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 much water should be added to 100 ml of1MH_(2)SO_(4) solution to make it exactly 0.1 N ? |
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Answer» 1200 ml Applying `N_(1)V_(1)=N_(2)V_(2)` `2xx200=0.1xxV_(2)` or `V_(2)=(1xx200)/(0.1)=2000` ml Amt. of water to be ADDED `=2000-100` `=1900` ml. |
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| 2. |
How much volume of oxygen will be required for complete combustion of 40 mL of acetylene (C_(2)H_(2)) and how much volume of carbon dioxide will be formed? All volumes are measrued at NTP. |
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Answer» Solution :`{:(2C_(2)H_(2),+,5O_(2),=,4CO_(2),+,2H_(2)O),("2vol",,"3vol",,"4vol",,),(40mL,,(5)/(2)xx40mL,,(4)/(2)xx40mL,,),(40mL,,100mL,,80,,):}` So, for complete combustion of 40mL of acetylene, 100 mL of OXYGEN are required and 80mL of CARBON dioxide is formed. |
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| 3. |
How much urea must be dissolved in 10^(-2)m^(3) water to yield a pressure is 2.03xx10^(5)Nm^(-2) at 300K? |
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Answer» 48.30 g `pi=CRT` `2xx10^(5)=m/60xx1000=80.186xx60xx0.01g` in `0.01m^(3)=48.3gimplies1J=1Nm` |
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| 4. |
How much urea (molar mass = "60 g mol"^(-1)) should be dissolved in 50 g of water so that its vapour pressure at room temperature is reduced by 25%. Calculate molality of the solution obtained. |
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Answer» `"Molality of the solution "=(55.556//"60 mol")/(50 g)xx1000g KG^(-1)=18.52m` |
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| 5. |
How much time would it take in minutes to deposit 1.18 g of metallic copper on a metal object when a current of 2.0 A is passed through the electrolytic cell containing Cu^(2+) ions? [Molar mass of Cu = 63.5 g mol^(-1), 1F = 96,500 Cmol^(-1)] |
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Answer» SOLUTION :`m = Z xx I xx t` `1.18 = (63.5)/(2 xx 96500) xx 2 xx t` `t = (1.18 xx 2 xx96500)/(2 xx 63.5)` `= 1793.23` SEC `= (1793.23)/(60) = 29.88` min. |
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| 6. |
How much type of colloidal systems can be formed on the basis of physical state of dispersion medium and dispersion phase ? |
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Answer» Two |
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| 7. |
How much time would be needed to deposit 0.25 g of metallic nickel. (Atomic mass = 58.5) on a metal object using a current of 1A during electroplating? |
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Answer» Solution :`Ni^(++) + 2e^(-) to Ni` 2F or `2 xx 96500` C of electricity is required to deposit 58.5 g of Ni. 58.5 g of Ni REQUIRES `2 xx 96500` C 0.25 g Ni requires `=(2 xx 96500 xx 0.25)/58.5 = 824.8` C Amount of electricity = CURRENT in amperes `xx` time in SECONDS Time in seconds `=("Amount of electricity")/("Current in amperes")` `=824.8/1 = 824.8 sec = 13.74` MINUTES. |
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| 8. |
How much time in minutes passes between 99% to 99.9% completion of a first order reaction with half life 0.3010"min"^(-1)? |
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Answer» `DELTA=(t_(2)-t_(1))=(3-2)=1,Deltat=1` |
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| 9. |
How much time (in hours ) would it take to distribute one avogardo number of wheat grains if 10^20 grains are distributed each second ? |
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Answer» 0.1673 Time taken to distribute `6.022 XX 10^23`grains ` = (1)/(10^20) xx 6.022 xx 10^23 ` seconds ` = (6.022 xx 10^23)/(10^20 xx 60 xx 60) ` hours = 1.673 hours |
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| 10. |
How much time (in hours) is required for a current of 2 amp to decompose electrolytically 18g of water. |
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Answer» |
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| 11. |
How much time is required for completion of a zero order reaction? |
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Answer» `(2[R_(o)])/(K)` |
