94420 + Interview Questions in Chemistry in Class 12

1.	Molecular formula C_(4)H_(4)O_(4) can have four isomers A,B,C and D: A is dicarboxylic acid giving racemic tartaric acid with alk. B is dicarboxylic acid giving meso tartaric acid with alk. KMnO_(4). C is also dicarboxylic acid giving another monobasic acid on heating. D is cyclic ester Identify A ,B, C andD
Answer» SOLUTION :

Discussion

2.	Molecular formula C4H_10O has ........ isomeric ethers ,
Answer» 4 3 7 5 Answer :B

Discussion

3.	Molecular formula C_(3)H_(6)O_(2) on acid hydrolysisgives A and B. The compound 'A' reduce Tollens reagent. The possible structure of C_(3)H_(6)O_(2) will be
Answer» `CH_(3)- CH_(2) - COOH` `CH_(3) - COOCH_(3)` `H - COOC_(2)H_(5)` `CH_(3) - overset(O) overset(\|\|)C - CH_(3)` Answer :C

Discussion

4.	Molecular formula C_(3)H_(9)N canshown. (1) chain isomerism (2) positionisomerism (3) functionalisomerism (4) metamerism .
Answer» 1,3 3,4 1,2 2,3 Solution :Molecularformula`C_(3)H_(9)N` can SHOW positionandfunctionalisomerism. `CH_(3) - CH_(2) - NH_(2)and CH_(3) underset(NH_(2)) underset(\|)(CH)- CH_(3)`arepositionisomers. `1^(@),2^(@),3^(@)`- AMINES are functional isomers.

Discussion

5.	Molecular formula C_4H_10O exhibits
Answer» CHAIN isomerism position isomerism metamerism all of these Answer :D

Discussion

6.	Molecular formula C_(3)H_(6)O_(2) have two structures A & B. Structure A releases CO_(2) gas with NaHCO_(3) but B does not. Compound B is fruity smelling liquid. Write the structures & IUPAC name of A and B.
Answer» SOLUTION :N//A

Discussion

7.	Molecular formula C_(3)H_(6)O represents
Answer» ALDEHYDES ketones both 'a' and 'b' aldehydes and alcohols Answer :C

Discussion

8.	Molecular formula C_(2)H_(7)N represents
Answer» `2^(0) and 3^(0)`AMINES `1^(0) and 2^(0)` amine only `1^(0)` amines only `2^(0)` amine Answer :B

Discussion

9.	Molecularformula C_(2)H_(7)N showswhich typesof isomerism ?
Answer» POSITION Functional CHAIN OPTICAL Answer :B

Discussion

10.	Molecular formula C_3H_3O shows
Answer» chain and OPTICAL isomers position and FUNCTIONAL isomers functional and METAMERS chain and position isomers Answer :B

Discussion

11.	Molecular formula C_2H_6O shows
Answer» FUNCTIONAL isomerism metamerism position isomerism optical isomerism Answer :A

Discussion

12.	Molecular formula C_2H_6O represents
Answer» alcohols and acids alcohols and ETHERS only alcohols only ethers SOLUTION : Molecular FORMULA `C_2H_6O` are ethers and alcohols. i) `C_2H_5OH`, ii) `CH_3OCH_3`

Discussion

13.	Molecular formula C_(2)H_(4)O_(2) shows
Answer» CHAIN ISOMERISM POSITION isomerism FUNCTIONAL isomerism metamerism Answer :C

Discussion

14.	Molecular formula C_(2)H_(4)O_(2) represents
Answer» ACIDS ESTERS ALDEHYDES both 'a' and 'B' ANSWER :D

Discussion

15.	Molecular formula 'A' (C_94)H_(8)O) reacts with CH_(3)MgI gives 3- methyl 2-butanol . The structure of A is
Answer» `CH_(3)CH_(2)CH_(2)CHO` `C_(2)H_(5)COCH_(3)` `(CH_(3))_(2)CHCHO` `C_(2)H_(5)OCH_(3)` Answer :C

Discussion

16.	Molecular formula and empirical formula are related as :
Answer» `("Molecular FORMULA")/("EMPIRICAL formula")=`Atomicity Molecular formula `=("Empricial formula"xx"Molecular MASS")/("Empirical formula mass")` Molecular formula `=("Molecular mass")/("Empirical formula mass")` `("Molecular formula")/("Empirical formula")=("Empirical formula mass")/("Molecular mass")` ANSWER :B

