94420 + Interview Questions in Chemistry in Class 12 Page 1837 InterviewSolution

91801.	(A) 1 mole H_(2)SO_(4)contains same mass of oxygen and sulphur. (R) 1 mole of H_(2)SO_(4) represents 98g mass.
Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). If both (A) and (R) are correct but (R) is not the correct explanation of (A). If (A) is correct but (R) is WRONG. If (A) is wrong but (R) is correct. Answer :D

Discussion

91802.	(A) 1 lit of dry air contains 10ml of noble gas mixture(R) The atmosphere abundance of noble gases in dry air is nearly 1%
Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :A

Discussion

91803.	A 1 litre solution of pH=1 diluted upto 10 times.What volume of a solutions with pH =2 is to be added in diluted solution so that final pH remains '2'.
Answer» 1 litre 10 litre 100 litre 25 litre Solution :After DILUTION `[H^+]=10^(-2) implies pH=2` Let V litre solution of pH=2 is added in original solution so that pH REMAINS fixed. `:. [H^+]=(10^(-2)xx10+Vxx10^(-2))/(10+V)=10^(-2)` This result is INDEPENDENT of volume TAKEN.

Discussion

91804.	50% of a first order reaction was found to complete in 16 minute. When will 75% of the same reaction complete:
Answer» 32 minute 16 minute 8 minute 4 minute Answer :B

Discussion

91805.	75% of the first order reaction is completed in 30 minutes. Calculate rate constant of the reaction.
Answer» Solution :t= 30 min, `[R]_circ = 100,[R]` = 25 `K= 2.303 /t LOG``([R]_circ )/([R] )``K= 2.303 /t log``([R]_circ )/([R] )` t= `2.303/30 log ``100/25` K= 0.0767`xx` log 4 = 0.0767 `xx` 0.6021 `k= 0.0461 min^-1`

Discussion

91806.	A 1-g mixture of cuprous oxideand cupric oxide was quantitatively reducedto 0.839 g of metallic copper . Whatwas t he weightof cupricoxidein the original sample ? (Cu = 63.5, O = 16)
Answer» Solution :Letthe WEIGHTOF `CuO be x G`. The weight of `Cu_(2) O` will be (1 - x) g . Asthe Cu atomsareconserve,applyingPOAC for Cu atoms, molesof Cu in CuO + moles of Cu in `Cu_(2)O` = moles of Cu in theproduct `1 xx ` moles of `CuO + 2 xx "moles of " Cu_(2) O` = moles of Cu `(x)/( 79 . 5) + 2 xx (1 - x)/( 143) = (0.839)/( 63.5)[{:(CuO = 79 . 5),(Cu_(2) O = 143):}] ` x = 0.55 g

Discussion

91807.	75% of a first order reaction was completed in 32 minutes, 50% of the reaction will be completed in
Answer» 24 minute 16 minute 8 minute 4 minute Answer :B

Discussion

91808.	A 1 L of water sample has 0.1 g fluoride concentration. What is the concentration of fluoring in terms of ppm level ?
Answer» 250 100 400 1000 Solution :PPM of `F^(-)`ions =`W_(2)/"MASS of solution"xx10^(6)=0.1/10^(3)xx10^(6)` =100

Discussion

91809.	75 % of a first order reaction is completed in 30 minutes.What is the time require for 93.75% of the reaction (in minutes)…
Answer» 45 120 90 60 Answer :D