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| 12. |
How much sulphur is present in organic compoundif on analysis 0.53 gmof thiscompound gives 1.158 gm of BaSO_(4). |
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Answer» `10%` `= (32)/(233) xx (" MASS of" BaSO_(4))/(" mass of ORGANIC compound") xx 100` `= (32)/(233) xx (1.158)/(0.53) xx 100 = 30%`. |
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| 13. |
How much sulpher is present in the rubber used for tyres ? |
| Answer» SOLUTION :`5 %` | |
| 14. |
How much sulphur is added during vulcanization to get rubber for tyre ? |
| Answer» SOLUTION :`5 %` | |
| 15. |
How much sulpher is present in the rubber used for bettery case ? |
| Answer» SOLUTION :`30 %` | |
| 16. |
How much sugar (C_(12)H_(22)O_(11)) will be required if each person on the earth is given 100 molecules of sugar? The population of the earth is 3xx10^(10). |
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| 17. |
How much sodium chloride be dissolved in water to make 1 litre of 0.1 F solution ? |
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Answer» VOLUME OS solution = L Formality of solution = 0.1 F=0.1 (Formula mass)/LITRE `"Formality (F)"=("Mass os NaCI/Gram formula mass")/("Volume of solution in litres")` `"0.1 (Formula mass/litre)"=W/(58.5 "g (Formula mass)"^(-1)xx1L)` `W=0.1 ("Formula mass/L")xx58.5 g("Formula mass")^(-1)xx1L` ` W=0.1xx58.8 g= 5.85 g`. |
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| 18. |
How much sodium acetate should be added to a 0.1 M solution of CH_(3)COOH to give a solution of pH = 5.5 (pK_(a) of CH_(3)COOH = 4.5) |
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Answer» 0.1 M `log.(["Salt"])/("Acid") = 5.5 - 4.5 = 1` `(["Salt"])/(0.1) = "antilog 1" = 10, ["Salt"] = 1` . |
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| 19. |
How much should the pressure be increased in order to decrease the volume of a gas 5% at a constant temperature ? |
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Answer» <P>0.05 New volume =0.95 V `therefore P_(1)V=P_(2)xx0.95 V` `P_(2)=(P_(1))/(0.95)=1.0526P_(1)` `therefore` INCREASE in pressure `=1.0526P_(1)-P_(1)` `=0.0526P_(1)` `=5.26%" of "P_(1)` HENCE, (B) is the correct answer. |
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| 20. |
How much quantity of electricity present on 1 electron in coulomb ? |
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Answer» `6.02xx10^(23)` |
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| 21. |
How much quantity of electricity is required to liberated 1 g equivalent weight of an element ? |
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Answer» 96,500 Faradays |
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| 22. |
How much quantity of electricity has to be passed through 200 mL of 0.5 M CuSO_(4) solution to completely deposite copper ? |
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Answer» 96500 C `Cu^(2+) + 2E^(-) to Cu` To deposit 0.1 mole of `Cu^(2+)` , electricity required = `2XX 9650` C |
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| 23. |
How much Pt is deposited on cathode when 0.80F current is pass through 1.0 M solution of Pt^(4+) ? |
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Answer» 1.0 MOLE `4F=1` mole Pt `THEREFORE 0.80F=(?)` `=(1xx0.80)/(4)=0.20` mole. |
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| 24. |
How much propanol is required for dehydration to get 2.24 litre of propene at N.T.P. If yield is 100% |
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Answer» Solution :`C_(3)H_(8)O+H_(2)SO_(4)rarrC_(3)H_(6)+H_(2)O+H_(2)SO_(4)` Molecular weight of propanol `=60` from the equation given above we can see that from dehydration of `1` mole or `60` GRAM of propanol we get `1` mole (`22.4`lit) of PROPENE as PRODUCT. `because 22.4` litre of `C_(3)H_(6)` can be get from dehydration of `60g` of propanol. `therefore 1` litre of propene can be get from dehydration of `(60)/(22.4)g` of propanol |
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| 25. |
How much PCl_(5) must be added to one litre volume reaction vessel at 250^(@)C in order to obtain a concentration of 0.1 mole of Cl_(2),K_(c) for PCl_(5)hArrPCl_(3)+Cl_(2) is 0.0414" mol dm"^(-3) at 250^(@)C. |