Discussion

17.	Molecular formual C_(2)H_(7)Nshownwhichtypeofisomerism ?
Answer» 1 2 3 4 Answer :B

Discussion

18.	Molecular forces that bind polymer chains are __________.
Answer» dipole-dipole interaction hydrogen bonding van der Waal's FORCES either (B) or (C) ANSWER :D

Discussion

19.	Molecuels ions and their magnetic properties are given below {:("Molecular/ion","Magnetic property"),(C_(6)H_(6),"Antiferromagnetic"),(CrO_(2),"Ferrimagnetic"),(MnO,"Ferromagnetic"),(Fe_(3)O_(4),"Paramagnetic"),(Fe^(3+),"Diamagnetic"):} The correctly matched pairs in the above is
Answer» i-5,ii-3iii-2,iv-1,v-4 `i-3,ii-5,iii-1,iv-4,v-2` i-5,ii-3,iii-1,iv-2,v-4 i-5,ii-3,iii-1,iv-4,v-2 Solution :`C_(6)H_(6)` is diamagnetic (i-5) `CrO_(2)` is FERROMAGNETIC (ii-3) `MnO` is antiferromagnetic (iii-1) `Fr_(3)O_(4)` is FERRIMAGNETIC (iv-2) `FE^(3+)` is PARAMAGNETIC with 5 unpaired electrons (v-4).

Discussion

20.	Molecular folmula of salvarsan is
Answer» `C_(11)H_(11)N_(3)O_(2)S` `C_(11)H_(12)CL_(2)N_(2)O_(5)` `C_(11)H_(12_)As_(2)N_(2)O_(2)` C_(12)H_(13)N_(5)O_(2)S` Answer :C

Discussion

21.	Molecular crystals exist in:
Answer» CRYSTALLINE state Amorphous state Non-crystalline state All Answer :D

Discussion

22.	Molecular association is highes in
Answer» n-propyl AMINE trimethyl amine ETHYL METHYL amine equal in all Answer :A

Discussion

23.	Molecular attraction and size of the moleculesin a gas are negligible at :
Answer» CRITICAL point HIGH pressure High TEMPERATURE and LOW pressure Low temperature and high pressure Answer :C

Discussion

24.	Molecuar formula of chloroxylenol is
Answer» `C_(10)H_(18)O` C_(8)H_(9)CIO` `C_(12)H_(6)C_(14)O_(2)S` C_(11)H_(12)CL_(2)N_(2)O_(%)` ANSWER :B

Discussion

25.	Molecualar formula of prontosil is
Answer» `C_(11)H_(11)N_(3)O_(2)S` `C_(11)H_(12)CI_(2)N_(3)O_(2)S` `C_(12)H_(12)As_(2)N_(2)O_(2)` `C_(12)H_(13)N_(5)O_(2)S` ANSWER :C

Discussion

26.	Molecualarformula C_(4)H_(11)N
Answer» chainisomerism positionisomerism funcitonal ISOMERISM all of these ANSWER :D

Discussion

27.	Molecluar foumula of BHA is
Answer» `C^(15)H^(24)O` `C^(12)H^(15)O` `C^(15)H^(12)O^(2)` `C^(11)H^(16)O^(2)` ANSWER :D

Discussion

28.	Mole fracton of helium in a saturated solution at 20^(@)C is1.2xx10^(-6). Find out the pressure of helium above the solution. Given that at 20^(@)C, Henry's Law Constat (k_(H))=144.97 K bar.
Answer» SOLUTION :`rho_(He)=x_(He)xxK_(H)1.2xx10^(6)XX(144.97xx10^(3)"bar")=0.174 "bar"`

Discussion

29.	Mole of Chlorobenzene and bromo benzene is 0.1 and 0.2 respectively. The vapour pressure of chlorobenzene and bromo benzen is 0.350 bar and 0.500 bar respectively. So find out total vapour pressure of solution prepared by mixing of chlorobenzene and bromo benzene.
Answer» ANSWER :0.450 BAR

Discussion

30.	Mole fration of a solute in 2.0 molal aqueous solutions is :
Answer» 1.77 1.87 0.0347 0.347 Solution :`"Mole fraction of solute"(X_(B))=n_(B)/(n_(B)+n_(A))` `"No, of moles of solute"(n_(B))=2 MOL` `"No. of MOELS of WATER"(n_(A))=((1000g))/((18g mol^(-1)))` =55.55 mol `X_(B)=((2 mol))/((57.55 mol))=0.0347`

