52671 + Interview Questions in Chemistry in Class 11 Page 1030 InterviewSolution

51451.	2 litres of nitrogen completely react with 6 litres of hydrogen under suitable conditions to form 4 litres of NH_3. If all the volumes are measured at the same temperature and pressure, show that the data given are in accordance to the Gay Lussac's law of combining volumes.
Answer» Solution :The reaction mentioned in the question takes place as follows : `underset("2 litres")((N_(2)(g)) +underset("6 litres")(3H_(2)(g)) to underset("4 litres")(2NH_(3)(g))` The volumes of the reacting gases are in the ratio 2 : 6, i.e., 1: 3. The product ammonia is ALSO a gas and its VOLUME also bears a simple whole number ratio with the volumes of reactants. The ratio of the volumes of all reactants and product in this reaction is thus 1:3:2 Since, the ratio is a simple whole number ratio, the data are in ACCORDANCE to the Gay Lussac.s law of combining volumes.

Discussion

51452.	2 L of hydrogen and 2.5 L of chlorine are allowed react in diffused light. Write the volume composition of the component gases after reaction.
Answer» Solution :Reaction between and chlorine leading to the production of hydrogen chloride is given as `H_(2)+Cl_(2)rarr2HCl` `{:("1 vol","1 vol","2 vol","Gay - Lussac.s volumes"),(2,2.5L,0L,"at START"),(0,0.5L,4L,"after the reaction"):}` The volume of the mixture is 4. 5L Volume of unreacted `Cl_(2)=0.5L` Volume of `HCl` PRODUCED = 4 L

Discussion

51453.	2-iodo-2-methylpropane belongs to …………………. type.
Answer» SOLUTION :`3^@`HALOALKANES

Discussion

51454.	2-Hexyne gives trans-2-hexene on treatment with
Answer» `Li//NH_3` `Pd//BaSO_4` `LiAlH_4` `Pt//H_2` SOLUTION :Na/liq. `NH_3` or Li/Liq. `NH_3` GIVES trans-products.

Discussion

51455.	2-hexyne give trans-2-hexene on treatment with
Answer» `Li//NH_3` `Pd//BasO_4` `LiAlH_4` `Pt//H_2` Solution :Alkynes on TREATMENT with ALKALIMETALS in liquid ammonia GIVES TRANS hydrogenation product.

Discussion

51456.	2-Hexyne gives trans-2-Hexene on treatment with :
Answer» `Pt//H_(2)` `Li//NH_(3)` `Pd//BaSO_(4)` `Li//AlH_(4)` ANSWER :B

Discussion

51457.	2 grams of Helium diffuses from a porous plate in 4min. How many grams of CH_4 would diffuse through the same plate in same time under similar conditions?
Answer» 4 G 16 g 8 g 2 g Solution :`(w_1)/(t_1) xx (t_2)/(w_2) = sqrt((M_1)/(M_2)) IMPLIES 2/(w_2) = sqrt((4)/(16))`. `W_(CH_4) = 4 gm`.

Discussion

51458.	2 grams of a mixture of sodium carbonate and sodium bicarbonate on strong heating liberated carbondioxide, which occupie. 0.224 L at STP. What is the weight of Na_(2)CO_(3) in the original mixture?
Answer» ANSWER :0.32g

Discussion

51459.	2gm of hydrogen is present in a closed vessel at S.T.P. If the same quantity of another gas 'X' when introduced into the vessel the pressure becomes 1.5 atm. The gas 'X' would be
Answer» `CH_(4)` `SO_(2)` He `O_(2)` ANSWER :C

Discussion

51460.	2 g of NaOH per 250 ml of solution is added to a buffer solution of buffer capacity 0.2 . Then the change in pH is
Answer» `0.5` `1` `1.5` `2.0` ANSWER :B