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| 26. |
How much PCl_(5) must be added to one litre volume reaction vessel at 250^(@)C in order to obtain a concentration of 0.1 mole of Cl_(2),K_(c) for PCl_(5)hArrPCl_(5)+Cl_(2) is 0.0414" mol dm"^(-3) at 250^(@)C. |
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| 27. |
How much oxygen is required for complete combustion of 560 g of ethene (Mw = 28 g/mol) ? |
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Answer» 6.4 KG `(560)/(28) mol ` |
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| 28. |
How much of sulphur is present in an organic compound, if 0.53g compound gave 1.158g of BaSO_(4) analysis? |
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Answer» 0.1 |
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| 29. |
How much of NaOH is required to neutralise 1500cm^(3) of 0.1 N HCl ? (Na = 23) |
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Answer» Solution :`1500cm^(3)" of of 0.1 N HCl"-=(0.1)/(1000)xx1500="0.15 g eq."` It will neutralise `NaOH=0.15" g eq."=0.15xx40g=6g` |
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| 30. |
How much of NaOH is required to neutralise 1500 cm^(3)of 0.1 N HCI? (Na = 23) |
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Answer» 40 G It will NEUTRALISE `NaOH = 0.15 g eq. = 0.15 xx 40 g = 6 g` |
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| 31. |
How much of NaOH is required to neutralise 1500 cm^(3) of 0.1N HCl? (Na= 23) |
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Answer» 4g E.W (NaOH) = 40M V= 1500 `cm^(3)`, N= 0.1 N, W= ? `rArr W= (1500 xx 0.1 xx 40)/(1000) rArr W= 6gm` |
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| 32. |
How much of Fe can be theoretically obtained by the reaction of 1 kg of Fe_(2)O_(3)? |
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| 33. |
How much moles of oxidizing agents are reduced by complete reaction of 63.5 gm of Cu with concentrated HNO_(3) solution ? (Atomic weight of Cu=63.5 gm mol^(-1)) |
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Answer» 8 |
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| 34. |
How much minimum volume (in ml) of ((5)/(51)) M aluminium sulphate solution should be added to excess calcium nitrate to obtain atleast 1 gm of each salt in the reaction Al_(2) (SO_(4))_(3) + 2Ca(NO_(3))_(2) rarr 2Al(NO_(3))_(3) + 3CaSO_(4) |
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| 35. |
How much NaOH is required to prepare 10% w/w 500 gram solution of NaOH ? |
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Answer» 0.5 GRAM `THEREFORE 10=(100xx"mass of solute")/(500)` `therefore` Mass of solute `= (10xx500)/(100)=50` gram. |
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| 36. |
How much molecular mass of NaCl is obtained experimentally using colligative properties? |
| Answer» SOLUTION :As CONCENTRATION of particles in the solution is doubled, value of COLLIGATIVE property is double than expected value and hence molecular mass is HALVED, i.e. OBSERVED molecular mass = 58.5/2 = 29.25 | |
| 37. |
How much methanol should be added to water, to make 150 mL. 2 M solution of CH_(3)OH ? |
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Answer» 9.6 gram `therefore 150 mL 2M CH_(3)OH= ?` `= (32xx150xx2)/(1000)=9.6` gram. |
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| 38. |
How much marble of 96.5% purity would be required to prepare 10 litres of carbon dioxide at STP when the marble is acted upon by dilute hydrochloric acid? |
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Answer» 10 L of `CO_(2)` at STP will be obtained from pure `CaCO_(3)=(100)/(22.4)xx10=44.64g` `"Impure marble required "=(100)/(96.5)xx44.64=46.26g` |
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| 39. |
How much marble mainly [CaCO_(3)] " of " 96 . 5%purity would be required to prepare 10 litres of CO_(2) at STP when the marble is acted upon by dilute HCl ? |
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Answer» |
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| 40. |
