94420 + Interview Questions in Chemistry in Class 12 Page 1834 InterviewSolution

91651.	A + 2B+ 3C AB_2 C_3 Reaction of 6.0 g of A, 6.0 xx 10^(23)atoms of B, and 0.036 mol of C yields 4.8 g of compoundAB_(2)C_(3). If the atomic mass of A and C are 60 and 80 amu, respectively. The atomic mass of B is:(Avogadro number= 6 xx 10^(23))
Answer» 70 amu 60 amu 50 amu 40 amu Solution :`{:(,A,+,2B,+,3C,rarr,AB_(2)C_(3)),(t = 0,0.1,,1,,0.036,,0):}` C is limiting regent therefore MOLES of `AB_(2)C_(3)`will be 0.012. Mass of 0.012 mol `AB_(2)C_(3)`is 4.8 gm Mass of 1 mol `AB_(2)C_(3)`is `(4.8)/(0.012) = 400` Molar mass of`AB_(2)C_(3)`is 400 gm `A + 2B + 3C = 400 gm` `60 + 2B + 240 = 400 rArr B = 50` ATOMIC mass of B is 50u

Discussion

91652.	A 27^(@) one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atmto 10 atm , the value of DeltaE and q are (R = 2 cal )
Answer» `0, -965.84 cal` `-965.58 cal ,-865.58 cal ` `+ 865.58 cal , -865.58 cal ` `+965.84 cal ,+ 865.58 cal ` Solution :lsothermaly (At CONSTANTTEMPERATURE) and reversibly work is ` W= 2.303 n RTlog(P_(2)/(P_(1)))= 2.303 xx 1xx2 xx 300 log(10/2)` 965.84 at constant TEMPERATURE , `DeltaE=0` `DeltaE-Q+ W,q--W=-965.84 cal`

Discussion

91653.	A 25.0 mL sample of waste water is obtained to analyze for Pb^(2+) ions. This sample is evaporated to dryness and redissolved in 2.0 mL of H_(2)O, mixed with 2.0 mL of a buffer solution and 2.0 mL of a solution of dithizone then diluted to 10.0 mL. The absorbance of the coloured Pb^(2+) dithizone complex is compared with the Beer-Lambert plot below.The absorbance of a proton of the final solution is 0.13. What is the concentration of Pb^(2+) ions in the waste water in ppm ?
Answer» `2.9` `7.2` 18 36 Answer :A

Discussion

91654.	A 250.0 mL sample of a 0.20M Cr^(3+) is electrolysed with a current of 96.5 A. If the remaining [Cr^(3+)] is 0.1 M, the duration of process is:
Answer» 75 sec 150 sec 225 sec 25 sec Answer :A

Discussion

91655.	A 250-mL sampleof 0.20 M hydrochloric acid to be made by diluting the approximate amount of the concentrated reagent 11.7 M .What volume of the latter should be used ?
Answer» SOLUTION :`4.27` ML

Discussion

91656.	(A) 25% of the first order reaction will be completed in 50% of the half - life period. (R ) Half-life of first order reaction depends on concentration of reactant
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 :D

Discussion

91657.	A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. =45) is dissolved in 100 ml water and titrated with 0.5 M HCl when(1/5)^(th) of the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given: All data at 25^@C & log 2= 0.3. Select correct statement(s).
Answer» `K_b` of base is less than `10^(-6)` Concentration of SALT at equivalent POINT is 0.25 M VOLUME of HCl is used at equavalent point is 100 ml Answer :B::C

Discussion

91658.	A 22.4 litre flask contains 0.76 mm of ozone at 25^(@)C. Calculate: (i) the concentration of oxygen atoms needed so that the reaction O+O_(3)rarr 2O_(2) having rate constant equal to 1.5xx10^(7) litre mol^(-1) sec^(-1) can proceed with a rate of 0.15 mol litre^(-1) sec^(-1). (ii) the rate of formation of oxygen under this condition.
Answer» SOLUTION :N//A

Discussion

91659.	a) 24 g of a non-volatile, non-electrolyte solute is added to 600 g of water. The boiling point of the resulting solution is 373.35K. Calculate the molar mass of the solute ("Given boiling point of pure water = 373 K and Kb for water=0.52 K kg mol"^(-1)).
Answer» SOLUTION :GIVEN `w_(2)=24g` `w_(1)=600g` `DeltaT_(B)=0.35 K` `m_(2)=?` `"Molar mass of the SOLUTE, "M_(2)=(1000KbW_(2))/(DeltaTbxxw_(1))` `therefore M_(2)="59.42g mol"^(-1)`

