This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
How many molecules of hydrogen is required to produce 4 moles of ammonia? |
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Answer» 15 moles To get 2 moles of AMMONIA, 3 mole of `H_(2)` is required. `therefore` To get 4 moles of ammonia = `(3)/(cancel2)xxcancel(4)^(2) = 6` moles of `H_(2)` is required. |
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| 2. |
How many molecules of CO_(2)will needed to obtain 1.8 g of glucose according to given reaction. Reaction : 6CO_(2) + 6H_(2) O rarr C_(6) H_(12) O_(6) + 6O_(2) [M. wt C_(6)H_(12)O_(6) = 180 g mol^(-1)] (C=12, H=1, O=16) |
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Answer» `0.6xx6.022xx10^(23)` ` =0.01 ` mole glucose ...(1) Reaction : `6CO_(2) + 6H_(2) O RARR C_(6)H_(12)O_(6) + 6O_(2)` `:.` 6 mole `CO_(2)` is needed to GET 1 mole glucose `= (0.01xx6)/(1)` `= 0.06` mole `CO_(2)` is needed ...(ii) `=0.06xx6.022xx10^(23)` molecular of `CO_(2)`. ...(iii) |
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| 3. |
How many moles of hydrogen is required to produce 4 moles of ammonia? |
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Answer» 15 moles To get 2 moles of AMMONIA, 3 mole of `H_(2)` is required. `:.` To get 4 moles of ammonia = `3/cancel(2) XX cancel(4^(2))` = 6 moles of `H_(2)` is required. |
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| 4. |
How many molecules of CO_(2)is in 44 g CO_(2) ? |
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Answer» `6xx10^(23)` |
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| 5. |
How many molecules of an ideal gas are there is 1xx10^(-3)dm^(3)at STP ? |
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Answer» SOLUTION :`22.4 DM^(3)` of an IDEAL GAS has `6.023xx10^(23)` molecules at STP ? `1xx10^(-3) dm^(3)` of an ideal gas contains `=(6*023xx10^(23))/(22.4)xx1xx10^(3)=2.69xx10^(19)` molecules. |
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| 6. |
How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl? Note:The number of molecules in a sample is related to moles of compound (1 mol HCl = 6.023 xx 10^(23) HCl molecules). Therefore if you first convert grams HCl to moles, then you can convert moles to number of molecules). |
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Answer» SOLUTION :`3.46 g "HCl" xx (1 " mol HCl")/(36.5 g "HCl")xx (6.023 xx 10^(23)" HCl molecules")/(1 " mol HCl")` `= 5.71 xx 10^(22)` HCl molecules |
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| 7. |
How many molecules are present in 5.6xx10^(-8)cc of Cl_(2) at stP? |
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| 8. |
How many molecules are present in 32 g of methane? |
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Answer» `2xx6.023xx10^(23)` 16 g contains `6.023 xx 10^(23)` molecules. 32 g of methane will contain =`(6.022xx10^(23))/CANCEL(16)xxcancel(32^(2))=2xx6.023xx10^(23)` |
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| 10. |
How many molecules and atoms of phosphorus are present in 0.1 mole of P_(4) molecules ? |
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| 11. |
How many molecules are present in 1 equivalent of hydrogen? |
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Answer» Solution :1 EQUIVALENT of H CONTAINS 1 MOLE of H. `THEREFORE` 1 equivalent of H contains 1/2 mole of `H_2`. |
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| 12. |
How many mole of FeCl_(3) can be prepared by the reaction of 10g pf KMnO_(4)10.07mole of FeCl_(2) and 500mL of 3M HCl follwing the reaction: 5FeCl_(2) + KMnO_(4) + 8HCI rarr 5FeCl_(3) + KCl + MnCl_(2) + 4H_(2)O |
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| 13. |
Find the number of moles of the solute present in 600ml of 0.05M solution. |
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| 14. |