Discussion

91810.	A 1 - g sample of KClO_(3) washeated under suchconditions thata part ofit decomposedaccording tothe equation(i) 2KCiO_(3) = 2 KCl + 3O_(2)and theremainingunderwentchange accurdingto the eqation : (ii) 4KCIO_(3)= 3KCIO_(4) + KCIIf theamount of O_(2)evelved was 146.8mL, at NTP, calculate the percentage byweight of KClO_(4) in the residue .
Answer» Solution :`KCIO_(3) to KCI + O_(2)` APPLYING POAC for O atoms in the eqn. (i), ,oles of O in `KCIO_(3) = ` moles of O in `O_(2)` `3 xx` moles of `KCIO_(3) = 2XX ` moles of ` O_(2)` `3 xx (wt . if KCIO_(3))/( "mol. wt. of " KCIO_(3)) = 2 xx ("volumeatNTP (mL))/( 22400)` Wt. of `KCIO_(3)= (2 xx 146 . 8 xx 122.5)/( 3 xx 22400)` = 0.5358 g Againapplying POAC for K atoms, moles of K atoms in `KCIO_(3)` = moles of K atoms in KCI `("wt. of " KCIO_(3))/( "mol. wt. of " KCIO_(3)) = ("wt. of KCI")/( "mol. wt. of KCI")` Wt. of KCI ` = (0. 5358)/(122.5) xx 74.5 = 0.3260g"". . . (i)` In the secondreaction : the amount of `KCIO_(3)` left = 1 - 0.5358 = 0.4642 g We have, `KCIO_(3) to KCIO_(4) + KCI` 0.4642 g. Applying POAC for O atoms, moles of O in `KCIO_(3)` = moles ofO in ` KCIO_(4)` `3 xx ` moles of `KCIO_(3) = 4 xx ` molesof `KCIO_(4)` `3 xx ("wt. of " KCIO_(3))/("mol. wt. of "KCIO_(3))= 4 xx ("wt. of " KCIO_(4))/( "mol. wt.of " KCIO_(4))` Wt. of `KCIO_(4) = (3 xx 0.4642 xx 138.5)/( 122.5 xx 4)` = 0.3937 g. . . (ii) WT. of KCIproducedby second reaction = wt. of `KCIO_(3)` - wt. of `KCIO_4)` = 0.4642 - 0.3937 = 0.0705 g. . . (iii) Now since on heating `KCIO_(3),O_(2)` shall ESCAPE out, the substance asresidue are KCIproducedby THEREACTION (i) and (ii) and `KCIO_(4)` Wt. of residue = (i) + (ii) + (iii) = 0.3260 +0.3937 +0.0705 = 0.7902 g `:. %of KCIO_(4)` in the residue `= (0.3937)/(0.7902) xx 100` `= 49.8%`

Discussion

91811.	75% of a first-order reaction was completed in 32 minutes. When was 50% of the reaction completed
Answer» `24` min `4` min `16` min `8` min Solution :LET original amount, `N_(0) = 100` Since `75%` COMPLETED, so final amount `N = 100 - 75 = 25` As we know `(N_(0))/(N) = 2^(n)`,where `n =` no. of half LIVES or,`(100)/(25) = 2^(n)` or,`4 = 2^(n)` or, `2^(2)` or, `2^(2) = 2^(n)` `:. n=2` Since total time `= n xx t_(1//2)` `32` minutes `= 2 xx t_(1//2)` `:. t_(1//2) = 16` minutes

Discussion

91812.	A 1 g sample of H_(2)O_(2) solution containing x% H_(2)O_(2) by mass requires x cm^(2) of a KMnO_(4) solution for complete oxidation under acidic conditions. Calculate the normality of KMnO_(4) solution.
Answer» ANSWER :0.588 N

Discussion

91813.	75 % of a first order reaction was completed in 32 minutes when was 50% of the reaction complete……
Answer» 16 min 24 min 8 min 4 min Answer :A

Discussion

91814.	(A) 1 Avogram is equal to 1 amu. (R) Avogram is reciprocal of Avogadro's number.
Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). If both (A) and (R) are correct but (R) is not the correct explanation of (A). If (A) is correct but (R) is WRONG. If (A) is wrong but (R) is correct. Answer :C

Discussion

91815.	75% of a first order reaction was completed in 32 min . When would 50% of the reaction completed
Answer» 24 min 16 min 8 min 64 min Solution :The first order RATE constant is given as `k = (2.303)/(t) "log" (a_(0))/(a_(0) - x) ` _______(i) also , half life `t_(1//2) =(2.303 "log"2)/(K)` ______(ii) EQUATING K from equation (i) and (ii) `therefore (2.303"log"2)/(t_(1//2)) = (2.303)/(32) "log" (100)/(100-75)` or , `("log"2)/(t_(1//2)) = (1)/(32)` log 4 . or `("log"2)/(t_(1//2)) = (1)/(32) xx 2` log 2 `therefore t_(1//2) = 16 ` MINUTES

Discussion

91816.	A 1^(@) aliphatic nitro compound when treated with H_(2)SO_(4) gives
Answer» an aldehyde a carboxylic acid an alcohol None of these. Solution :`RCH_(2)NO_(2) OVERSET(H_(2)SO_(4))(rarr) R-overset(O)overset("\|\|")(C)-NHOH overset(H_(2)O)(rarr) R-overset(O)overset("\|\|")(C)-OH+NH_(2)OH`