Discussion

91660.	A 2.18 g sample contains a mixture of XO and X_(2)O_(3). It reacts with 0.015 moles of K_(2)Cr_(2)O_(7) to oxidize the sample completely toform XO_(4)^(-) " and " Cr^(3+) . If 0.0187 mole of XO_(4)^(-)is formed , what is the atomic mass of X ?
Answer» SOLUTION :`XO + K_(2)Cr_(2)O_(7) to Cr^(3+) + XO_(4)^(-)` `X_(2)O_(3) + K_(2)Cr_(2)O_(7) to Cr^(3+) + XO_(4)^(-)` Let, WT : of `K_(2)Cr_(2)O_(7) ` consumed by themixture `= 0.015 xx 6` Equivalents of `XO = x/(x + 16) xx 5` Equivalents of `X_(2)O_(3) = (2.18 -x)/(2x + 48) xx 8` ` :. x/(x + 16) xx 5 + (2.18 -x)/(2x - 14 8) xx 8 = 0.015 xx 6` Since 1 mole of XO gives 1 mole `XO_(4)^(-)` and 1 mole of `X_(2)O_(3)` gives 2 moles of `XO_(4)^(-)`, ` :. x/(x + 16) + (2x(2.18 -x))/(2x + 48) = 0.0187` On solving , `x = 99`

Discussion

91661.	(A) 2,2-Dimethylpropanal undergoes Cannizzaro reaction with conc. NaOH. (R) Cannizzaro reaction is a disproportionation reaction.
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) is false Answer :B

Discussion

91662.	A 200W, 110 V incadescent lamp was connected with an electrolytic cell containing ZnCl_(2) . What weight of metal will be deposited on passing current for 1 hour? (At. Wt. of Zn = 65.4)
Answer» Solution :Current strength in ampere = Watt/Volt = `200//110 = 1.818` Time = `1 xx 60 xx 60 = 3600 "sec"`, QUANTITY = `1.818 xx 3600 = 6545 ` coul. Zince is divalent. `193000` coul are required to deposite 65.4g of zinc. WEIGHT of metal deposite USING 6545 coud= `(6545 xx 65.4)/(193000) = 2.219 g` .

Discussion

91663.	A 200 W, 110 V in candescent lamp is connected in series with cells containing aqueous ZuCI . solution. What is the time required to deposit 1.109 g of metal ?
Answer» Solution :CURRENT strength in AMPERE `= "Watt/Volt"` `=200/110=1.818` Time `=1xx60xx60=3600` SEC, Quantity of electricity `=1.818xx3600` `=6545` coul. Zinc is divalent. 193000 coul are required to deposit 65.4 G of zinc. Weight of metal DEPOSITED using 6545 coul `=(6545xx65.4)/(193000)=2.219` g.

Discussion

91664.	A 200-watt, 110-volt incandescent lamp is connected in series with an electrolytic cell of negligible resistance containing a solution of zinc chloride. What weight of zinc will be deposited from the solution on passing the current for 30 minutes ? (Zn = 63.5, 1 faraday = 96500 coulombs)
Answer» ANSWER :1.075 G

Discussion

91665.	A 2000 gm sample of CaCO_(3) is 80 % pure. Find weight ( in gram ) of pure CaCO_(3)
Answer» ANSWER : `(##ALN_NC_CHM_MC_E01_023_A01##)` 1600 GM

Discussion

91666.	A 200 W bulb emits monochromatic light of wavelenght 1400 A and only 10%of the energy is emitted as light. The number of photons emitted by the bulb per second will be
Answer» `1.4` x `10^(18)` `1.4` x `10^(20)` `1.4`x`10^(19)` `1.4` x `10^(21)` ANSWER :C

Discussion

91667.	A 20.0 mL mixture of CO,CH_(4)and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cm^(3) . A further contraction of 14.0 cm^(3)occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
Answer» Solution :`CO(G) + 1/2 O_(2)(g) rarr CO_(2)(g)` `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(L)` 'X' is the VOLUME of CO and y is+ the volume of `CH_(4)` Thus, 1/2 x + y = 13........ (i) x + y =14........ (ii) By solving eq. (i) & (ii), we get x = 2 cc , y = 12 cc Thus, % `CH_(4) = 60 , % CO = 10 , % He = 30 `