Discussion

51461.	2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields : (a)C_2H_5CH_2undersetunderset(CH_3)(\|)oversetoverset(CH_3)(\|)C-OCH_3 (b)C_2H_5CH_2undersetunderset(CH_3)(\|)C=CH_2 (c )C_2H_5CH=undersetunderset(CH_3)(\|)C-CH_3
Answer» (a) and (C ) (c ) only (a) and (B) All of these ANSWER :D

Discussion

51462.	2- butyne on chlorination gives ………………..
Answer» 1-chlorobutane 1,2-dichlorobutane 1,1,2,2-tetrachlorobutane 2,2,3,3 tetrachklorobutane ANSWER :d

Discussion

51463.	2-butyne converts in to dimethyl glyoxal by
Answer» BAEYER's REAGENT `SeO_2` REDUCTIVE OZONOLYSIS All of these SOLUTION :Baeyer reagent , `SeO_2` , reductive ozonolysis

Discussion

51464.	2-butene is an example of .............. compound.
Answer» SOLUTION :UNSATURATED OPEN CHAIN

Discussion

51465.	2-Butanone is heated with PCl_5 and the product so obtained is heated with alcoholic KOH , the final product obtained is
Answer» 1-BUTYNE 2-butyne 1,2-butandiene 1,3-butadiene Solution :`UNDERSET("2- Butanone")(CH_3CH_2COCH_3)underset(-POCl_3)OVERSET(PCl_5)to CH_3CH_2 - C Cl_2 CH_3 underset(-2HCl)overset(Alc KOH, Delta)to underset("2 - Butyne")(CH_3C =-= C CH_3`

Discussion

51466.	2-Bromopentane is treated with alcoholic KOH solution. Whatt will be the major product formed in this reaction and what is the type of elimination called?
Answer» Pent-1-ene, `beta`-Elimination<BR>Pent-2-ene, `beta`-elimination Pent-1-ene, Nucleophilic substitution Pent-2-ene, Nucleophilic substitution. Solution :`underset("2-Bromopentane")(CH_(3)CH_(2)-CH_(2)-) underset(Br)underset(\|)(C)H-CH_(3) underset(-HBR" "(beta-"elimination"))overset("alc. KOH")to CH_(3)-underset("2-Pentene")(CH_(2)-CH=CH-CH_(3))`.

Discussion

51467.	2-Acetoxy benzoic acid is used as an:
Answer» antimalarial antidepressant antiseptic antipyretic ANSWER :D

Discussion

51468.	2, 3-pentadiene is optically active since it
Answer» CONTAINS one chiral carbon atom contains TWO chiral ATOMS but the molecule as a WHOLE is achiral does not contain any chiral carbon atom but the molecule as a whole is chiral none of the above Solution :does not contain any chiral carbon atom but the molecule as a whole is chiral

Discussion

51469.	1^(ul@) alcohols are poor starting material for synthesis of 1-Alkene. Explain ?
Answer» Answer :`1^@` carbocation `R-overset(o+)CH_2` would REARRANGE and 2-alkene would RESULT. Even if 1-alkene is ALSO formed but , it would TEND to rearrange in ACIDIC medium to 2-alkene.

Discussion

51470.	1^(st) law does not tell about
Answer» LAW of conservation of energy Workdone whether +ve (or ) -ve Feasibility of a process `Delta E` at CONST T Solution :`1^(st)` law does not give feasibility

Discussion

51471.	1n which of the following compounds the rule , of octet is not obeyed ?
Answer» `H_(2)`O `PCl_(5)` `NH_(3)` `CH_(4)` ANSWER :B

Discussion

51472.	1mL of unknown solution of H_(2)SO_(4) is diluted upto 100mL and then its 25mL is titrated with 10mL of 0.2 M NaOH. The excess acid required 10mL of 0.1 MBa(OH)_(2) solution. Find molarity of original H_(2)SO_(4) solution.
Answer» ANSWER :8

Discussion

51473.	1m^(3) of C_(2) H_(4) at STP is burnt in oxygen, according to the thermochemical reaction : C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(l) , Delta H = -1410 kJ mol^(-1). Assuming 70% efficients, determine how much of useful heat is evolved in the reaction.
Answer» Solution :22.4 L of `C_(2)H_(4)` at STP produces 1410 kJ of energy 1000 L of `C_(2)H_(4)` at STP produces `(1410)/(22.4) xx 1000` (FULL efficiency) At 70% efficiency `= (1410)/(22.4)xx 1000 xx (70)/(100) = 44.02 xx 10^(3) kJ`.