Discussion

91817.	75% of a first order reaction is completed in 30 minutes . What is the time required for 93.75 % of the reaction (in minutes )
Answer» 45 120 90 60 Solution :`K = (2.303)/(t) "log" ((a)/(a-x)) , k = (2.303)/(30) "log" ((100)/(100-75))… (i)` `k = (2.303)/(t)"log" ((100)/(100-93.75))` PUT the value of k from above equation (i) we get the value of t THEREFORE `therefore` t = 60 min .

Discussion

91818.	A 1^(@) aliphatic nitro compound is first converted into carbanion salts with NaOH. When this salt is treated with H_(2)SO_(4) the product formed is
Answer» an aldehyde an alcohol a carboxylic acid None of these. Solution :`R-CH_(2)-NO_(2) overset(M^(+)OH^(-))(RARR) R-overset(-)(C)H-NO_(2)M^(+) overset(H_(2)SO_(4))(rarr) RCHO+MNO_(3)`

Discussion

91819.	A: 1^(@), 2^(@), 3^(@) Amines can be distinuished by diethyl oxalage. R: 1^(@) amines from N-alkyloxamide solid product 2^(@) amine form oxamic ester which is liquid 3^(@) amine do not react.
Answer» Both A and R are true and R is correct explanation of A Both A and R are true but R is not correct explanation of A A is true R is false. A is false, R is true Solution :Explanation:

Discussion

91820.	A 1. 0 g sample of Fe_(2)O_(3) solid of 55. 2 % purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100. 0 mL. An aliquot of 25. 0 mL of this solution requires 17. 0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.
Answer» Solution :Mass of `Fe_(2) O_(3) - ( 55.2) /( 100) xx 1 = 0.552 G ` moles of`Fe_(2)O_(3)= ( 0.552)/( 160) = 3.45 xx 10^(-3)` Equivalents of the oxidant `- ( n xx 0.0167 xx 17)/( 1000)- 2.84 xx 10^(-4)n ` (n is the 'n' factor of the oxidant ) Since on adding Zinc dust to the `Fe_(2) O_(3)` solution all the `Fe^(+3)` will become `Fe^(+2)` moles of `Fe^(2+)` in 100mL. `= 3.45 xx 10^(-3) xx 2 = 6.9 xx 10^(-3)` `:.` Equivalents of `Fe^(2+)` in the 25 mL , that is reacting with oxidant `= ( 6.9 xx 10^(-3))/(4)= 1.725 xx 10^(-3)` `:.` according to the Law of Equivalents `:. n = ( 1.72 xx 10^(-3))/(2.84 xx 10^(-4)) = 6.07 ~~6`

Discussion

91821.	75%. of a first order reaction is complete is 30 min. Calculate (a) half life, (b) rate constant and (c) time required for 99.9% completion of the reaction.
Answer» Solution :Time required for `75%` completion is 2 HALF - lives = 30 min. (a) Half - LIFE `(t_(1//2))=15` min (B) Rate CONSTANT, `k=(0.693)/(t_(1//2))=(0.693)/(15)=0.046" min"^(-1)` (or) `7.7xx10^(-4)s^(-1)` (c) Time required for `99.9%` of the reaction (t) `t=(2.303)/(k)"log"(a)/(a-x)=(2.303)/(0.046)"log"(100)/(0.1)=149` min

Discussion

91822.	A 0.6% urea solution would be isotonic with :
Answer» 0.1 M glucose 0.1 M KCl 0.6% glucose SOLUTION 0.6 % KCl solution Solution :ISOTONIC solutions are those which have same CONCENTRATION.

Discussion

91823.	75% of a first order reaction completed in 32 minutes then, the time required to complete 99.9% of a reaction in seconds is
Answer» 9600 96500 16 1660 Answer :A

Discussion

91824.	A 0.6% solution of urea ( molar mass =60) would be isotonic with
Answer» 0.1 M glucose 0.1 M KCl 0.6% glucose solution 0.6% NACL solution Solution :Isotonic solutions have same CONCENTRATION. Concentration of urea solution `=(0.6//60)/(100)xx1000=0.1 M` This concentration is same as 0.1 M glucose.