Discussion

91668.	A 200 gm sample of CaCO_(3) having40 % purity is heated. Find moles of CO_(2) obtained.
Answer» ANSWER : `(##ALN_NC_CHM_MC_E01_025_A01##)` `0.8` MOLES

Discussion

91669.	A 20% (W/W) solution of NaOH is 5 M. The dens of the solution is
Answer» `1 G. ML^(-1)` `2 g . Ml^(-1)` `0.5 g.ml^(-1)` `0.25 g. ml^(-1)` ANSWER :A

Discussion

91670.	A 2.0 molal solution of sodium chloride in water cuses an elevation in boiling point of water by 1.88 K. What is the vlue of Van't Hoff factor ? What does does it signigy ? (K_(b) =0.52K kg mol^(-1)).
Answer» SOLUTION :`i=(DeltaT_(B))/(K_(f)K_(bxxm))` `DeltaT_(b)=1.88K, K_(b)=0.52" K kg mol"^(-1), m=2.0" MOLAL"=2.0" mol kg"^(-1)` `i=((1.88K))/((0.52" K kg mol"^(-1))xx(2.0"mol kg"^(-1)))=1.8076` `"The volue of signifies that when of sodium chloride DISSOCIATES in water, it furmnishes 1.8076 molr "Na^(+)" and "CI^(-)" ions".`

Discussion

91671.	A 20 mL,ureasolutionof 2%(w/V) is mixed with80mL of glucosesolution of 4%(w/V)at 300 K. Calculate the osmoticpressureof the solution .
Answer» 6.02 atm 1.642 atm 4.378 atm 3.01 atm Solution :` pi = w/(MV) RT ` ` = [(w/M)_("UREA") + (w/M)_("glucose")]xx (RT)/V ` Now, ` w_(urea) "in"20 mL= (2 xx 20)/(100) = 0.4 G ` `w_(" glucose") "in" 80 mL = ( 4 xx 80)/100 = 3.2 g ` ` therefore= [ (0.4)/60 + (3.2)/(180)]xx ( 0.0821 xx 300 xx 1000)/(100) ` ` (thereforeV = 20 + 80 = 100 mL= 100/1000 L) ` ` pi = 6.02 atm `

Discussion

91672.	A 20 ml solution containing 0.2 g impureH_(2)O_(2)reacts completely with 0.316 g ofKMnO_(4)in acidic medium. What would be volume strength ofH_(2)O_(2)solution? [Given: Molar mass ofH_(2)O_(2)=34 & KMnO_(4)= 158 g/mol]
Answer» Solution :Meq of `H_(2)O_(2)=` Meq of `KMnO_(4)` `(W xx1000)/(17)=(0.316)/(158//5)xx1000` `W_(H_(2)O_(2))=0.18 gm`. Normality of `H_(2)O_(2)=(w)/(E )xx(1000)/("Volume (in ML)")=(0.17)/(17)xx(1000)/(20)=0.5` Volume strength `= N xx5.6=0.5xx5.6=2.8`

Discussion

91673.	A 20 litre container at 400 K contains CO_(2)(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of slid SrO). The volume of the container is now decreased by moving the movable pistonfitted in the container . The maximum volume of the container, when pressure of CO_(2) attains its maximum value, will be [Given that : SrCO_(3)(s) hArr SrO(s) +CO_(2)(g), Kp=1.6 atm]
Answer» 10 litre 4 litres 2 litre 5 litre Solution :`SrCO_(3)hArrSrO(s)+CO_(2)(g)(s) ` `K_(p)= P_(CO_(2))` maximum pressure of `CO_(2)= 1.6 ATM` `P_(1)V_(1)=P_(2)V_(2)` `0.4 xx 20 = 1.6 V_(2)IMPLIES V_(2)= 5L`.

Discussion

91674.	""_(90)^(234)Th disintegrates to give ""_(82)^(206)Pb as the final product. How many alpha- and beta-particles are emitted during the process?
Answer» Solution :Since with the emission of an `ALPHA`-particle, the mass number of the product DECREASES by four units. Thus, number of `alpha`-particles emitted `=(234-206)/(4)=7` Now with the emission of 7 `alpha`-particles, the atomic number decreases by 14 units, i.e., the resulting atomic number will be `90-14=76`. But since Pb has an atomic number of 82, therefore, `6 BETA`-particles will be emitted in order to increase the atomic number from 76 to 82. Thus `""_(90)^(234)Th underset(6beta)overset(7alpha)rarr ""_(82)^(206)Pb`