Discussion

51474.	1g Ra^(226) is placed in an evacused tube whose volume is 5 cc, Assuming that each Ra nucleusyieldsfor He atoms which are retainedin the tube, what willbe the pressureof He preoducedat 27^(@)C after the end of 1590 year? (t_(1//2) for Ra is 1590 year)
Answer» ANSWER :`43.54` ATM

Discussion

51475.	1g of moist sample of KCI and KCIO_(3) was dissolved in water to make 250mL solution, 25mL of this solution was treatedwith SO_(2) to reducechlorirate to chloride and excess of SO_(2) was removedby boiling. The totalchloride was precipitaed as silver chloride. the weight of precipitate was 0.1435g. In anotherexperiment, 25mL of original solution was heated with 30mL of 0.2N ferrous sulphate solution and unreachedferrous sulphate required 37.5mL of 0.8Nsolution of an oxidant for complete oxidation. Caculate the molar ratio of chlorate int eh given mixture. Fe^(2+) reacts with CIO_(3)^(-) accroding to equaction, CIO_(3)^(-) + 6Fe^(2+) ++ 6H^(+) rarr CI^(-) + 6Fe^(3+) + 3H_(2)O
Answer» ANSWER :`(1)/(2), (1)/(2)`

Discussion

51476.	1g of Mg is burnt in a vessel containing 0.5 g of oxygen. The remaining unreacted is
Answer» 0.25g of Mg 0.1g of Mg 0.1g of `O_2` 0.75g of Mg Answer :A

Discussion

51477.	1 g of methane diffused in 20 sec. under certain conditions. Under the same conditions sqrt(20)g of a hydrocarbon (A) diffused in 40 sec. A 10 mg of sample of (A) took up 8.40 ml of H_(2) gas measured at 0°C and 760 mm pressure: Identify the incorrect statement.
Answer» One of the isomers of (A) exhibits GEOMETRICAL isomerism One of the isomers of (A) gives white PRECIPITATE with Tollen's reagent One of the isomers of (A) is asymmetric One of the isomers on OZONOLYSIS gives only ACETONE Solution :ON ozonolysis it does not give only acetone.

Discussion

51478.	1 g of methane diffused in 20 sec. under certain conditions. Under the same conditions sqrt(20)g of a hydrocarbon (A) diffused in 40 sec. A 10 mg of sample of (A) took up 8.40 ml of H_(2) gas measured at 0°C and 760 mm pressure: The number of n. bonds present in the compound A is/are
Answer» 2 1 3 4 Solution :`SQRT((16)/(M_A)) = 1/20 xx 40/(sqrt(20)) , M_A = 80` `10 xx 10^(-3) GM to `8.4 mol of `H_2`.

Discussion

51479.	If one of the open chain isomer of (A) on ozonolysis gives only formaldehyde & glyoxal, then its monocyclic isomer on ozonolysis gives
Answer» only glyoxal only METHYL glyoxal FORMALDEHYDE & glyoxal glyoxal & methyl glyoxal Solution :Ozonoliysis reaction.

Discussion

51480.	1g of a sample of NaOH was dissolved in 50mL 0.33M alkaline solution of KMnO_(4) and refulxed till all the cycanide was converted into OCN^(-). The reaction mixture was cooled and tis 5mLportion was acidified by adding H_(2)SO_(4) in excess an dtehn titrated to end point against 19.0mL of 0.1M FeSO_(4) solutio. Calculate%purity of NaCN sample.
Answer» ANSWER :`93%`

Discussion

51481.	1g hydrogen atoms are excited by radiations. The study of spectrum indicates that 27% of the atoms are in 3rd energy level and 15% of atoms are in 2nd energy level and the rest in the ground state. Ionization potential of hydrogen is 13.6 eV. Calculate (a) number of atoms present in energy level 1st, 2nd and 3rd (b) total energy released in joules when all atoms return to ground state.
Answer» SOLUTION :By mole concept, 1g of hydrogen atoms CONTAIN `= 6.02 xx 10^(23)` atoms `:.` 1.8g of hydrogen atoms contain H-atom `= 6.02 xx 10^(23) xx 1.8 = 10.84 xx 10^(23)` Atoms present in 3rd energy level `= (27)/(100) xx 10.84 xx 10^(23) = 292.68 xx 10^(21)` Atoms present in 2nd energy level `= (15)/(100) xx 10.84 xx 10^(23) = 162.6 xx 10^(21)` Atoms present in 1st energy level `= (58)/(100) xx 10.84 xx 10^(23) = 628.72 xx 10^(21)` Ionization potential of H = 13.6 eV means that `E_(1) = -13.6 eV, E_(2) = - (13.6)/(2^(2)) eV, E_(3) = - (13.3)/(3^(2)) eV` Energy released from all atoms when electrons return from 3rd level to 1st level `=(E_(3) - E_(1)) xx 292.68 xx 10^(21) = (-(13.6)/(9) + 13.6) xx 292.68 xx 10^(21) eV` `= 3.537 xx 10^(24) eV` Energy released from all atoms when electrons return from 2nd level to 1st level `= (E_(2) - E_(1)) xx 162.6 xx 10^(21) = ((-13.6)/(4) + 13.6) xx 162.6 xx 10^(21) eV` `= 1.659 xx 10^(24) eV` `:.` Total energy released `= (3.537 + 1.659) xx 10^(24) eV` `= 5.196 xx 10^(24) eV = (5.196 xx 10^(24)) xx (1.602 xx 10^(19)) J` `= 8.3239 xx 10^(5) J = 832.4 kJ`

Discussion

51482.	1g atom of Ra^(226) is placed in an evacuated tube of volume 5 liter. Assuming that each ._(88)Ra^(226) nucleusis an alpha- emitterand all the contentsare present in tube, calculate the total pressure of gases and partial pressure of He collected in tube at 27^(@)after the end of 800 year. Neglect volume occupied by undecayed Ra.
Answer» ANSWER :`2.887` ATM, `1.443` atm

Discussion

51483.	1)C_("Graphite") + O_(2(g)) rarr CO_(2(g)) , DeltaH = -94 K.Cals 2) C_("Diamond") + O_(2(g)) rarr CO_(2(g)) , DeltaH =-94.5 K.Cals From the above data the heat of transition of C_("diamond") rarr C_("Graphite")
Answer» `-50 Cal` `-100 Cal ` `-500 Cal` `100 Cal` ANSWER :C

Discussion

51484.	.4 sample of H_(2)O_(2) solution containing H_(2)O_(2) by weight requires x ml of KMnO_(4) solution for completed oxidation under acidic condition. The formality of KMnO_(4) solution is
Answer» 0.2 M 0.11 M 0.011 M 0.25 M Solution :Meq of `H_(2)O_(2)` = meq of `KMnO_(4)` `=(1XX(y//100))/("Eq.wt")xx1000` Meq of `KMnO_(4)=N(y)=(1xx(y//100))/(34//2)xx1000` `=N=(10)/(17)` Molarity = `("Normality")/(5)=(10//17)/(5)=(2)/(17)=0.117M`

Discussion

51485.	18.4 gms of a mixture of calcium carbonate and magnesium carbonate on heating gives 4.0 gms of magnesium oxide. The volume of CO_(2) produced at STP in this process is
Answer» `1.12` lit `4.48` lit `2.24` lit `3.36`lit Solution :`MgCO_(3) +CaCO_(3) RARR MGO +CaO +2CO_(2)`