Discussion

31.	Mole fraction of x M aqueous urea solution is : [Given : density of solution is d gm//ml ]
Answer» `(18x)/(1000d - 42 x)` `(18 x)/(1000d - 60x)` `(18x)/(1000d - 18x)` None of these Solution :x MOLES of Urea in 1000 ml of solution mass of solution = 1000 d gm mass of solute `60 x gm` mass of solvent `= (1000d - 60x) gm` MOLE fracction of urea `= (x)/(x + ((1000d - 60x)/(18)))` `= (18x)/(18x + 1000d - 60x)` `x_("urea") = (18x)/(1000d - 42X)`

Discussion

32.	Mole fraction of solute in some solution is 1//n . If 50% of solute molecules dissociate into two parts and remaining 50% get dimerised , new mole fraction of solvent becomes 4//5 . Find value of n.29. 0.1 M KI and 0.2 M are mixed in 3:1 volume ratio. The depression in freezing point of the resulting solution will be 0.1 x (Assume K_(f) of and molality =molarity). Then x=_________
Answer»

Discussion

33.	Molefractionof urea in itsaqueous solution is 0.25 .Whatis themassof ureapresentin 200 gureasolution? (Molecularmass of ureais 60 g mol^(-1))
Answer» Solution :GIVEN : Mole fractionof UREA = 0.25 Mass ofureain 200 gsolution = ? Molefractionof water= 1- 0.25 =0.75 Massof urea= 0.25 `xx 60 =15 g` Massof water= 0.75 `xx 100= 75 g` `:.` Massof solution= 15+75 = 90 g `:.` 90 g solutioncontains15 gurea `:.200`g solutionwillcontain `(200 xx 15)/(90) ` = 33.33 g urea

Discussion

34.	Mole fraction of nitrogen in air is
Answer» 0.14 0.28 0.5 0.78 Answer :4

Discussion

35.	Mole fraction of K_(2)CO_(3) in a mixture of K_(2)CO_(3) and KHCO_(3) is 0.5. What will be the volume of 0.1 N HCl required to neutralize 1.252 g of the mixture?
Answer» Solution :As MOLE fraction of `K_(2)CO_(3)` is 0.5, this mean the mixture contains equal number of moles of `K_(2)CO_(3)` and `KHCO_(3)`. Suppose no. of moles of each is x. `"No. of g eq. in x moles of "K_(2)CO_(3)=2X.` `"No. of g eq. in x moles of "KHCO_(3)=x` `"Total g equivalents of "K_(2)CO_(3) and KHCO_(3)=2x+x=3X` Suppose volume of 0.1 N HCl REQUIRED to neutralize the mixture = V mL `therefore"No. of g equivalents present in V mL of 0.1 HCl "=(0.1 xxV)/(1000)"g eq."` As acids and bases are neutralized in equivalent amounts, `3x=(0.1xxV)/(1000) or V=3xx10^(4)x` `"Molecular mass of "K_(2)CO_(3)=2xx39+12+48=138` `"Molecular mass of "KHCO_(3)=39+1+12+48=100` `therefore x" mole of "K_(2)CO_(3)="138x gandx moles of "KHCO_(3)=100xg` `therefore 138x+100x=1.252 or x=(1.252)/(238)=5.26xx10^(-3)` `therefore V=3xx10^(4)xx5.26xx10^(-3)=157.8mL`

Discussion

36.	Mole fraction of glucose in 18% (wt./wt.) solution of glucose is
Answer» 0.18 0.1 0.017 0.021 Solution :GLUCOSE =18 g, Solution =100g :. WATER = 100-18=82g MOLE freaction of glucose `=(18//180)/(18//180+82//18` `=0.1/(0.1+4.56)=0.1/4.66=0.021

Discussion

37.	Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer, than answer the following based upon above data. Calculate the equilibrium constant for dimerisation fo acetic acid.
Answer» 3.39 5.62 6.78 `8.00` ANSWER :A

Discussion

38.	Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer, than answer the following based upon above data. The degree of association according to above data acetic acid should be
Answer» 0.6 0.48 0.92 0.8 Answer :B

Discussion

39.	Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer, than answer the following based upon above data. what should be the molality of the solution formed
Answer» 0.462m 0.4m 0.262m 0.626m Answer :C