Discussion

91675.	A 20 L flask contains 4.0 gm of O_(2) & 0.6 gm of H_(2) at 100^(@)C. If the contents are allowed to react to form water vapors at 100^(@)C,find the contents of flask and there partial pressures.
Answer» Solution : `H_(2)` reacts with `O_(2)` to form water `[H_(2)O_((g))]` `2H_(2(g))+O_(2(g)) rarr 2H_(2)o_((g))` `rArr 2` MOLES of `H_(2) = ` 1 mole of `O_(2)` = 2 moles of `H_(2)O` Here masses of `H_(2)` and `O_(2)` are given, so one of them can be in excess. So first check out which of the reactants is in excess. Now, Moles of `O_(2)= 4//32 = 0.125` and Moles of `H_(2) = 0.6 //2 = 0.3` SINCE 1 mole of `O_(2)= 2 `moles of `H_(2)` `rArr 0.125` moles of `O_(2) -= 2 xx 0.12 5 ` moles of `H_(2)` i.e., 0.25 moles of `H_(2)` are USED , so `O_(2)` reacts completelywhereas `H_(2)` is in excess. `rArr `Moles of `H_(2)` in excess `= 0.3 - 0.25 = 0.05` moles. Also, 2 moles of `H_(2) -= 2` moles of `H_(2)O` `rArr 0.25` moles of `H_(2) -= 0.25` moles of `H_(2)O` are produced. `rArr` Total moles after the reaction =0.05 ( moles of `H_(2) + 0.25` ( moles of `H_(2) O)=0.3` `rArr ` The total PRESSURE `P_("Total")` at the end of reaction is given by `:` `P_("Total")= (nRT)/( V) = (0.3 xx 0.0821 xx 373)/( 20) = 0.459 atm` Now partial pressure of A = mole FRACTION of `A xx P_("total")` `rArr P_(H_(2)) = 0.05 // 0.3 xx 0.459 = 0.076 ` atm `rArr P_(H_(2)O) = 0.25 // 0.3 xx 0.459 =0.383 ` atm

Discussion

91676.	""_(90)^(232)Th disintegrates to ""_(y)^(x)Pb by emitting six alpha-and four beta-particles. Find x and y
Answer» ANSWER :208, 82

Discussion

91677.	A 20 gm mixture of Ca(OH)_(2) and CaCl_(2) require 50ml 2M HCl for complete reaction Then what will be the mass % of Ca(OH)_(2)
Answer» 0.037 0.185 0.37 0.0925 Solution :`n_(Cal(OH)_(2))=a n_(CaCl_(2))=B` `74a+111b=20`. . . .(i) `Ca(OH)_(2)+2HClrarrCaCl_(2)+2H_(2)O` `n_(Ca(OH)_(2))=a=(n_(HCL))/(2)=(50xx2)/(1000xx2)=(1)/(20)=(0.1)/(2)` `W_(Ca(OH)_(2))=(7.4)/(2)gm=3.7` MASS `% Ca(OH)_(2))=(3.7)/(20)xx100=18.5%`

Discussion

91678.	900 ml of pure & dry O_2 is subjected to O_2silent electric discharge, so that after a time 10 min. volume of ozonized oxygen is found to be 870 ml. Now, average rate of reaction in this interval is (in ml/min)
Answer» 3 9 90 60 Answer :B

Discussion

91679.	A 2.0-g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO_2 ceases. The volume of CO_2at 750 mmHg pressure and at 298 K is measured to be 123.9 mL. 1.5 g of the same sample requires 150 mL of (M/10) HCl for complete neutralization. Calculate the percentage composition of the components of the mixture.
Answer» SOLUTION :26.5%, 42.0%, 31.5%