How much litres of dihydrogen gas will be produced at STP in the reaction of ethanol with 12 gram of Mg ? (Mg = 24 gram//mol) |
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Answer» 11.2 LITER |
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| 41. |
How much is the oxidising power of the MnO_4^(-) (1M)|Mn^(2+) (1M) couple decreased if the H^+ concentration is decreased from 1M to 10^-4 M at 25^@C |
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Answer» Solution :The HALF cell REACTION is `MnO_4^(-) +8H^+ +5e=Mn^(2+)+4H_2O` Let us suppose that only the `H^+` concentration deviates from 1mole`//` litre `E_(MnO overline4. Mn^(2+))=E_(MnO_4^-,Mn^(2+))- 0.0591/2 log""([Mn^(2+)])/([MnO_4^-][H^+]^8)` `E-E^@=-0.0118 log""((1))/((1)(10^-4)^8)=-0.38` volt The couple `MnO_4^-//Mn^(2+)` has thus moved down by 0.38 volt from its standard value i.e to a position of less oxidising POWER. |
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| 42. |
How much ice will separate if a solution containing 25g of ethylene glycol [C_(2)H_(4)(OH)_(2)] in 100g of water is cooled to -10^(@)C, ? K_(f)(H_(2)O)=1.86 |
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| 43. |
How much heat will be required at constant pressurent to form 1.28 kg of CaC_(2) from CaO(s) & C(s)? {:("Given :",Delta_(f)H^(@) (CaO","s),=,-152 kcal//mol),(,Delta_(f)H^(@) (CaC_(2)","s),=,-14 kcal//mol),(,Delta_(f)H^(@) (CO","g),=,-26 kcal//mol):} |
| Answer» Answer :D | |
| 44. |
How much heat is produced in buring a mole of CH_(4) under standard conditions if reactants and products are brought to 298 K and H_(2)O(l) is formed ? What is the maximum amount of useful work thatcan be accomplished under standard conditions by this system ? CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(l) {:(DeltaHf^(0):-74.8,0,-393.5,-285.85kJ),(DeltaGf^(0):-50.8,0,-394.4,-236.8kJ):} |
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Answer» Solution :`DeltaH^(@)=(-393.5)+(2xx-285.85)-(-74.8)-2xx0` `=-890.4kJ`/MOLE Now the free ENERGY CHANGE for a process `DeltaG` equals the maximum useful work that can be done by the system at constant TEMPERATURE and pressure. `:. W_(max)=DeltaG=(2xx-236.8)+(-394.4)-(50.8)-2xx0` `=-817.2kJ`/mole `CH_(4)`. |
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| 45. |
How much heat is required to change 5 gram ice (0^@C) to steam at 100^@C ? Latent heat of fusion and vaporization for water are 80 cal/g and540 cal/g respectively . Specific heat of water is 1 cal/ g/k. |
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Answer» 7200 cal |
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| 46. |
How much heat is required to change 10 g of ice at 0^@Cto steam at 100^@C? |
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Answer» SOLUTION :`Delta H("total") = Delta H_("FUSION") + Delta H_("heating") + Delta H_(vap)` ` = 10(80) + 1 XX 100 + 540 ) ` CAL |
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| 47. |
How much gram of acid should be present in 100 mL volume of solution to obtain decimolar solution having 200 gm/mole molecular mass of diabasic acid ? |
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Answer» 20 gm `M=("MOLE")/("Liter")=(w)/(200)xx(1)/(100)xx1000 =0.1` `therefore 0.1=(w)/(20) therefore 2=2` gm. |
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| 48. |
How much H_(2)SO_(4) is required to prepare 0.1 M 1 liter H_(2)SO_(4) solution ? The density of 98% W/W H_(2)SO_(4) is 1.80 gm/mL. |
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Answer» 22.20 mL `M=(98xx1.80xx10)/(98)=18M` `M_(1)V_(1)=M_(2)V_(2)` `18xx V_(1)=0.1xx1000` `V_(1)=(0.1xx1000)/(18)=5.55 mL` |
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| 49. |
How much glucose, C_(6)H_(12)O_(6) must be dissolved per litre of the solution that yields osmotic pressure os 2.72 atmospheres at 298 K ? R=0.0821 L atm K^(-1). |
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Answer» `V=1.0L, pi=2.72 ATM, R=0.0821" L atm K"^(-1)mol^(-1), T=298K, W_(B)=?` `piV=n_(B)RT or n_(B)=(piV)/(RT),W_(B)/M_(B)=(piV)/(RT)` `W_(B)=(M_(B)xxpixxV)/(RT)=((180" g mol"^(-1))xx(2.72" atm")xx(1L))/((0.0821"L atm K"^(-1)mol^(-1))xx(298 K))=20.0116 g.` |
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