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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.
| 92351. |
3 Faraday of electricity are passed through molten Al_2O_3, aqueous solution of CuSO_4 and molten Nacl taken in three different electrolytic cells. The amount of Al, cu and Na deposited at the cathodes will be in the ratio of : |
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Answer» 1 MOLE : 2 mole : 3 mole |
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| 92352. |
3 faraday electricity was passed through the three electrolytic cells connected in series containing Ag^(+)Ca^(2+) and Al^(+3) ions respectively. The molar ration in which the three metl ions are liberated at the electrodes is : |
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Answer» `1:2:3` |
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| 92353. |
3-ethyl pentan-3-ol is obtained by C_2H_5MgBr and what? |
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Answer» pentan-2-one |
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| 92354. |
3 amp of current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pd^(n+) was deposited at the cathode. Find n. (Pd = 106.4) |
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Answer» |
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| 92355. |
3^(@) alcohol can not be oxidised even with strong oxidising agent because - |
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Answer» C' with -OH GROUP does not having H atom `R- underset(R..)underset(|)OVERSET(R.)overset(|)C-OHunderset(Na_2Cr_2O_7//H_2SO_4)overset([O])to "No oxidation"` |
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| 92356. |
3 BaCl_(2) + 2Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2)+6NaCl Maximum amount of Ba_(3)(PO_(4))_(2) formed when 2 moles each of Na_(3)PO_(4) and BaCl_(2) react is |
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Answer» 4 mol Here `BaCl_(2)` is the limiting reactant No. of moles of `Ba_(3)(PO_(4))_(2)` formed `=(1)/(3)xx"N0. of moles of" BaCl_(2) = (2)/(3)` mol |
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| 92357. |
3, 5 - Dibromotoluene can be best synthesised by |
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Answer»
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| 92358. |
3/4 th of first order reaction was completed in 32 min 15/16 the part will be completed in |
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Answer» `24` MIN |
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| 92359. |
3, 3-dimethyl butan-2-ol on dehydration gives |
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Answer» 3, 3-dimethyl but-2-ene |
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| 92360. |
2Zn + O_(2) rightarrow 2ZnO, Delta G^(@) = -616 J 2Zn + S_(2) rightarrow 2ZnS, Delta G^(@) = -293 JS_(2) + 2O_(2) rightarrow 2SO_(2), Delta G^(@) = -408 J Delta G^(@) for the following reaction 2ZnS + 3O_(2) rightarrow 2ZnO+ 2 SO_(2)is |
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Answer» `-713J` |
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| 92361. |
2xx10^(-3) g of green algae absorbs 7 xx 10^(-4) moles of CO_(2) per hour by photosynthesis as per the following equation: 6CO_(2) + 5nH_(2)O underset("Chlorophyll")overset(hr)to (C_(6)H_(1)O_(5))_(n) + 6nO_(2) If all the carbon of CO_(2) is converted into starch, then how long will it take for algae to increase its mass by 100%? |
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Answer» `6.34` HOURS |
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| 92362. |
2X(g)+Y(g)+3Z(g)rarrProducts Choose the correct statement (s). |
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Answer» If `75%` of X undergoes reaction in 20 SEC, `50%` of X will react in 10 sec if `[Z]gtgt[X]` |
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| 92363. |
2SO_(3)hArr2So_(2)+O_(2). IF K_(c)=100 alpha=1, half of the reaction is completed, the concentration of SO_(3)and SO_(2) are equal, the concentration of O_(2) is |
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Answer» `0.001 M` |
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| 92364. |
2SO_(2)(g) + O_(2)(g) iff (V_(2)O_(5)) is an example of |
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Answer» neutralisation reaction It is an example of hetrogeneous catalysis as the REACTANTS and catalysts are in DIFFERENT phases as `V_(2)O_(5)` is a solid. |