Discussion

91680.	A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO_(2) cases. The volume of CO_(2) at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5 g of the same sample required 150 ml of M/10 HCl for complete neutralization. Calculate the percentage composition of the mixture.
Answer» SOLUTION :Suppose `Na_(2)CO_(3)=xg, NaHCO_(3)=yg` Then `Na_(2)SO_(4)=2-(x+y)g` On heating only `NaHCO_(3)` will decompose to give `CO_(2)` as follows : `underset(2(84)g)(2NaHCO_(3))rarrNa_(2)CO_(3)+H_(2)O+underset("22400 ml at STP")(CO_(2))` `"y g NaHCO"_(3)" will give CO"_(2)=(22400)/(168)xx"y ml at STP"` Actual `CO_(2)` produced at STP MAY be calculated as follows : `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(760xxV_(1))/(273)=(750xx123.9)/(298),V_(1)=112.0ml` HENCE, `(22400)/(160)y=112 or y=0.84g` `"1.5 g of the mixture requires M/10 HCl = 150 ml"` `therefore "2.0 g of the mixture will require M/10 HCl"=(150)/(1.5)xx2.0="200 ml = 0.02 mole HCl"` `underset(106G)(Na_(2)CO_(3))+underset("2 moles")(2HCl)rarr2NaCl+H_(2)O+CO_(2)` `underset(84g)(NaHCO_(3))+underset("1 mole")(HCl)rarr NaCl+H_(2)O+CO_(2)` `xg Na_(2)CO_(3)" required HCl"=(2)/(106)xx"x moles"` `"0.84 g "NaHCO_(3)" require HCl"=(1)/(84)xx"0.84 mole = 0.01 mole"` `"Hence,"(2x)/(106)+0.01=0.02 or x=0.53g` `%" of "Na_(2)CO_(3)=(0.53)/(2)xx100=26.5%` `%" of "NaHCO_(3)=(0.84)/(2)xx100=42.0%` `%" of "Na_(2)SO_(4)=100-(26.5+42.0)=31.5%`

Discussion

91681.	90% of a first order reaction was completed in 3 min. When (in minutes) will99.9% of the reaction be complete?
Answer» ANSWER :9

Discussion

91682.	A 20 C,C_(4) mixture of CO_(1) CH_(4) and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The decrease in volume is found to be 13 c.c. A further contraction of 14 c.c. occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
Answer» SOLUTION :50% 20% 30%

Discussion

91683.	A 20 mL mixture of CO, CH_4, and Helium (He) gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13 mL. A further contraction of 14 mL occurs when the residual gas is treated wityh KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
Answer» SOLUTION :50% 20%, 30%

Discussion

91684.	90% of first order process X to Y is completed in a time equals to 99% of another first order process Y to Q. If K value of Yto Q is 0.09 sec^(-1), K value of X toY will be
Answer» 0.27 0.3 0.03 0.045 Answer :D

Discussion

91685.	A 2 litre container contains 4 moles ofN_(2)O_(5). On heating to 100^(@)C, N_(2)O_(5) undergoes complete dissociation to NO_(2) and O_(2). If rate constant for decomposition of N_(2)O_(5) is 6.2xx10^(-4)"sec"^(-1) select the correct statemertns:
Answer» The MOLE ratio before and after dissociation is 4:2 The time REQUIRED to complete 40% of reaction is 824 sec `t_(1//2)` of `N_(2)O_(5)` is 1117.7 sec and it is independent of temperature. If volume of container is doubled, the rate of decomposition becomes halfof the initial rate Solution :`4"mole" ""0""0` `0"8""2` a. `4:10=2:5` B. `Kxxt=2.303log((C_(0))/(C_(t)))` `6.2xx10^(-4)t=2.303log(100/60),t=(2.303xx0.2219)/(6.2xx10^(-4))=0.0824xx10^(4)=824` sec c. `(t_(1/2))=0.693//(6.2xx10^(-4))=0.11177xx10^(4)=1117.7`

Discussion

91686.	90 mL of oxygen is required for complete combustion of unsaturated 20mL gaseous hydrocarbon, hydrocarbon is ?
Answer» Solution :`("Volume of Hydrocarbon")/("Volume of" O_(2))=(2)/(3n)` (for Alkene) `("Volume of Hydrocarbon")/("Volume of" O_(2))=(2)/(3n-1)` (for alkyne) By PUTTING the values in above fomulae we can find the hydrocarbon for which `n` is natural number. ` (20)/(90)=(2)/(3n) n=3` So hydrocarbon is Propene `[C_(3)H_(6)]`.