Discussion

51486.	18.4g of a mixture of CaCO_(3) and MgCO_(3) on heating gives 4.0g of magnesium oxide. The volume of CO_(2) produced at STP in this process is
Answer» 1.12 L 4.48 LIT 2.24 L 3.36 L ANSWER :B

Discussion

51487.	1.84 g of a mixture of CaCO_(3) and MgCO_(3) is strongly heated till no further loss in mass takes place. The residue is found to weigh 0.96 g. Calculate the percentage of each component in the mixture.
Answer» Solution :Both `CaCO_(3) and MgCO_(3)` upon HEATING LOSE `CO_(2)` and the mass of residue is that of CaO and MgO. Let us calculate the same from the mixture and quate with the given mass `(0.96g)` Let the mass of `CaCO_(3)` in the mixture =xg `:.` Mass of `MgCO_(3)` in the mixture `(1.84 - x)g` Step I. Calculation of mass of residue (CaO) from x g of `CaCO_(3)` The chemical equation for the REACTION is : `{:(CaCO_(3),overset("Heat")(rarr),CaO+CO_(2)),(40+12+48,,40+16),(=100,,=56g):}` 100 g of `CaCO_(3)` upon heating from CaO (residue) = 56 g `:.` x g of `CaCO_(3)` upon heating form CaO (residue) `= (56)/(100) xx xg` Step II. Calculation of mass of residue (MgO) from `(1-x)`of `MgCO_(3)` `{:(MgCO_(3),overset("Heat")(rarr),MgO+CO_(2)),(24+12+48,,24+16),(=84g,,=40g):}` 84 g of `MgCO_(3)` upon heating from MgO (residue) = 40 g `(1-x)` of `MgCO_(3)` upon heating from MgO (residue) `= (40)/(84) xx(1-x)g` Step III. Calculation of percentage composition of the mixture Total mass of the residue `= [(56)/(100)xxx+(40)/(84)(1-x)]g` But the mass of residue actually formed = 0.96 g EQUATING the two, we get `(56x)/(100)+(40(1-x))/(84)=0.96` `4704x+7360-4000x=0.96xx100xx84=8064` `704x=8064-7360=704,x=(704)/(704)=1` Mass of `CaCO_(3)` in the mixture = 1g Mass of `MgCO_(3)` in the mixture `= 1.84 - 1 = 0.84`g Percentage of `CaCO_(3)` in the mixture `= (1)/(1.84) xx 100 = 54.35 %` Percentage of `MgCO_(3)` in the mixture `= (0.84)/(1.84) xx 100 = 45.65 %`

Discussion

51488.	180ml of hydrocarbon having a molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120ml of So, to diffuse is
Answer» 2 min 1.5 min 1 min 75 min Solution :`(lambda_1)/(lambda_2) = SQRT((M_2)/(M_2)) , ((180/1.5))/((120/t)) = sqrt(64/16) = 2` `implies t = (120 XX 2)/(180) xx 1.5 = 2 ` min.

Discussion

51489.	180 mL of a hydrocarbon diffuse through a porous membrane in 15 minutes, while 120 mL of SO_2 under identical conditions diffuse in 20 minutes. What is the molecular mass of the hydrocarbon ?
Answer» SOLUTION :Rate of diffusion of hydrocarbon `(r_(1)) = 180//15 = 12` mL/MIN Rate of diffusion of `SO_(2)(r_(2)) = 120/20 = 6 mL`/min Molecular mass of `SO_(2) (M_(2)) = 32 + (2 xx 16) = 64` According to Graham.s LAW of diffusion, `r_(1)/r_(2) = sqrt(M_(2)/M_(1))` Substituting the values, we get `12/6 = sqrt(64/M_(1))` or `M_(1) = (6 xx 6 xx 64)/(12 xx 12) = 16` Hence, the molecular mass of the hydrocarbon is 16.

Discussion

51490.	18.0 g of water completely vaporises at 100°C and 1 bar pressure and the enthalpy change in the process is 40. 79 "kJ mol"^(-1). What will be the enthalpy change for vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?
Answer» SOLUTION :Quantity of water = 18.0 g, pressure 1 bar As we know that, `18.0 g H_(2) O = 1 "MOLE" H_(2) O` ENTHALPY change for vapourising l mole of `H_(2) O = 40.79 "kJ mol"^(-1)` `therefore` Enthalpy change for vapourising 2 moles of `H_(2) O = 2 xx 40.79 "kJ" = 81.358 "kJ"` Standard enthalpy of vaporisation at 100°C and 1 bar pressure, `Delta_("vap") H^(@) = + 40.79 "kJ mol"^(-1)`

Discussion

51491.	18.0 g of water completely vapourises at 100^(@)C and 1 bar pressure and the enthalpy change for the process is 40.79 kJ mol^(-1). (i) What will be the enthalpy change for vapourising 2 moles of water under the same conditions? (ii) What is the standard enthalpy of vapourisation for water?
Answer» SOLUTION :18.0 of water = 1 MOL Standard enthalpy of VAPORISATION at `100^(@)C` and 1 bar PRESSURE = `40.79 kJ"mol"^(-1)`. Enthalpy change in vaporisation of 2 mol of water `= 2 XX 40.79 = 81.58` kJ.

Discussion

51492.	18.0 g of water completely vaporises at 100^(@)C and 1 bar pressure and the enthalpy change in the process is40.79 kJ mol^(-1). What will be the enthalpychange for vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?
Answer» Solution :`18.0` g of WATER `=` 1 mole `:. ` Standard ENTHALPY of vaporisation at`100^(@)C` and 1 bar pressure `=40.79 kJ MOL^(-1)` Enthalpy change in vaporisationof 2 moles of water `= 2 xx 40.79 kJ = 81. 58 kJ`

Discussion

51493.	1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO_4 solution . Calculated the percentage purity of the sample.
Answer» Solution :The BALANCED chemical equation is : `2MnO_4^(-)+5C_2O_4^(2-)+16H^(+)rarr2Mn^(2+)+10CO_2+8H_2O` Applying morality equation , `((M_1V_1)/n_1)_(KMnO_4)-=((M_2V_2)/n_1)_(C_2O_4^(2-))` `1/50xx30/2-=(M_2xx20)/5` `M_2=(30xx5)/(50xx2xx20)=0.075M` MOLECULAR wt. of `C_2O_4^(2-) =88` Wt of `C_2O_4^(2-) = 0.075 xx88=6.6g` AMOUNT of `C_2O_4^(2-) ` in 250 mL `=(6.6)/1000xx250=1.65g` % Purity `=(1.65)/(1.80)xx100=91.7%`

Discussion

51494.	1.80 g of a metal oxide required 833 mL of hydrogen at NTP to be reduced to its metal. Find the equivalent weights of the oxide and the metal.
Answer» SOLUTION :24.2, 16.2

Discussion

51495.	18 gms of glucose contains
Answer» 0.6 gram-atoms of carbon 0.6 grammolecules of Hydrogen 0.6 grammolecules of `CO_(2)` 1.2 gramatoms of Hydrogen SOLUTION :Molecular WEIGHT of glucose = 180 180 GRAMS glucose - `6.023xx10^(23)` MOLECULES - 72 grams carbons - 12 grams hydrogen

Discussion

51496.	18 g of water contain:
Answer» 1 g ATOM of hydrogen 2 g atoms of hydrogen 3 g atoms of hydrogen None of the above. Answer :B

Discussion

51497.	18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………….
Answer» <P> Solution :`0.1 ` `(P^(@) -P)/(P ^(@)) = x _(2) impliesx_(2) =` No. of moles of GLUCOSE `= (18)/(180) =0.1` `therefore (P^(@) -P)/(P^(@)) =0.1`

Discussion

51498.	18 C-H and 7 C-C sigma bonds are in
Answer» Cyclohexane 3, 3-dimethylpentane 2, 3, 3-trimethylpentane n-heptane Answer :C

Discussion

51499.	1.78N ofH_(2)O_(2) solution is :
Answer» 10 VOLUMES of `H_(2)O_(2)` 3.03% `((W)/(V))` of `H_(2)O_(2)` 2.56M `H_(2)O_(2)` 1 mL `H_(2)O_(2)` liberates 10 mL of `O_(2)` at STP Answer :A::B::D

Discussion

51500.	1.725 g of a metal carbonate is mixed with 300 mL of (N)/(10)HCl. 10 mL " of" (N)/(2) sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.
Answer» Solution :`10 mL "of" (N)/(2) NaOH` solution `=10 mL " of" (N)/(2)HCL` solution `-=50 mL " of" (N)/(10)HCl` solution Volume of `(N)/(10)HCl` used for neutralisation =300-50=250 mL 250 mL of `(N)/(10)HCl-=250 mL " of" (N)/(10)` METAL carbonate solution Let the equivalent mass of metal carbonate be E. Mass of metal carbonate present in solution `=(NxxExxV)/(1000)=1.725` `=(1xxExx250)/(10xx1000)=1.725` `=(E )/(40)=1.725` `E=40xx1.725=69`

Discussion

Explore topic-wise InterviewSolutions in .

2 litres of nitrogen completely react with 6 litres of hydrogen under suitable conditions to form 4 litres of NH_3. If all the volumes are measured at the same temperature and pressure, show that the data given are in accordance to the Gay Lussac's law of combining volumes.

2 L of hydrogen and 2.5 L of chlorine are allowed react in diffused light. Write the volume composition of the component gases after reaction.

2-iodo-2-methylpropane belongs to …………………. type.

2-Hexyne gives trans-2-hexene on treatment with

2-hexyne give trans-2-hexene on treatment with

2-Hexyne gives trans-2-Hexene on treatment with :

2 grams of Helium diffuses from a porous plate in 4min. How many grams of CH_4 would diffuse through the same plate in same time under similar conditions?

2 grams of a mixture of sodium carbonate and sodium bicarbonate on strong heating liberated carbondioxide, which occupie. 0.224 L at STP. What is the weight of Na_(2)CO_(3) in the original mixture?

2gm of hydrogen is present in a closed vessel at S.T.P. If the same quantity of another gas 'X' when introduced into the vessel the pressure becomes 1.5 atm. The gas 'X' would be

2 g of NaOH per 250 ml of solution is added to a buffer solution of buffer capacity 0.2 . Then the change in pH is

2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields : (a)C_2H_5CH_2undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-OCH_3 (b)C_2H_5CH_2undersetunderset(CH_3)(|)C=CH_2 (c )C_2H_5CH=undersetunderset(CH_3)(|)C-CH_3

2- butyne on chlorination gives ………………..

2-butyne converts in to dimethyl glyoxal by

2-butene is an example of .............. compound.

2-Butanone is heated with PCl_5 and the product so obtained is heated with alcoholic KOH , the final product obtained is

2-Bromopentane is treated with alcoholic KOH solution. Whatt will be the major product formed in this reaction and what is the type of elimination called?

2-Acetoxy benzoic acid is used as an:

2, 3-pentadiene is optically active since it

1^(ul@) alcohols are poor starting material for synthesis of 1-Alkene. Explain ?

1^(st) law does not tell about

1n which of the following compounds the rule , of octet is not obeyed ?

1mL of unknown solution of H_(2)SO_(4) is diluted upto 100mL and then its 25mL is titrated with 10mL of 0.2 M NaOH. The excess acid required 10mL of 0.1 MBa(OH)_(2) solution. Find molarity of original H_(2)SO_(4) solution.