Discussion

91825.	75% of a first order process is completed in 30 min. The time required for 93.75% completion of same process (in hr) ?
Answer» 1 120 2 `0.25` ANSWER :A

Discussion

91826.	A 0.6% solution of urea (molecular mass =60) would be isotonic with
Answer» 0.1 M glucose 0.1 M KCl 0.6% glucose 0.6 % KCI Solution :0.6% UREA `SOL^(n)=0.6/60 XX 1/100 xx 1000 = 0.1 M.`

Discussion

91827.	75 cc of gas was collected over mercury in a tube closed a the top by a porous plug. On standing in air for some time, and when the mercury level became constant again, the volume was found to be 123 cc. What is the molecular weight of the gas ?"" (1 litre of air weighs 1.293 g at NTP)
Answer» Solution :Molecular weight of air = weight of 1 MOLE of air = WT. of 22.4 LITRES of air at NTP `= 1.293 xx 22.4` `= 28.96` From the GIVEN question it is clear that the time during which 75 cc of the gas diffuses out and the time during which 123 cc of air diffuses in are the same. Thus `(V_(1))/(V_(2)) = sqrt((M_(2))/(M_(1)))` `(75)/(123) = sqrt((28.96)/(M))` M = 78

Discussion

91828.	A 0.561 m solution of anunknown elecrolyte depresses the freezing point of water by 2.93^(@)C. What is van't Hoff factor for the electrooyte ? The freezing point depression constant (K_(f)) for water is 1.86 kg mol^(-1).
Answer» Solution :`i=(DeltaT_(f))/(K_(f)xxm)` `DeltaT_(f)=2.93^(@)C, 2.93 K, m = 0.561 " mol KG"^(-1)` `K_(f)=1.86^(@)C" kg mol"^(-1)=1.86" K kg mol"^(-1)` `i=((2.93K))/((1.86" K kg mol"^(-1))XX(0.561" mol kg"^(-1)))=2.81`

Discussion

91829.	7.5 A current is passed for 200 second through AgNO_(3) solution. If experimentally 1.08 g Ag is obtained then calculate the cell capacity. [Ag=108u]
Answer» ANSWER :0.6433

Discussion

91830.	A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water be 1.86^(@)C per mol, the mixture will freeze at
Answer» `0.93^(@)C` `-0.93^(@)C` `1.86^(@)C` `-1.86^(@)C` ANSWER :B

Discussion

91831.	74.5g of a metallic chloride contains 35.5 g of chlorine . The equivalent mass of the metal is:
Answer» 19.5 35.5 39 78 Answer :C

Discussion

91832.	A 0.5% aqueous solution of KCl was found to freeze at 272.76 K. Calcualte the van't Hoff factor and the degree of dissociation of KCl at this concentration K_(f)(H_(2)O)=1.86^(@)Cm^(-1)
Answer» ANSWER :`1.92,0.92`

Discussion

91833.	A 0.5 dm^3 flask contains gas A and 2 dm^3 flask contains gas B at the same temperature. If density of A = 4 g/dm^3, that of B = 2 g/dm^3 and molar mass of A is half of that of B, then the ratio of pressure exerted by gas is
Answer» `P_A/ P_B = 0.5` `P_A/ (P_B) = 2` `P_A/ (P_B) = 4` `P_A/ (P_B) = 0.25` ANSWER :C

Discussion

91834.	7.35 g of a dibasicacid was dissolved in water and diluted to 250 mL. 25 mL of this solution was neutralised by 15 mL of N NaOH solution . Calculateeq. wt and mol.wt of the acid .
Answer» Solution :Let the equivalent weight of the acid be E. Equivalent of acid ` = (7.35)/E "" ….(Eqn.4i)` m.e of the acid ` = (7.35)/E xx 1000 = 7350/E "" …(Eqn.3)` Now , 250 mL of the acid CONTAINS `7350/E xx 1000 =7350/E m.e ` ` :.25 mL ` of the acid contains `(735)/E ` m.e . Again, m.e of 25 mL of the acid = m.e of NAOH... (Eqn.2) ` 735/E = 1 xx 15 ` ` E = 735/15 = 49` ` :. ` eq. wt of acid= 49 . ` :. ` eq. wt of acid = 49 . ` :.` molecular weight of the acid = eq.wt `xx` basicity ` = 49 xx2 ` ` = 98 `