Discussion

91687.	9.0 g of H_2 O is vaporized at 100^@C and 1 atm pressure. If the latent heat of vaporization of water is x J/g , then Delta S is given by ______ .
Answer» `x/373` `(18X)/100` `(18x)/373 ` `1/2 XX(18x)/373` ANSWER :D

Discussion

91688.	A 2 - g samplecontainingNa_(2) CO_(3) and NaHCO_(3) loses 0 . 248 g when heated to 300^(@) C, the temperature atwhich NaHCO_(3) decomposesto Na_(2) CO_(3) , CO_(2) and H_(2) O . Whatis thepercentage of Na_(2) CO_(3) in thegiven mixture ?(Na = 23, C = 12 , O = 16 and H = 1 )
Answer» Solution :On heating `Na_(2) CO_(3)and NaHCO_(3), Na_(2)CO_(3)`remains unchangedwhile `NaHCO_(3)`changesinto `Na_(2) CO_(3) , CO_(2) and H_(2) O`. The loss in weightis due to remove of `CO_(2) and H_(2)O` which escape out on heating . `:.`WT. of `NaHCO_(3) = (2 . 00 - x) g` Since `Na_(2) CO_(3)`in the products contains x g of unchanged reactant `Na_(2) CO_(3)`and rest produced from `NaHCO_(3)` the wt.of `Na_(2) CO_(3)`produced by `NaHCO_(3)`only= (1.752- x) g Now, we have , ` {:(NaHCO_(3) to Na_(2)CO_(3) + (H_(2)O + CO_(2) )uarr),("(2 . 0 - x) g(1 . 752 - x) g "):}` Applying POAC for Na atoms, moles of Na in ` NaHCO_(3) ` = MOLESOF Na in `Na_(2)CO_(3)` ` 1 xx ` moles of `NaHCO_(3) = 2xx ` moles of `Na_(2) CO_(3)` `(2. 0 - x)/( 84) = 2 xx (1 . 752 - x)/( 106) [{:(NaHCO_(3) = 84 ),(Na_(2) CO_(3) = 106):}]` `x = (82)/( 62) = 1.328 g` `:. % of Na_(2) CO_(3) = (1.328)/( 2 . 0) xx 100 = 664 %`

Discussion

91689.	90 gm mixture of H_(2) and O_(2)is taken In stoichiometric ratio and gives H_(2)O with 50% yeild. The produced mass of H_(2)O (in gm) is :
Answer» 45 gm 36 gm 20 gm 90 gm Answer :A

Discussion

91690.	A + 2 B to C + D .if - (d [A])/(dt ) = 5 xx 10^(-4) mol l^(-1) s^(-1) , then - (d[B])/(dt) is
Answer» `2.5 XX 10^(-4)` mol `l^(-1) s^(-1)` `5.0 xx 10^(-4)` mol `l^(-1) s^(-1)` `2.5 xx 10^(-3)` mol `l^(-1) s^(-1)` `1.0xx 10^(-3)` mol `l^(-1) s^(-1)` Solution :` A + 2 B to C + D` `(-d[A])/(dt) = 5 xx 10^(-4) implies (-1)/(2) (d[B])/(dt) = 5 xx 10^(-4)` `(d[B])/(dt) = 10 xx 10^(-4) = 10^(-3)`

Discussion

91691.	9 volumes of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion yielded on burning 4 volumes of CO_(2). 6 volumes of water vapour and 2 volumes of N_(2), all volumes measured at the same temperature and pressure. If the compound has only C, H and N, What is the molecular formula of the compound A?
Answer» SOLUTION :(7 vol, `C_(2)H_(6)N_(2)`)

Discussion

91692.	"A "+2" B" to 3" C"+2" D". The rate of disappearance of B is 1xx10^(-2)" mol lit"^(-1)"sec"^(-1). What will be (i) Rate of the reaction (ii) Rate of change in concentration of A and C ?
Answer» ANSWER :(i) `0.5xx10^(-2)" mol LIT"^(-1)"SEC"^(-1)` (ii) `-(d[A])/(dt)=0.5xx10^(-2)" mol lit"^(-1)"sec"^(-1),+(d[C])/(dt)=1.5xx10^(-2)" mol lit"^(-1)"sec"^(-1)`

Discussion

91693.	A 1st order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete 80%.
Answer» SOLUTION :`K=2.303/t"LOG"[A]_0/[A]=2.303/20log)[A]_0/[(80/100)[A]_0] =2.303/20"log"(5/4)=2.303/20xx0.0969=0.011min^-1` Now ,`t=2.303/20"log"[A]_0/[A]impliest=2.303/0.011xxlog5=2.303/0.011xx0.699=144.2min.`

Discussion

91694.	""_(8)^(x)X atom is isotone to ""_(9)^(17)Y atom. The value of x is
Answer» 8 16 9 17 Answer :B

Discussion

91695.	A 1m solution can be more concentrated than a 1M solution if
Answer» DENSITY of SOLVENT is more than 1 density of solvent is equal to 1 density of solvent is less than 1 it is never possible Solution :A level information