1m^(3) of C_(2) H_(4) at STP is burnt in oxygen, according to the thermochemical reaction : C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(l) , Delta H = -1410 kJ mol^(-1). Assuming 70% efficients, determine how much of useful heat is evolved in the reaction.

1g Ra^(226) is placed in an evacused tube whose volume is 5 cc, Assuming that each Ra nucleusyieldsfor He atoms which are retainedin the tube, what willbe the pressureof He preoducedat 27^(@)C after the end of 1590 year? (t_(1//2) for Ra is 1590 year)

1g of Mg is burnt in a vessel containing 0.5 g of oxygen. The remaining unreacted is

1 g of methane diffused in 20 sec. under certain conditions. Under the same conditions sqrt(20)g of a hydrocarbon (A) diffused in 40 sec. A 10 mg of sample of (A) took up 8.40 ml of H_(2) gas measured at 0°C and 760 mm pressure: Identify the incorrect statement.

1 g of methane diffused in 20 sec. under certain conditions. Under the same conditions sqrt(20)g of a hydrocarbon (A) diffused in 40 sec. A 10 mg of sample of (A) took up 8.40 ml of H_(2) gas measured at 0°C and 760 mm pressure: The number of n. bonds present in the compound A is/are

If one of the open chain isomer of (A) on ozonolysis gives only formaldehyde & glyoxal, then its monocyclic isomer on ozonolysis gives

1)C_("Graphite") + O_(2(g)) rarr CO_(2(g)) , DeltaH = -94 K.Cals 2) C_("Diamond") + O_(2(g)) rarr CO_(2(g)) , DeltaH =-94.5 K.Cals From the above data the heat of transition of C_("diamond") rarr C_("Graphite")

.4 sample of H_(2)O_(2) solution containing H_(2)O_(2) by weight requires x ml of KMnO_(4) solution for completed oxidation under acidic condition. The formality of KMnO_(4) solution is

18.4 gms of a mixture of calcium carbonate and magnesium carbonate on heating gives 4.0 gms of magnesium oxide. The volume of CO_(2) produced at STP in this process is

18.4g of a mixture of CaCO_(3) and MgCO_(3) on heating gives 4.0g of magnesium oxide. The volume of CO_(2) produced at STP in this process is

1.84 g of a mixture of CaCO_(3) and MgCO_(3) is strongly heated till no further loss in mass takes place. The residue is found to weigh 0.96 g. Calculate the percentage of each component in the mixture.

180ml of hydrocarbon having a molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120ml of So, to diffuse is

180 mL of a hydrocarbon diffuse through a porous membrane in 15 minutes, while 120 mL of SO_2 under identical conditions diffuse in 20 minutes. What is the molecular mass of the hydrocarbon ?

18.0 g of water completely vaporises at 100°C and 1 bar pressure and the enthalpy change in the process is 40. 79 "kJ mol"^(-1). What will be the enthalpy change for vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?

18.0 g of water completely vapourises at 100^(@)C and 1 bar pressure and the enthalpy change for the process is 40.79 kJ mol^(-1). (i) What will be the enthalpy change for vapourising 2 moles of water under the same conditions? (ii) What is the standard enthalpy of vapourisation for water?

18.0 g of water completely vaporises at 100^(@)C and 1 bar pressure and the enthalpy change in the process is40.79 kJ mol^(-1). What will be the enthalpychange for vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?

1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO_4 solution . Calculated the percentage purity of the sample.

1.80 g of a metal oxide required 833 mL of hydrogen at NTP to be reduced to its metal. Find the equivalent weights of the oxide and the metal.

18 gms of glucose contains

18 g of water contain:

18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………….

18 C-H and 7 C-C sigma bonds are in

1.78N ofH_(2)O_(2) solution is :

1.725 g of a metal carbonate is mixed with 300 mL of (N)/(10)HCl. 10 mL " of" (N)/(2) sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.