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| 92365. |
2SO_(2)(g) + O_(2) (g) overset(NO_(2)(g))to Is this reaction an examplefor Homogeneous or Heterogeneous catalysis. |
| Answer» SOLUTION :HOMOGENEOUS CATALYSIS | |
| 92366. |
2SO_(2)+O_(2)hArr2SO_(3), if the volume of the reaction vessel is doubled, the rate of forward reaction will be |
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Answer» `1//4` th of INITIAL value |
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| 92367. |
2SO_2+O_2 hArr 2SO_3 Starting with 2 mol of SO_2 and 1 mol of O_2 in 1-L flask, mixture required 0.4 mol of MnO_4^- in acidic medium . Find the K_c value |
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Answer» SOLUTION :`5 SO_2+2MnO_4 to 2Mn^(2+)+5SO_4^2` 2 MOL of `MnO_4^(-)-= 1 mol SO_2` `therefore 0.4 ml "of" Mno_4^(-)-=1 mol SO_2` `{:(2SO_2,+,O_2,hArr,2SO_3),(2,"",1,"",0),((2-2x),"",(1-x),"",2x):}` (2-2x)=1 mol (as DETERMINED by `MnO_4^-`) `therefore [SO_3]=(2x)/V=1M, [SO_2^-]=(2-2x)/V=1M` `[O_2^-]=((1-x))/(V=0.5 M,K_c=([SO_3]^2)/([SO_2]^2[O_2])=2` |
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| 92368. |
2RCl+Sioverset("Cu power")underset("570 K")rarr R_(2)SiCl_(2)overset(H_(2)O)rarr R_(2)Si(OH_(2))overset("Polymerisation")rarr X, Then X will be |
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Answer» CYCLIC silicone |
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| 92369. |
2PbS + 3O_(2) rarr 2PbO + 2SO_(2) Name the process. |
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Answer» Roasting |
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| 92370. |
2P overset(-H_(2)O) to Q overset(-[O])toR If P is parent phosphoric acid then according to given information the correct statement is/are: |
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Answer» <P>Q is pyro form and R is hypo form of givenn present oxy ACID P 
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| 92371. |
2NO_(2)hArr 2NO+O_(2),K=1.6xx10^(-12)NO+1/2O_(2)hArrNO_(2),K'=? |
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Answer» `K'(1)/(K^(2))` `K=1.6xx10^(-12)` `NO+1/2O_(2)hArrNO_(2)""...(ii)` REACTION (ii) is half of reaction (i) `K=([NO]^(2)[O_(2)])/([NO_(2)]^(2))""...(i)` `K=([NO_(2)])/([No][O_(2)]^(1//2))""...(ii)` On multiplying (i) and (ii) `KxxK'=([NO]^(2)[O_(2)])/([NO_(2)]^(2))xx([NO_(2)])/([NO][O_(2)]^(1//2))` `=([NO][O_(2)]^(1//2))/([NO_(2)])=1/K` `KxxK'=1/K,K=(1)/(K^(2)),K'=(1)/(sqrt(K)).` |
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| 92372. |
2NO + H_(2) to N_(2)O + H_(2)O , The reaction , follows third orderkinetics. Write (a) ratelaw and (b) unitsof rateconstant. What happens to the rate if the volumeof vessel is reduceto one - halfat constnattemperature ? |
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Answer» Solution :(a) Rate LAW is given as, rate `=[NO)^(2)[H_(2)]` (b) Units of rate constant `=L^(2)" MOL"^(-2)s^(-1)` It volume of the vessel is halved, concentration of each REACTANT is doubled. The rate will be Rate `=k[2NO)^(2)[2H_(2)]=8k[NO)^(2)[H_(2)]` Rate will be 8 times more than the ORIGINAL rate of the reaction. |
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| 92373. |
2NH_(3)hArrN_(2)+3H_(2).the vessel is such that the volume remains effectively constant where as pressure increases to 50 atm. Calculate the percentage of NH_(3) actually decomposed |
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Answer» `65%` `{:("Initial mole",a,0,0),("Mole at equlibrium",(a-2x),x,3x):}` Initial pressure of `NH_(3)` of mole = 15 atm at `27^(@)C` The pressure of 'a' mole of `NH_(3)=P atm at 347^(@)C` `therefore15/300=P/620` `thereforep=31atm` At constant volume and at `347^(@)C, `mole `prop` pressure `a prop31` (before equlibrium) `thereforea+2x prop50` (after equlibrium) `THEREFORE(a+2x)/(a)=50/31` `thereforex=(19)/(62)a` `therefore% of NH_(3)"decomposed"=(2x)/(a)xx100` `=(2xx19a)/(61xxa)xx100=61.29%` |
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| 92374. |
2NH_(3) to N_(2) + 3H_(2) If at the reaction, 18 mole of H_(2) is produced then find moles of NH_(3) initially taken |
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Answer» 12 |
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| 92375. |
2NH_(3) to N_(2) + 3H_(2) Find moles of H_(2) produced from 4 moles of NH_(3) |
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Answer» 6 |
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| 92376. |
2NaOH+SO_(2) to A+H_(2)O B+H_(2)O+SO_(2) to 2NaHSO_(3). A and B are |
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Answer» `A-Na_(2)SO_(3), B-Na_(2)SO_(3)` |
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| 92377. |
2N_(2)O_(5)rarr NO_(2)+O_(2), (d[N_(2)O_(5)])/(dt)=k_(1)[N_(2)O_(5)], (d[NO_(2)])/(dt)=k_(2)[N_(2)O_(5)] and (dO_(2))/(dt)=k_(3)[N_(2)O_(5)], the relation between k_(1), k_(2) and k_(3) is |
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Answer» `2k_(1)=4k_(2)=k_(3)` |
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| 92378. |
2N-HCl will have the sae molar concentration is |
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Answer» `0.5 N-H_(2)SO_(4)` As given, 25 ml of 0.2 N HCl neutralises remaining 1/2 of `Na_(2)CO_(3)`. So number of milliequivalents of HCl=no. of Milliequivalents of NaOH `250xx0.1" N HCl"=250xx0.1" N NaOH"` Which is equal to 1g of NaOH. |
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| 92379. |
2m urea solution is diluted from 2 kg to 5 kg by addition of water. Calculate molality of diluted solution (Molecular weight of urea = 60 g/ mol) |
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Answer» 0.4 m `THEREFORE M=(240xx1)/(5xx60)=0.8 M` |
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| 92380. |
2M solution of Na_(2)CO_(3) is boiled in a closed container with excess of CaF_(2) .Very little amount of CaCO_(3) and NaF are formed.If the solubility product (K_(sp)) of CaCO_(3) is x and molar solubility of CaF_(1) is y.Find the molar concentration of F^(-) in resulting solution after equilibrium is attained. |
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Answer» SOLUTION :`{:(,Na_(2)CO_(3)+,CaF_(2)(s)hArr,2NaF(AQ)+,CaCO_(3)(s)),(t=0,2,-,0,-),(t=eq,2-a,-,2a,-):}` where is very small For `CaCO_(3),K_(sp)=x=[Ca^(2+)][CO_(3)^(2-)]=[Ca^(2+)]xx2( :.CO_(3)^(2-)` mainly COMING from `Na_(2)CO_(3)` `[Ca^(2+)]=(x)/(2)` For `CaF_(2),K_(sp)=4y^(3)=((x)/(2))[F^(-)]^(2)RARR [F^(-)]=sqrt((8y^(3))/(x))` |
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| 92382. |
2HIoverset("Au")(to)H_(2)+I_(2) following n^(th) order kinetics. If rate of the reaction is given by rate =-(1)/(2)(d[HI])/(dt) and concentration of HI drops from 1.5M to 0.3M in 7.3 sec, value K is |
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Answer» `8XX10^(-2)"M -s"^(-1)` |
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| 92383. |
2HgtoHg_(2)^(++),E^(@)=+0.855V HgtoHg^(2+),E^(@)=0.799V Equilibrium constant for the reaction Hg+Hg^(2+)toHg_(2)^(++) at 27^(@)C is |
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Answer» 89 `=0.855-0.799=0.056V` `E_(cell)^(@)=(0.0591)/(2)"LOG"K_(c)` or `logK_(c)=(0.056xx2)/(0.059)=1.8983` `K_(c)`=ANTILOG `1.8983=79:12=79` |
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| 92384. |
2HI_((g))hArrH_(2(g))+I_(2(g)) The equilibrium constant of the above reaction is 6.4 at 300 K. If 0.25 mole of H_(2) and I_(2) are added to the system, the equilibrium constant will be |
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Answer» 0.8 |
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| 92385. |
2HI_((g))hArrH_(2(g))+I_(2(g)), When inert gas added, effect on its K_(p)(at V="const") |
| Answer» ANSWER :C | |
| 92386. |
2HCHOoverset(50% NaOH)rarr CH_(3)OH+HCOONa This reaction is called : |
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Answer» ALDOL condensation |
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| 92389. |
2HCHO + conc. NaOH to CH_(3)OH+HCOONa is: |
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Answer» CROSS ALDOL condensation |
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| 92390. |
2gm of O_(2) at 27^(@)C and 76mm of Hg pressure has volume |
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Answer» 1.5 lit |
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| 92391. |
2g of benzoic acid dissolved in 25g of C_(6)H_(6). Shows a depression in freezing point equal to 1.62K. Molal depression constant of C_(6)H_(6) is 4.9K kg "mol"^(-1). What is the percentage association of acid if it forms double molecules in solution? |