Discussion

91696.	(A) 1g O_(2) and 1 g O_(3) have equal number of atoms (R) Mass of 1 mole atom is equal to its gram-atomic 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 :B

Discussion

91697.	8g of the radioactive isotope, cesium -137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 g of it remained. So the half-life period of the isotope is
Answer» 37.5 days 30 days 23 days 50 days Solution :t = Feb 1 to July 1 `= 28 + 31 + 30 + 31 + 30 = 150` days `lamda = (2.303)/(150) "log" (8)/(0.25) = (2.303)/(150) log 2^(5) = (0.693)/(30) "DAY"^(-1)` `t_(1//2) = (0.693)/(0.693//30)` = 30 days.

Discussion

91698.	A 15% (V/V) solution of ethylene glycol, an antifreeze is used in cars for cooling the engine.
Answer» SOLUTION :FALSE, it is 35%

Discussion

91699.	A 1.458 g of Mg reacts with 80.0 ml of a HCl solution whose pH is -0.477.The change in pH when all Mg has reacted.(Assume constant volume. Mg=24.3g//"mol".)(log 3=0.47, log2= 0.301)
Answer» -0.176 `+0.477` -0.2345 0.3 Solution :`MG(aq)+2HCL(aq)toMgCl_2(aq)+H_2` `(1.458 g)/(24.3 g)` millimoles of HCl =`3xx80=240` millimoles millimoles of HCl after REACTION =240-60x2=120 New Molarity `=120/80=1.5 M` `pH=-log[H^+]=-log1.5=-0.176` CHANGE is pH=-0.176-(-0.477)=0.3

Discussion

91700.	8g of oxygen has the same number of molecules as in:
Answer» 11g `CO_(2)` 22g `CO_(2)` 7g CO 14g CO Answer :A::C::D

Discussion

Explore topic-wise InterviewSolutions in .

A + 2B+ 3C AB_2 C_3 Reaction of 6.0 g of A, 6.0 xx 10^(23)atoms of B, and 0.036 mol of C yields 4.8 g of compoundAB_(2)C_(3). If the atomic mass of A and C are 60 and 80 amu, respectively. The atomic mass of B is:(Avogadro number= 6 xx 10^(23))

A 27^(@) one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atmto 10 atm , the value of DeltaE and q are (R = 2 cal )

A 250.0 mL sample of a 0.20M Cr^(3+) is electrolysed with a current of 96.5 A. If the remaining [Cr^(3+)] is 0.1 M, the duration of process is:

A 250-mL sampleof 0.20 M hydrochloric acid to be made by diluting the approximate amount of the concentrated reagent 11.7 M .What volume of the latter should be used ?

(A) 25% of the first order reaction will be completed in 50% of the half - life period. (R ) Half-life of first order reaction depends on concentration of reactant

a) 24 g of a non-volatile, non-electrolyte solute is added to 600 g of water. The boiling point of the resulting solution is 373.35K. Calculate the molar mass of the solute ("Given boiling point of pure water = 373 K and Kb for water=0.52 K kg mol"^(-1)).

A 2.18 g sample contains a mixture of XO and X_(2)O_(3). It reacts with 0.015 moles of K_(2)Cr_(2)O_(7) to oxidize the sample completely toform XO_(4)^(-) " and " Cr^(3+) . If 0.0187 mole of XO_(4)^(-)is formed , what is the atomic mass of X ?

(A) 2,2-Dimethylpropanal undergoes Cannizzaro reaction with conc. NaOH. (R) Cannizzaro reaction is a disproportionation reaction.

A 200W, 110 V incadescent lamp was connected with an electrolytic cell containing ZnCl_(2) . What weight of metal will be deposited on passing current for 1 hour? (At. Wt. of Zn = 65.4)

A 200 W, 110 V in candescent lamp is connected in series with cells containing aqueous ZuCI . solution. What is the time required to deposit 1.109 g of metal ?

A 200-watt, 110-volt incandescent lamp is connected in series with an electrolytic cell of negligible resistance containing a solution of zinc chloride. What weight of zinc will be deposited from the solution on passing the current for 30 minutes ? (Zn = 63.5, 1 faraday = 96500 coulombs)

A 2000 gm sample of CaCO_(3) is 80 % pure. Find weight ( in gram ) of pure CaCO_(3)

A 200 W bulb emits monochromatic light of wavelenght 1400 A and only 10%of the energy is emitted as light. The number of photons emitted by the bulb per second will be

A 200 gm sample of CaCO_(3) having40 % purity is heated. Find moles of CO_(2) obtained.