How many mole of electrons are involved in the reduction of one mole of MnO_(4)^(-)ion in alkaline medium to MnO_(3)^(-)? |
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Answer» 2 |
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| 15. |
How many mL of 1 M H_(2)SO_(4) acid solution is required to neutralise 10 mL of 1 M NaOH ? |
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Answer» 5 mL |
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| 16. |
How many mL of 0.125 M Cr^(3+)must be rected with 12.00 mL of 0.200 M MnO_(4)^(-) if theredox products are Cr_(2)O_(7)^(2-) andMn^(2+) ? |
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Answer» Solution :Reduction equation OXIDATION equation `2Cr^(3+)+7_(2)OrarrCr_(2)O_(7)^(2-)+14H^(+)+6e^(-)xx5` balanced redox equation applying molarity equation `therefore (0.125xxV)/(10)=(0.2000xx12.00)/(6)` or `V_(1)=(0.200xx12.00)/(6)xx(10)/(0.125)=32 mL` |
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| 17. |
How many mL is one m^(3) ? |
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Answer» SOLUTION :`1m^(3) = (10dm)^(3) = 10^(3) DM^(3) = 10^(3)L , 1L = (10cm)^(3) = 10^(3)cm^(3) = 10^(3)mL` THEREFORE `1m^(3) = (100cm)^(3) = 10^(6)cm^(3) = 10^(6)mL` One `m^(3)` of GAS is `10^(6)`mL. |
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| 18. |
How many millilitres of 0.5 M H_(2)SO_(4) are needed to dissolve 0.5 g of copper (II) carbonate ? |
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Answer» `N_(1)= "Normality of " H_(2)SO_(4)=0.5xx2=1N` `V_(1)= " VOL of " H_(2)SO_(4)` `N_(2)`= Normality of copper (II) carbonate `=(0.5xx2)/(123.5)N` `V_(2)`=VOLUME of copper (II) carbonate =1000 ML Thus, `1.0xxV_(1)=(0.5xx2)/(123.5)xx1000` or `V_(1)=8.09 mL` |
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| 19. |
How many milliliters of 0.125 M KMnO_4 are required to react completely with 25.0 mL of 0.250 MFeSO_4 solution in the acidic medium ? |
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Answer» Solution :The balanced chemical EQUATION is : `MnO_4^(-) +5Fe^(2+) +8H^(+) rarr MN^(2+) +5Fe^(3+) +4H_2O` From the balanced equation it is clear `{:(1 mol KMnO_4 =, 5 mol FeSO_4),(158,5xx152):}` Moles of `FeSO_4` present in 25 mL of 0.250 M solution : `=(0.250)/1000xx25=0.00625` mol Let us determine the number of moles of `KMnO_4` that must react 5 mol of `FeSO_4` react with `KMnO_4` = 1mol 0.00625 mol of `FeSO_4` will react with `KMnO_4` `=1/5xx0.00625=0.00125 mol` Now, we are the calculate the volume of 1.25 M `KMnO_4` solution which contain 0.00125 mol . According to definition of morality , 0.125 M solution means that `0.125 mol KMnO_4` is present in = 1000 mL 0.00125 mol of `KMnO_4` is present in `=1000/(0.125)xx0.00125=10 mL` This can also be CALCULATED as : Morality `=("Moles"xx1000)/(Volume)` `0.125=(0.00125xx1000)/(volume)` `:. Volume = (0.00125 xx1000)/(0.125) = 10 ` mL |
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| 20. |
How many milliliters fo a 0.05 M KMnO_4 solution are required to oxidize 2.0 g FeSO_4 in a dilute acid solution ? |
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Answer» ` 32. 56 mL` `overset(+7)(M)nO_(4)^(-) rarroverset(+2)(M)n^(2+), n_("factor")=5` `Fe^(2+) rarr Fe^(3+), n_("factor")=1` Normality of `KMnO_(4)=(5)(0.05)` `=0.25 N` Volume of `KMnO_(4)=V` milliliters Thus, milliequivalents of `KMnO_(4)=NxxV` `=0.25 V` Equivalents of `FeSO_(4)=("Mass"_(FeSO_(4)))/("Gram EQUIVALENT mass"_(FeSO_(4)))` `["Note that eq. wt. of "FeSO_(4)=("Formula weight")/("Change in O.N.")=152/1]` Milliequivalent of `FeSO_(4)=2/152xx1000` ACCORDING to the law of equivalence, `"Milliequivalents"_(KMnO_(4))="Miliequivalents"_(FeSO_(4))` `0.25 V= 2/(152) xx 1000` ` V= (2xx 1000)/(152 xx 0. 25) ` ` = 52. 63 mL`. |
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| 21. |
How many milligrams of pure sulphuric acid in 250mL aqueous solution has a pH value 3 ? |
| Answer» SOLUTION :`12.25 MG` | |
| 22. |
How many milliliters of 0.025 M K_2Cr_2O_7 are required to react completely with 25.0 mK of 0.20 M solution of FeSO_4 ? |
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Answer» Solution :The BALANCED chemical equation is `3Fe^(2+)+Cr_2O_7^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+)+7H_2O` 1 mol of `K_2Cr_2O_7 =6 " mol of "FeSO_4` Moles of `FeSO_4` PRESENT in 25.0 mL of 0.20 M solution `(0.20)/1000xx25.0 =0.005 mol` Now 6 mol of `FeSO_4` REQUIRE `K_2Cr_2O_7` = 1mol `=1/6xx0.005 = 0.000833 ` mol Let US calculate the volume of `K_2Cr_2O_7` solution containing 0.000833mol 0.025 mol of `K_2Cr_2O_7` are present in =100 mL 0.000833 mol of `K_2Cr_2O_7` are present in `=(1000xx0.000833)/(0.025)=33.3mL` |
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| 23. |
How many milligrams of bromine are present in a bottle containing 1/10 kg ? |
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Answer» `1//10kg=0.1kg=((10^(6)mg))/((1kg))xx(0.1kg)=10^(5)`mg |
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| 24. |
How many milligrams of calcium should be added to decrease the concentration of 200ml of 1.02N HCl to 1.0N HCl? |
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| 25. |
How many milli-gram of iron (Fe^(2+))are equal to 1mL of 0.1055N K_(2) Cr_(2)O_7equivalent ? |
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Answer» 5.9 mg `w = (0.1055xx56)/1000 = 5.9` mg |
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| 27. |
How many methyl groups present in neopentance. |
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| 28. |
How many methods can be used for the preparation of the isopropyl benzene? (v). Benzene +H_(3)C-overset(CH_(3))overset(|)(C)=CH_(3)overset(H_(2)SO_(4))to |
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| 29. |
How many methyl groups are present in 2, 5-Dimethyl-4-ethylnonane ? |
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Answer» 2 The no. of `CH_(3)` groups is 4. |
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| 30. |
How many methods can be used for the preparation of iodobenzene? (i). C_(6)H_(6)+I_(2) (ii). C_(6)H_(6)+I_(2)overset(HNO_(3))to |
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Answer» `C_(6)H_(5)N_(2)+CL^(-)+KI` |
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| 31. |
How many “methyl groups” are present in 2,3-dimethyl-4-ethyl heptane |
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Answer» 2 |
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| 32. |
How many metamers of 3-pentanone are possible? Write their structures and IUPAC names. Can these be regarded as position isomers as well ? |
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Answer» Solution :Two metamers (I and II) of 3-pentanone are POSSIBLE. `underset("3-Pentanone")(CH_(3)CH_(2)-OVERSET(O)overset(||)(C)-CH_(2)CH_(3)) "" underset("2-Pentanone (I)")(CH_(3)-overset(O)overset(||)(C)-CH_(2)CH_(2)CH_(3))``{:(""O""CH_(3)),("||""|"),(CH_(3)-C-CH-CH_(3)),("3-Methyl-2-butanone (II)"):}` Both these can ALSO be regarded as position isomers of 3-pentanone because they differ in the position of the KETO group. |
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| 33. |
How many metal cations get precipitated, when (NH_(4))_(2)S solution is added into the aqueous solutions of the following cations? Ag^(+),Cu^(2+),Cr^(3+),Mn^(2+),Zn^(2+),Bi^(3+) |
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| 34. |
How manymaximumnumberofelectrons in 3porbitalhavespinquantumnumbers= +(1)/(2)? |
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Answer» 0 |
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| 36. |
How many main states of matter ? |
| Answer» Solution :(i) Solid (ii) LIQUID and (iii) Gas. In additional care PLASMA and BEC (BOSE Einstein Condense) is KNOWN as states of MATTER. | |
| 37. |