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Answer» Given, w=2g,W=25g,DeltaT=1.62,K_(f)=4.9` `therefore 1.62=(1000xx4.9xx2)/(25xxm)` or `m_("EXP")=241.98` `becauseFor association, `nC_(6)H_(6)COOOHhArr(C_(6)H_(5)COOH)_(n)` `{:(1,0),(1-alpha,alpha//n):}` Total MOLES at equilibrium `=1-alpha+(alpha//n)` `=1-alpha+(alpha//2)=1-(alpha//2)` `n=2` (for dimer formation) `(m_(N))/(m_("exp")=1-(alpha)/(2)` or `1-(alpha)/(2)=(122.0)/(241.98)` `therefore alpha=0.992` or `99.2%` |
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| 92392. |
2g of benzoic acid dissolved in 25g of benzene produces a freezing-point depression of 1.62^(@). Calculate the molecular weight. Compare this with the molecular weight obtained from the formula for benzoic acid, C_(6)H_(5)COOH. (K_(f)=4.90) |
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Answer» Solution :Molality `=(DeltaT_(f))/(K_(f))` ……….(Eqn. 7) `=(1.62)/(4.9)` Suppose the molecular weight of benzoic acid is M `:.` moles of benzoic acid `=(2)/(M)` `:.` molality (moles /1000g)`=(2)/(M)XX(1000)/(25)=(80)/(M)` Thus `(80)/(M)=(1.62)/(4.9)`, `M=241.97` The actual value of the molecular weight of benzoic acid, from its formula `C_(6)H_(5)COOH`, is 122. The observed (experimental) value i.e., `241.97` is much greater than the NORMAL value because of the association of benzoic acid in benzene |
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| 92393. |
2g of a non-electrolyte solute dissolved in 200 g of water shows an elevation of boiling point of 0.026^(@)C. K_(b) of water is 0.52 K . "mole"^(1) kg then the molecular weight of solute is x xx100. What is the value of 'x'? |
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Answer» |
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| 92394. |
2g of a gas X are introduced into an evacuated flask kept at 25^(@)C. The pressure is found to be 1 atm. If 3g of another gas Y are added to the same flask, the total pressure becomes 1.5 atm. Assuming that ideal behaviour, the molecular mass ratio of M_(x) and M_(y) is |
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Answer» `1.5 = 1.0 + P'_(Y) `or `P'_(Y)=0.5 ATM` for gas `X, P'_(X) XX V = (2)/(M_(X))RT` for gas `Y, P'_(Y) xxV = (3)/(M_(Y))RT` `implies (P'_(X))/(P'_(Y))=(2)/(3)(M_(Y))/(M_(X))` or `(M_(Y))/(M_(X))= (3)/(2) xx (P'_(X))/(P'_(Y))=(3xx1.0)/(2xx0.5)=3` |
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| 92395. |
2F of electricity is passed through 20 L of a solution of aquous solution of KCI. Calculate the pH of the solution. |
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Answer» IF = 1eq of `H_(2)(g)` = 1 eq of `Cl_(2) -=` 1 eq of `OH^(-)` IONS 2F = 2 eq of `OH^(-)` `[OH^(-)] = (2eq)/("Volume in L") = 2/(20 L) = 10^(-1) N` or M `therefore pOH =- log(10^(-1)) =1` `pH = 14-1 = 13` |
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| 92396. |
2D,3L,4D,5D-6-Pentahhydroxyhexanal can give. |
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Answer» Tollen's TEST 
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| 92397. |
2CuFeS_(2) + O_(2) to Cu_(2)S+ 2FeS + SO_(2) Which process of metallurgy of copper is represented by above equation ? |
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Answer» Concentration `( C _6H _ 5 )_3overset ( * )C gt(C _6H_ 5 )_ 2overset (* ) C Hgt(CH_3)_3 overset (* )C gt(CH_3 )_2 overset (* )CH ` Thestabilisation offirsttwo is duetoresonance andlasttwois due to inductiveeffect. |
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| 92398. |
2CO(g)+ O_(2)(g) to 2CO_(2)(g) , 2NO(g)+ O_(2)(g) to 2NO_(2)(g) . Whichis relatively faster ? Why ? |
| Answer» SOLUTION :`2NO_((g))+O_(2(g)) to 2NO_((g))` is relatively faster. In both the reactions, the stoichiometry is same and number of bonds transformed is also same. The number of electrons shared in CO MOLECULE is `6` and in NO molecule is `5`. Breaking bond in NO molecule is relatively EASY than that in CO molecule. Further NO has unpaired electron and thus more REACTIVE. | |
| 92399. |
2CHCl_(3) + O_(2) overset(x)to 2COCl_(2) + 2HCl In the above reaction , X stands for: |
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Answer» An oxidant |
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| 92400. |
2CH_(3)COOH underset(300^(@)C)overset(MnO)toA, product 'A' in the reaction is:- |
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Answer» `CH_(3)CH_(2)CHO` |
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