A 20% (W/W) solution of NaOH is 5 M. The dens of the solution is

A 2.0 molal solution of sodium chloride in water cuses an elevation in boiling point of water by 1.88 K. What is the vlue of Van't Hoff factor ? What does does it signigy ? (K_(b) =0.52K kg mol^(-1)).

A 20 mL,ureasolutionof 2%(w/V) is mixed with80mL of glucosesolution of 4%(w/V)at 300 K. Calculate the osmoticpressureof the solution .

A 20 ml solution containing 0.2 g impureH_(2)O_(2)reacts completely with 0.316 g ofKMnO_(4)in acidic medium. What would be volume strength ofH_(2)O_(2)solution? [Given: Molar mass ofH_(2)O_(2)=34 & KMnO_(4)= 158 g/mol]

""_(90)^(234)Th disintegrates to give ""_(82)^(206)Pb as the final product. How many alpha- and beta-particles are emitted during the process?

A 20 L flask contains 4.0 gm of O_(2) & 0.6 gm of H_(2) at 100^(@)C. If the contents are allowed to react to form water vapors at 100^(@)C,find the contents of flask and there partial pressures.

""_(90)^(232)Th disintegrates to ""_(y)^(x)Pb by emitting six alpha-and four beta-particles. Find x and y

A 20 gm mixture of Ca(OH)_(2) and CaCl_(2) require 50ml 2M HCl for complete reaction Then what will be the mass % of Ca(OH)_(2)

900 ml of pure & dry O_2 is subjected to O_2silent electric discharge, so that after a time 10 min. volume of ozonized oxygen is found to be 870 ml. Now, average rate of reaction in this interval is (in ml/min)

90% of a first order reaction was completed in 3 min. When (in minutes) will99.9% of the reaction be complete?

90% of first order process X to Y is completed in a time equals to 99% of another first order process Y to Q. If K value of Yto Q is 0.09 sec^(-1), K value of X toY will be

A 2 litre container contains 4 moles ofN_(2)O_(5). On heating to 100^(@)C, N_(2)O_(5) undergoes complete dissociation to NO_(2) and O_(2). If rate constant for decomposition of N_(2)O_(5) is 6.2xx10^(-4)"sec"^(-1) select the correct statemertns:

90 mL of oxygen is required for complete combustion of unsaturated 20mL gaseous hydrocarbon, hydrocarbon is ?

9.0 g of H_2 O is vaporized at 100^@C and 1 atm pressure. If the latent heat of vaporization of water is x J/g , then Delta S is given by ______ .

A 2 - g samplecontainingNa_(2) CO_(3) and NaHCO_(3) loses 0 . 248 g when heated to 300^(@) C, the temperature atwhich NaHCO_(3) decomposesto Na_(2) CO_(3) , CO_(2) and H_(2) O . Whatis thepercentage of Na_(2) CO_(3) in thegiven mixture ?(Na = 23, C = 12 , O = 16 and H = 1 )

90 gm mixture of H_(2) and O_(2)is taken In stoichiometric ratio and gives H_(2)O with 50% yeild. The produced mass of H_(2)O (in gm) is :

A + 2 B to C + D .if - (d [A])/(dt ) = 5 xx 10^(-4) mol l^(-1) s^(-1) , then - (d[B])/(dt) is

"A "+2" B" to 3" C"+2" D". The rate of disappearance of B is 1xx10^(-2)" mol lit"^(-1)"sec"^(-1). What will be (i) Rate of the reaction (ii) Rate of change in concentration of A and C ?

A 1st order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete 80%.

""_(8)^(x)X atom is isotone to ""_(9)^(17)Y atom. The value of x is

A 1m solution can be more concentrated than a 1M solution if

(A) 1g O_(2) and 1 g O_(3) have equal number of atoms (R) Mass of 1 mole atom is equal to its gram-atomic mass.

8g of the radioactive isotope, cesium -137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 g of it remained. So the half-life period of the isotope is

A 15% (V/V) solution of ethylene glycol, an antifreeze is used in cars for cooling the engine.

A 1.458 g of Mg reacts with 80.0 ml of a HCl solution whose pH is -0.477.The change in pH when all Mg has reacted.(Assume constant volume. Mg=24.3g//"mol".)(log 3=0.47, log2= 0.301)

8g of oxygen has the same number of molecules as in: