `(N_(A))` Number of molecuels, in 16 g of oxygen `=(16)/(32)xxN_(A)=(N_(A))/(2)` In `16g` of `CO=(16)/(28)xxN_(A)=(N_(A))/(175)` In 28 g of `N_(2)=(14)/(28)xxN_(A)=(N_(A))/(2)` In `1 g "of " H_(2)=(1)/(2)xxN_(A)=(N_(A))/(2)` So, `16 g "of" O_(2)=14g "of" N_(2)=1.0g "of" H_(2)` |
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| 17. |
Assuming full decomposition, the volume of `CO_(2)` released at STP on heating 9.85 g of `BaCO_(3)` (At mass Ba = 137) will beA. 0.84 LB. 0.24 LC. 4.06 LD. 1.12 L |
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Answer» Correct Answer - D `underset(197)(BaCO_(3))toBaO+underset(22.4L)(CO_(2))` `V_(CO_(2))` released `= (22.4)/(197)xx9.85 = 1.12L` |
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| 18. |
The volume in litres of `CO_(2)` liberated at STP when 10 grams of 90% pure limestone is heated cmpletely isA. `22.4`B. `2.24`C. `20.6`D. `2.016` |
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Answer» Correct Answer - D Sample of limestone is only 90 % pure. Therefore, 100 g limestone sample contains 90 g pure `:.` 10 g limestone sample will contain `(10xx90)/(100)` `9.00 g` pure As limestone is nothing but `CaCO_(3)` `:.` 10 g limestone sample will contain `(10xx90)/(100) = 9.00` g pure As limestone is nothing but `CaCO_(3)` `:.` The molecular mass of `CaCO_(3) = 100` 100 g of limestone gives 22.4 litre of `CO_(2)` at S.T.P. `:.` 9 g of limestone will give `(22.4xx9)/(100) = 2.016` litre at S.T.P. Hence, 10g impure sample will give 2.016 litre of `CO_(2)` at STP |
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| 19. |
1 mol `BaF_(2)+2` mol `H_(2)SO_(4) rarr` resulting mixture will be neutralised byA. 1 mol of KOHB. 2 mol of `Ca(OH)_(2)`C. 4 mol of KOHD. 2 mol of KOH |
| Answer» Correct Answer - B::C | |
| 20. |
which of the following represent redox reactions? I. `Cr_(2)O_(7)^(2-)+2overset(ө)OHrarr2CrO_(4)^(2-)+H_(2)O` II. `Zn+CuSO_(4)rarrZnSO_(4)+Cu` iii.`MnO_(4)^(ө)+3Mn^(2+)+4overset(ө)OHrarr5MnO_(2)+2H_(2)O` IV. `2Cu^(o+)rarrCu+Cu^(2+)`A. `Cr_(2)O_(7)^(2-)+2OH^(-) rarr 2 CrO_(4)^(2-)+H_(2)O`B. `2 CrO_(4)^(2-) +2H^(+) rarr Cr_(2)O_(7)^(2-)+H_(2)O`C. `2 MnO_(4)^(-)+3 Mn^(2+)+4 OH^(-) + 4OH^(-) rarr 5 MnO_(2)+2 H_(2)O`D. `2 Cu^(+) rarr Cu+Cu^(2+)` |
| Answer» Correct Answer - C::D | |
| 21. |
Oxygen is prepared by catalytic decomposition of potassium chlorine `(KClO_(3))`. Decomposition of potassium, chloride gives potassium chloride (KCl) and oxygen `(O_(2))`. How many moles and how many grams of `KClO_(3)` are required to produce 2.4 mole `O_(2)`?A. `196.0`B. `190.6`C. `169.0`D. `196.2` |
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Answer» Correct Answer - A The balanced equation is `underset(underset(2xx(39+35.5+3xx16)=245"g")(2"mol"))(2KClO_(3)(s)+)20_(2)(g)rarr 2KCl(s) + underset(underset(2.4 "moles")(3 "mol"))(3O_(2))(g)` `because` 3 moles of `O_(2)` Is produced by decomposition of `KClO_(3) = 245`g `because 2.4` moles of `O_(2)` will be produced by the decompostion of ltb rgt `KClO_(3)=(245xx2.4)/3=196.0 "g"` |
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| 22. |
The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction `2A+4Brarr3C+4D`, when 5 moles of A react with 6 moles of B, then (a) which is the limiting reagent? (b) calculate the amount of C formed?A. C, 4.5 molB. B, 4.5 molC. B, 3.5 molD. C, 4.0 mol |
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Answer» In the reaction, `2A + 4B to 3C + 4D` 2 moles of A react with 4 moles of B 5 moles of A will react with `4/2xx5=10` moles of B Since in the reaction only 6 moles of B are there, hence B is the limiting reagent. Now, 4 moles of B give 3 moles of C 6 Moles of B will give `3/4xx6=4.5` mol of C |
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| 23. |
Oxidation number if iodine in `IO_(3)^(-), IO_(4)^(-),KI and I_(2)` respectively areA. `-1, -1, 0, +1`B. `+3, +5, +7, 0`C. `+5, +7, -1, 0`D. `-1, -5, -1, 0` |
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Answer» Correct Answer - C `IO_(3)^(-) : x+3(-2)=-1` `x-6=-1 implies x=5` `IO_(4)^(-) : x+4(-2)=-1` `x-8=-1 implies x=7` `KI : 1+x=0 implies x=-1` `I_(2) : 2x=0 implies x=0` |
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| 24. |
On analysis a certain compound was found to contain iodine and oxygen in the ration of 254 g of iodine (at. mass 127) and 80 g oxygen (at. mass 16). What is the formula of the compound?A. `IO`B. `I_(2)O`C. `I_(5)O_(3)`D. `I_(2)O_(5)` |
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Answer» Correct Answer - D Moles of iodine `= (247)/(127) = 2` Moles of oxygen = 80/16 = 5 Formula `= I_(2)O_(5)` |
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| 25. |
Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:A. (a)`1.576xx10^(4)`B. (b)`1.576xx10^(3)`C. (c )`15.76`D. (d)`2.136xx10^(4)` |
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Answer» Correct Answer - c The enzyme must contain at least one atom of Se. `:. 0.5 g` of Se is present in `100 g` of enzyme `:. 78.4 g` of Se will be present in `(100xx78.4)/0.5` `=1.576xx10^(4) g` of enzymes |
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| 26. |
Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:A. (a)`1.568xx10^(3)`B. (b)`15.68`C. (c )`1.568xx10^(4)`D. (d)`2.136xx10^(4)` |
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Answer» Correct Answer - C Suppose the mol. Wt. of enzyme =x Given `100 g` of enzyme wt of `Se=0.5 g` `:.` In x g of enzyme wt. of Se` =0.5/100 xx x` Hence `78.4=(0.5xx x)/100` `:. X=15680=1.568xx10^(4)` |
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| 27. |
Caffiene has a molecular mass of `194`. If it contains `28.9%` by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is :A. `3`B. `4`C. `5`D. `6` |
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Answer» Correct Answer - B Mass of 2 molecular of caffine = 194 amu N present in one molecule `= (28.9xx194)/(100)` = 56 amu 56 amu pertains to 4 nitrogen (N) atoms |
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| 28. |
A peroxidase enzyme isolated from red blood cells was found to contain 0.28% selenium. The minimum molecular mass of the enzyme is (at. Mass of selenium = 78.96 u).A. `3.67xx10^(3)`B. `2.7xx10^(4)`C. `2.90xx10^(7)`D. `2.9xx10^(4)` |
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Answer» Correct Answer - B Selenium content = 0.29% `=2.9xx10^(-3)` Minimum mol. Mass `=("At. Mass of Se")/("Selenium content")` `=(78.96u)/(2.9xx10^(-3))=2.7xx10^(4)u` |
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| 29. |
An alkaloid contains 17.28% of nitrogen and its molecular mass is 162. The number of nitrogen atoms present in on molecule of alkaloid isA. fiveB. fourC. threeD. two |
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Answer» Correct Answer - D Amount of nitrogen present in one mole of akaloid i.e., in 162 g `= (162xx17.28)/(100)` `:.` No. of nitrogen atoms per molecule `=(162xx17.28)/(100x24)=2`(approx.) |
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| 30. |
What volume of oxygen at N.T.P is needed to cause the complete combustion of 200 mL of acetylene ? Also calculate the volume of carbon dioxide formed.A. 300, 400B. 500, 400C. 400, 300D. 400, 500 |
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Answer» Correct Answer - B `underset(underset(2xx22400" cm"^(3))(2" mol"))(2C_(2)H_(2))+ underset(underset(5xx 22400" cm"^(3))(5" mol"))(5O_(2)) rarr underset(underset(4xx22400" cm"^(3))(4" mol"))(4 CO_(2)) +2H_(2)O` `because 2 xx22400" cm"^(3)` Of acetylene requir `O_(2)` for complete combustion `=5xx 22400" cm"^(3)` `200" cm"^(3)` of acetylene will require `O_(2)` for complete combustion `=(5xx22400)/(2xx22400)xx200` Further, `2xx22400" cm"^(3)` of acetylene produce `CO_(2)` `= 4 xx 22400" cm"^(3)` `therefore 200" cm"^(3)` of acetylene will produce `CO_(2)` `=(4xx22400)/(2xx22400)xx200` `400" cm"^(3)` at STP |
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| 31. |
Match the following prefixes with their multiples: `{:(,"Prefixes"," Multiples"),( (i), "micro" ,10^(6)),( (ii), "deca" ,10^(9)),( (iii), "mega", 10^(–6)),( (iv), "giga ",10^(–15)),( (v) ,"femto" ,10 ):}` |
| Answer» `{:(,"prefix","Multiples"),((i),"micro",10^(-6)),((ii),"deca",10),((iii),"mega",10^(6)),((iv),"giga",10^(9)),((v),"Femto",10^(-15)):}` | |
| 32. |
What is the mass of carbon dioxide which contains the same number of molecules as are contained in 40 g of oxygen?A. 40gB. 55gC. 32gD. 44g |
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Answer» Molar mass of `O_2 = 32g mol^(-1)` 32g of `O_2=6.023xx10^23` molecules 40g of `O_2=(6.023xx10^23xx40)/(32)=7.529xx10^23` molecules Mass of `6.023xx10^23 "molecules of " CO_2=44g` Mass of `7.529xx10^23 " molecules of " CO_2` `=(44xx7.529xx10^23)/(6.023xx10^23)=55g` |
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| 33. |
Match the mass of elements given in coloumn I with the no. of moles given in column II and mark the appropriate choice. A. `(A) to (iv), (B) to (iii), (C) to (ii), (D) to (i)`B. `(A) to (i), (B) to (iii), (C) to (ii), (D) to (iv)`C. `(A) to (iii), (B) to (ii), (C) to (i), (D) to (iv)`D. `(A) to (ii), (B) to (i), (C) to (iv), (D) to (iii)` |
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Answer» (A) : 28g of He `=28/4=7` mol (B) : 46g of Na `=46/23=2` mol (C) : 60g of Ca `=60/40=1.5` mol (D) : 27g of Al `=27/27=1` mol |
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| 34. |
Match the prefixes present in coloumn I with their multiples in coloumn II and mark the appropriate choice. A. `(A) to (i), (B) to (ii), (C) to (iii), (D) to (iv)`B. `(A) to (ii), (B) to (i), (C) to (iv), (D) to (iii)`C. `(A) to (iv), (B) to (iii), (C) to (i), (D) to (ii)`D. `(A) to (iii), (B) to (iv), (C) to (ii), (D) to (i)` |
| Answer» Pico `=10^(-12), " femto " = 10^(-15) , " milli " = 10^(-3), " giga " = 10^9` | |
| 35. |
Match the coloumn I with coloum II and mark the appropriate choice. A. `(A) to (ii), (B) to (i), (C) to (iv), (D) to (iii)`B. `(A) to (i), (B) to (ii), (C) to (iii), (D) to (iv)`C. `(A) to (iv), (B) to (iii), (C) to (i), (D) to (ii)`D. `(A) to (iv), (B) to (iii), (C) to (ii), (D) to (i)` |
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Answer» `(A) : Zn + 2HCl to ZnCl_2 + H_2` 1 mol of Zn produces 2 g of `H_2` 0.5 mole of Zn will produce 1g of `H_2` (B) : `C_70H_22` Molar mass =862 Mass of atoms = `862//6.023xx10^23=1.43xx10^(-21)` g 35.3 g of `Cl_2=3.01xx10^23` molecules (D) : Molar mass of `SO_2` = 64 = 1 mole 64g of `SO_2=6.023xx10^23` molecules |
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| 36. |
2 mol of `H_(2)S` and 11.2 L of `SO_(2)` at N.T.P. react to form x moles of sulphur, x is `SO_(2)+2H_(2)S to 3S +2H_(2)O`A. `1.5`B. `3`C. `11.2`D. `6` |
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Answer» Correct Answer - A 22.4 L of `SO_(2)` at N.T.P. = 1 mol 11.2 L of `SO_(2)` at N.T.P. = 0.5 mol `underset(1mol)(SO_(2))+underset(2mol)(2H_(2)S)to3S+2H_(2)O` Here `SO_(2)` is the limiting reactant `:.` Moles of sulphur formed `= 3xx"moles of" SO_(2)` `= 3xx0.5mol` `= 1.5 mol` |
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| 37. |
4 moles each of `SO_(2)` and `O_(2)` gases are allowed to react to form `SO_(3)` in a closed vessel. At equilibrium, 25% of `O_(2)` is used up. The total number of moles of all the gases at equilibrium isA. `6.5`B. `7.0`C. `8.0`D. `2.0` |
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Answer» Correct Answer - B Moles of `O_(2)` used up = 25 % of 4 = 1 mol. `{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("Initial moles",4,,4,,0),("Moles at eqm.",4-2,,4-1,,),(,=2,,=3,,2):}` Therefore total number of moels of all gases at eqm. = 2+3+2 = 7 |
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| 38. |
A solution contain`1.2046xx10^(24)`hydrochloric acid molecules in one `dm^(3)`of the solution . The strength of the solution isA. 6NB. 2NC. 4ND. 8N |
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Answer» Correct Answer - B Number of `H^(+)` ions in 1 L `(1dm^(3))` `=1.2046xx10^(14) = (1.2046xx10^(24))/(6.02xx10^(23))=2mol` `:.` Normality of solution = 2N |
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| 39. |
In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1 L of the mixture.A. `1.613xx10^23`B. `6.023xx10^23`C. `1.61xx10^27`D. `1.61xx10^19` |
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Answer» Volume of gas in 1 L `=(0.06)/(100)=6xx10^(-4)` L ` =6xx10^(-4)` L No. of molecules of `CO_2=nxxN_A` `(6xx10^(-4))/(22.4)xx6.023xx10^23=1.61xx10^19` |
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| 40. |
A gases mixture contains oxygen and nitrogen in the ratio `1 : 4` by weight. Therefore, the ratio of the number of molecules is:A. (a)1:4B. (b)1:8C. (c )7:32D. (d)3:16 |
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Answer» Correct Answer - C Let mass of oxygen `=1 g`, Then mass of nitrogen `=4 g` Mol. Wt. of `N_(2)=28 g`, Mol. Wt. of `O_(2)=32 g` `28 g` of `N_(2)` has`=6.02xx10^(23)` molecules of nitrogen `4 g` of `N_(2)` has`=(6.02xx10^(23))/28xx4` molecules of nitrogen `=(6.02xx10^(23))/7` molecules of nitrogen `32 g` of `O_(2)` has`=6.02xx10^(23)` molecules of oxygen `:. 1 g of O_(2)` has`=(6.02xx10^(23))/32xx1=(6.02xx10^(23))/32` molecules of oxygen Thus, ratio of molecules of oxygen: nitrogen `=(6.02xx10^(23)//32)/(6.02xx10^(23)//7)=7:32`. |
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| 41. |