Discussion

91835.	72.5 g of C_6H_5OH (phenol) is dissolved in a solvent of K_f= 14. If the depression in freezing point is7K then find the % of phenol that dimerises [Give your answer divide by 25%]:-
Answer» ANSWER :C

Discussion

91836.	A 0.3 M HCl solution contains the following ions Hg^(++),Cd^(++),Sr^(++),Fe^(++). The addition of H_(2)S to above solution will precipitate
Answer» CD, Cu and Hg Cd, Fe and Sr Hg, Cu and Fe Cu, Sr and Fe Solution :The SECOND group radicals will precipitate because their solubility product is very LOW so SULPHATES will be precipitated.

Discussion

91837.	72.2gof C_(6)H_(5)OH (phenol) is dissolved in a solvent of K_(f)=14. If the depression in freezing point is 7K then find the % of phenol that dimerises [Give your answer divide by 25%]
Answer» SOLUTION :NA

Discussion

91838.	A 0.2m aqueous solution of KCl freezes at -0.68^(@)C .Calculate 'I' and the osmotic pressure at 0^(@)C. Assume the volume of solution to be that of pure H_(2)O and K_(f) for H_(2)O is 1.86
Answer» Solution :`(DeltaT_(f))_("normal")=K_(f)xxm`…………(Eqn 7) `=1.86xx0.2` `=0.372` We have `i=("observed colligative property")/("normal colligative property")`………..(Eqn 9) `=(0.68)/(0.372)=1.83` Again, `i=("observed osmotic pressure")/("normal osmatic pressure")` `:.` observed osmotic pressure `=ixx` normal osmotic pressure `=i xx c RT`............(Eqn 6) `=1.83xx0.2xx0.082xx273` `=8.2` ATM

Discussion

91839.	._(72)^(108)X overset(2 alpha)rarr overset(beta)rarr overset(gamma)rarr ._(Z)^(A)X'. Z and A are
Answer» 69172 172,69 180,70 182,68 Solution :`._(72)^(180)X OVERSET(2 ALPHA)rarr ._(68)^(172)P overset(beta)rarr ._(69)^(172)Q overset(gamma)rarr ._(69)^(172)X`

Discussion

91840.	A 0.288 g sample of an unknown monoprotic organic acid is dissolved in waterand titrated with a 0.115 M sodium hydroxide solution. After the addition of 17.54 inl. of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. (a) What is the molar mass of the organic acid? (b) What is the K_a value for the acid? The K_avalue could have been determined very easily if a pH measurement had been made after the addition of 16.92 mL of NaOH. Why?
Answer» SOLUTION :`(a) 74, (b) 1.3 xx 10^(-5)`

Discussion

91841.	718.2 g of sucrose is dissolved in 1kg of water . If 100+x is the difference between the nmelting and boiling point of the solutions, then report the value of 'x'.
Answer» ANSWER :5

Discussion

91842.	71 g of chlorine combines with a metal giving 111g of its chloride. The chloride is isomorphous with MgCl_(2):6H_(2)O. The atomic mass of the metal is:
Answer» 20 30 40 69 Answer :C

Discussion

91843.	T/FA 0.2 to 0.4 ppm solution of chlorine in water is a disinfectant
Answer» ANSWER :T

Discussion

91844.	A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):
Answer» `-0.45^(@)C` `-0.90^(@)C` `-0.31^(@)C` `-0.53^(@)C` SOLUTION :HX IONISES in solution `HXhArrH^(+)+X^(-)` `alpha=(i-1)/(n-1)or 0.2=(i-1)/(2-1)` `DeltaT_(f)=ixxK_(f)xxm` `=1.2 xx(1.86^(@)Cmol^(-1))xx(0.2 m)` =`0.45^(@)C` Freezing POINT of solution `=0^(@)C-45^(@)C=-45^(@)C`.