How many litres of water must be added to 1 litre of an aqueous solution of HCl with pH of 1 to create an aqueous solution of with pH of 2 ? |
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Answer» Solution : pH = 1 means `[H^(+)]=10^(-1) M = 0.1 M` pH = 2 means `[H^(+)]= 10^(-2)M = 0.01 M` `{:("Applying ",MxxV_(1),=,M_(2)xxV_(2)),(,"(Before dilution)",,"(After dilution)"):}` `0.1xx1=0.01xxV_(2) "or " V_(2)=10L` `:.` Volume of WATER to be added = 10 - 1 = 9 L |
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| 38. |
How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? |
| Answer» SOLUTION :`0.1 XX 1 = (1 + V) (0.01) RARR v = 9 L ` | |
| 39. |
How many litres of SO_(2) taken at NTP have to be passed through a solution of HCIO_(3) to reduce 16.9 g of it to HCI? |
| Answer» SOLUTION :13.44 LITRES | |
| 40. |
How many litres of liquid C Cl_(4) (d=1.5 g/cc) must be measured out to contain 1 xx 10^(25) C Cl_(4)molecules? |
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| 41. |
How many litres of CO_(2) at STP wil be formed when 100 of 0.1 M H_(2)SO_(4) reacts with excess of Na_(2)CO_(3)? |
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Answer» 22.4 |
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| 42. |
How many liters of helium gas at NTP has weight same as 3.0115xx10^(23) molecules of methane? |
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| 43. |
How many layer are adsorbed in chemical adsorption? |
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Answer» ONE |
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| 44. |
How many lattice points are there in one unit cell of each of the following lattices ? (i)Hexagonal close packing and cubic packing ltbr |
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Answer» Solution :LATTICE points in face centred cubic or face centred tetragonasl8 ( at corconers) + 6 ( at face centres) = 14.However, particles per UNIT cell = ` 8 xx 1/8 + 6 xx 1/2= 4` Lattice points in body centreed cube =8 ( at corners) + 1 ( at body centre) = 9 However, PATICLES , per unit cell ` = 8 xx 1/8 + 1=2` |
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| 45. |
How many lattice points are there in one unit cell of each of the following lattices ? (i) face centred cubic ,(ii) face centred tetragonal ,(iii) body centred cubic. |
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Answer» Solution :Lattice points in face centred cubic or face centred TETRAGONAL =8 (at corners) + 6 ( at face centres ) =14. However , PARTICLES PER unit cell =`8xx1/8+6xx1/2=4` Lattice points in body centred cube =8 (at corners ) + 1 ( at body centre) =9 However, particles per unit cell =`8xx1/8+1=2` |
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| 46. |
How many isotopes of Hydrogen ? Give its Names. |
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Answer» Solution :HYDROGEN has THREE ISOTOPES : (i) protium `(._1^1H)` , (ii) deuterium (`._1^2H` and D ) and (III) tritium (`._1^3H` or T) |
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| 47. |
How many joules of translational kinetic energy is associated with 4 grams of methane at 27^@C? |
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Answer» Solution :Number of MOLES of methane ` = ("weight of methane")/("GMW") = (4)/(16) = (1)/(4)` Kinetic energy is GIVEN as (K.E.)` = (3)/(2)NRT = (3)/(2) xx(1)/(4) xx 8.314 xx 300 = 935` joule |
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| 48. |
How many isomertic pentanoic acid undergo ? Hence Vohard zelinsky reaction on treatment with Br_(2)//"red" P |
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| 49. |
How many isomers with melocular formula C_6H_(12) gives Acetaldehyde in reductive ozonolysis asone of product |
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| 50. |
How many isomers of C_8 H_(10) when react with hot alkaline KMnO_4 give di-carboxylic acid as a product ? |
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