Which of the following paris of gases contains the same number of molecules?A. 16 g of `O_(2)` and 14 g of `N_(2)`B. 8 g of `O_(2)` and 22g of `CO_(2)`C. 32 g of `O_(2)` and 32 g of `N_(2)`D. |
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Answer» a) 16 g of `O_(2)` = 0.5 mol = `(N_(A))/2` molecules 14 g of `N_(2)` = 0.5 mol = (N_(A))/2 molecules |
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| 42. |
`10dm^(3)` of `N_(2)` gas and 10 `dm^(3)` of gas X at the same temperature contain the same number of molecules The gas X isA. COB. `CO_(2)`C. `H_(2)`D. NO |
| Answer» a) Number of moles of `N_(2)` and X shoul dbe equal. This can be so if X has some molecular weight as `N_(2)`. | |
| 43. |
How many significant figures should be there in the answer of `((1.79xx10^(5))(29.2-20.2))/(1.39)` ?A. `3`B. `1`C. `4`D. `2` |
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Answer» Correct Answer - D `(29.2 - 20.2) = 9.0` Thus, answer should be reported to two significant figures. |
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| 44. |
The answer of the calculation `(2.568 xx 5.8)/(4.168)` in significant figures will beA. 3.57B. 3.6C. 3.57D. 3.579 |
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Answer» b) `(2.568 xx 58)/(4.168)`=`3.5735` Thus, insignificant figures = 3.6 |
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| 45. |
Which of the following forms the largest number of compounds ?A. CarbonB. HydrogenC. OxygenD. Nitrogen |
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Answer» Correct Answer - A `C` shows the greatest tendency to catenate. |
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| 46. |
Significant figures in 0.00051 areA. (a)` 5`B. (b)`3`C. (c )` 2`D. (d)`4 ` |
| Answer» Correct Answer - C | |
| 47. |
In the final answer of the expression `((29.2-20.2)(1.79xx10^(5)))/1.37`. The number of significant figures isA. (a)`1 `B. (b)`2 `C. (c )`3 `D. (d)` 4` |
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Answer» Correct Answer - B `((29.2-20.2)(1.79xx10^(5)))/1.37=(9.0xx1.79xx10^(5))/1.37` Least precise terms, i.e., 9.0 has only two significant figures. Hence, final answer will have two significant figures. |
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| 48. |
Fractional distillation of crude petroleum is performed to obtainA. dieselB. petrolC. gasolineD. All of these |
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Answer» Correct Answer - D Because the difference in the boiling points of different is very-very small. |
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| 49. |
Which of the following laws of chemical combination differentiarte a compared from a mixture?A. Law of multiple proportionsB. Law of definite proportionsC. Law of reciprocal proportiosD. Law of combining volumes |
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Answer» Correct Answer - B Unlike a mixturre a compound has a fixed composition by mass. |
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| 50. |
Laws of chemical combinations are important because they provided the first scientific evidence forA. rate of reactionB. energy exachanged during a reactionC. existance of atomsD. merchanism of reaction |
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Answer» Correct Answer - C Dalton explained these laws through his atomic theory |
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