Discussion

91845.	70 torr will be same as which of the following
Answer» 9332.5 Pa 70 MM HG 0.0933 BAR All of these ANSWER :D

Discussion

91846.	A 0.239 g sample of unknown organic base is dissolved in water and titrated witha 0.135 M HCl solution. After the addition of 18.35 mL of acid, a pH of 10.73 is recorded. The equivalence point is reached when a total of 39.24 mL of HCl is added. The base and acid combine in a 1:1 ratio.(a) What is the molar mass of the organic base?(b) What is the K_bvalue for the base? The K_bvalue could have been determined very easily if a pH measurement had been made after the addition of 19.62 mL of HCl. Why?
Answer» SOLUTION :`(a) 45.12 (B) 4.72 XX 10^(-4)`

Discussion

91847.	70 gms of nitrogen gas was at 50 atm and 25^(@) C. (a) It was allowed to expand isothermally against a constant external pressure of one atmopshere. Calculate Deltau.Deltaq and Deltaw assuming that gas behaves ideally. (b) Also find out the maximum work that would be obtained if the gas expanded reversibly and isothermally to one atmosphere.
Answer» Solution :(a) AMOUNT of gas, `n=70/28 = 2.5` moles. INITIAL pressure, `P_(1) =50`atm Final pressure, `P_(2) = 1` atm Since the gas ideal, `DeltaU =0`, as the TEMPERATURE is constant. The work obtained `DeltaW = P_(2)(V_(2)-1V_(1)) rArr P_(2) [(NRT)/P_(2) -(nRT)/P_(1)]` `=(nRTP_(2)(P_(1)-P_(2)))/(P_(1)P_(2)) = (2.5 xx 2 xx 298 xx 1(50-1))/(50 xx 1) = 1.5` kcal. (b) the maximum work done for isothermal reversible expansion. `DeltaW = nRT ln P-(1)/P_(2) = 2.303 xx 2.5 xx 2 xx 298 log 50/1 = 5.8 kcal`

Discussion

91848.	A 0.2 g sample containing copper (II) was analysed iodometrically, where copper (II) is reduced to copper (I) by iodide ions. 2Cu^(2+) +4I^(-)rarr2 CuI+I_(2) If 20 mL of 0.1 M Na_(2)S_(2)O_(3) solution is required for titrationof the liberated iodine, then the percentage of copper in the sample will be :
Answer» `31.75%` `63.5%` `53%` `37%` ANSWER :B

Discussion

91849.	70% component of acid rain is
Answer» `HNO_3` `H_2SO_4` HCL `CO_2` ANSWER :B

Discussion

91850.	A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given K_(f)=1.86.^(@)CKgmol^-1 for water):
Answer» `-0.45^(@)C` `-0.90^(@)C` `-0.31^(@)C` `-0.53^(@)C` Solution :`DeltaT_(F)=0.2xx1.2xx1.86=0.45` therefore FREEZING POINT `=-0.45^(@)C`

Discussion

Explore topic-wise InterviewSolutions in .

(A) 1 mole H_(2)SO_(4)contains same mass of oxygen and sulphur. (R) 1 mole of H_(2)SO_(4) represents 98g mass.

(A) 1 lit of dry air contains 10ml of noble gas mixture(R) The atmosphere abundance of noble gases in dry air is nearly 1%

A 1 litre solution of pH=1 diluted upto 10 times.What volume of a solutions with pH =2 is to be added in diluted solution so that final pH remains '2'.

50% of a first order reaction was found to complete in 16 minute. When will 75% of the same reaction complete:

75% of the first order reaction is completed in 30 minutes. Calculate rate constant of the reaction.

A 1-g mixture of cuprous oxideand cupric oxide was quantitatively reducedto 0.839 g of metallic copper . Whatwas t he weightof cupricoxidein the original sample ? (Cu = 63.5, O = 16)

75% of a first order reaction was completed in 32 minutes, 50% of the reaction will be completed in

A 1 L of water sample has 0.1 g fluoride concentration. What is the concentration of fluoring in terms of ppm level ?

75 % of a first order reaction is completed in 30 minutes.What is the time require for 93.75% of the reaction (in minutes)…

75% of a first-order reaction was completed in 32 minutes. When was 50% of the reaction completed

A 1 g sample of H_(2)O_(2) solution containing x% H_(2)O_(2) by mass requires x cm^(2) of a KMnO_(4) solution for complete oxidation under acidic conditions. Calculate the normality of KMnO_(4) solution.

75 % of a first order reaction was completed in 32 minutes when was 50% of the reaction complete……

(A) 1 Avogram is equal to 1 amu. (R) Avogram is reciprocal of Avogadro's number.

75% of a first order reaction was completed in 32 min . When would 50% of the reaction completed

A 1^(@) aliphatic nitro compound when treated with H_(2)SO_(4) gives

75% of a first order reaction is completed in 30 minutes . What is the time required for 93.75 % of the reaction (in minutes )

A 1^(@) aliphatic nitro compound is first converted into carbanion salts with NaOH. When this salt is treated with H_(2)SO_(4) the product formed is

A: 1^(@), 2^(@), 3^(@) Amines can be distinuished by diethyl oxalage. R: 1^(@) amines from N-alkyloxamide solid product 2^(@) amine form oxamic ester which is liquid 3^(@) amine do not react.

75%. of a first order reaction is complete is 30 min. Calculate (a) half life, (b) rate constant and (c) time required for 99.9% completion of the reaction.

A 0.6% urea solution would be isotonic with :

75% of a first order reaction completed in 32 minutes then, the time required to complete 99.9% of a reaction in seconds is

A 0.6% solution of urea ( molar mass =60) would be isotonic with

75% of a first order process is completed in 30 min. The time required for 93.75% completion of same process (in hr) ?

A 0.6% solution of urea (molecular mass =60) would be isotonic with

75 cc of gas was collected over mercury in a tube closed a the top by a porous plug. On standing in air for some time, and when the mercury level became constant again, the volume was found to be 123 cc. What is the molecular weight of the gas ?"" (1 litre of air weighs 1.293 g at NTP)

A 0.561 m solution of anunknown elecrolyte depresses the freezing point of water by 2.93^(@)C. What is van't Hoff factor for the electrooyte ? The freezing point depression constant (K_(f)) for water is 1.86 kg mol^(-1).

7.5 A current is passed for 200 second through AgNO_(3) solution. If experimentally 1.08 g Ag is obtained then calculate the cell capacity. [Ag=108u]

A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water be 1.86^(@)C per mol, the mixture will freeze at

74.5g of a metallic chloride contains 35.5 g of chlorine . The equivalent mass of the metal is:

A 0.5% aqueous solution of KCl was found to freeze at 272.76 K. Calcualte the van't Hoff factor and the degree of dissociation of KCl at this concentration K_(f)(H_(2)O)=1.86^(@)Cm^(-1)

A 0.5 dm^3 flask contains gas A and 2 dm^3 flask contains gas B at the same temperature. If density of A = 4 g/dm^3, that of B = 2 g/dm^3 and molar mass of A is half of that of B, then the ratio of pressure exerted by gas is

7.35 g of a dibasicacid was dissolved in water and diluted to 250 mL. 25 mL of this solution was neutralised by 15 mL of N NaOH solution . Calculateeq. wt and mol.wt of the acid .

72.5 g of C_6H_5OH (phenol) is dissolved in a solvent of K_f​= 14. If the depression in freezing point is7K then find the % of phenol that dimerises [Give your answer divide by 25%]:-

A 0.3 M HCl solution contains the following ions Hg^(++),Cd^(++),Sr^(++),Fe^(++). The addition of H_(2)S to above solution will precipitate

72.2gof C_(6)H_(5)OH (phenol) is dissolved in a solvent of K_(f)=14. If the depression in freezing point is 7K then find the % of phenol that dimerises [Give your answer divide by 25%]

A 0.2m aqueous solution of KCl freezes at -0.68^(@)C .Calculate 'I' and the osmotic pressure at 0^(@)C. Assume the volume of solution to be that of pure H_(2)O and K_(f) for H_(2)O is 1.86

._(72)^(108)X overset(2 alpha)rarr overset(beta)rarr overset(gamma)rarr ._(Z)^(A)X'. Z and A are

718.2 g of sucrose is dissolved in 1kg of water . If 100+x is the difference between the nmelting and boiling point of the solutions, then report the value of 'x'.

71 g of chlorine combines with a metal giving 111g of its chloride. The chloride is isomorphous with MgCl_(2):6H_(2)O. The atomic mass of the metal is:

T/FA 0.2 to 0.4 ppm solution of chlorine in water is a disinfectant

A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):

70 torr will be same as which of the following

70% component of acid rain is

A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given K_(f)=1.86.^(@)CKgmol^-1 for water):

72.5 g of C_6H_5OH (phenol) is dissolved in a solvent of K_f= 14. If the depression in freezing point is7K then find the % of phenol that dimerises [Give your answer divide by 25%]:-