Discussion

40.	Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer, than answer the following based upon above data. K_(f) for benzene is
Answer» 3K `XX` molality `2K xx` molality `5K xx` molality `10K xx` molality Answer :C

Discussion

41.	Mole fraction of a solute in 2.5 molal aqueous solution is:
Answer» 0.43 0.043 4.3 43 Solution :0.043

Discussion

42.	Molarity of H_(2)SO_(4) is 0.8 M and its density os 1.06 g//cm^(3). What will be the concentration in terms of molality and molar fraction?
Answer» Solution :`"Molarity of "H_(2)SO_(4)=0.8 M = 0.8" MOL L"^(-1)` `"Mass of "H_(2)SO_(4)="No. of MOLES"xx"Moles mass"` `=(0.8"mol")xx(98.0" G mol"^(-1))=78.4 g` `"Mass of 1 L solution"=(1000cm^(3))xx(1.06" g cm"^(-3))=1060 g` `"Mass of water in solution"=(1060-78.4)=981.6 g= 0.9816 kg` `"Molality of solution (m)"=("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")=((0.8"mol"))/((0.9816" kg"))` `=0.815" mol kg"^(-1)=0.815 m` `"Mole fraction of"H_(2)SO_(4)=n_(H_(2)SO_(4))/(n_(H_(2)SO_(4))+n_(H_(2)O))` `=((0.8mol))/((0.8mol)+((98.6g))/(18" g mol"^(-1)))=0.8/(0.8+54.5)=0.8/55.3=0.014`

Discussion

43.	Molasses contains :
Answer» `70%` sugar `50%` sugar `60%` sugar `10%` sugar Answer :B

Discussion

44.	Molarity of the liquid HCl if density of the solution is 1.17 g/cc is
Answer» 36.5 18.25 32.05 `42.10` SOLUTION :DENSITY 1.17 g/cc MEANS CONC. `=1170gL^(-1)=1170//36.5M=32.05M`

Discussion

45.	Molarity of pure water is
Answer» `1M` 55.55m 27.73m 80.55m Answer :B

Discussion

46.	Molarity of H_(2)SO_(4) is 0.8 and its density is "1.06 g/cm"^(3). What will be its concentration in terms of molality and mole fraction ?
Answer» Solution :Molarity of `H_(2)SO_(4)=0.8` MEANS 0.8 MOLE `H_(2)SO_(4)` are present in 1 L of the solution `"0.8 mole "H_(2)SO_(4)=0.8xx98 g = 78.4g,"1 L "H_(2)SO_(4) " solution "=1000xx1.06g=1060 g` `THEREFORE"Mass solvent (water )"=1060-78.4 g = 981.6g=(981.6)/(18)" moles"="54.53 moles"` Thus, 981.6 g of water contain 0.8 mole of `H_(2)SO_(4)`. `therefore"Molality "=("0.8 mol")/("981.6 g")xx"1000 g kg"^(-1)="0.815 mol kg"^(-1)` Further, the solution contains 0.8 mole of SOLUTE in 54.53 moles of solvent. `therefore"Mole FRACTION of solute"=(0.8)/(0.8+54.53)=0.014`

Discussion

47.	Molarity of aqueous glucose (C_(6)H_(12)O)^(6) will be, if mole fraction of glucose is 0.4.
Answer» 10M 3.7M 0.4M 2M Solution :`n_("glucose")=0.4rArrw_("glucose")=0.4xx180=72(M WT. C_(6)H_(12)O_(6)=180)` `n_(H_(2)O)=0.6rArrW_(H_(w)O)=0.6xx18=18.8` `W_("solution")=72+10.8=82.8gm` `d_("solution")=(82.8)/(V_("solution"))=2.07 gm//mlrArrV_("solution")=(82.8)/(2.07)=40ml` Molarity `rArrM=(0.4)/(40)xx1000=10M`

Discussion

48.	Molarity of 4% NaOH sollution is
Answer» 0.1 M 0.5 M 0.01 M 1.0 M Solution :`M=(4XX10)/(40)=1 M`.

Discussion

49.	Molarity of a solution prepared by dissolving 75.5 g ofpure KOH in 540 ml solution is
Answer» 3.05 M 1.35 M 2.50 M 4.50 M SOLUTION :`M=(W xx1000)/(m.wt. XX V" in ML")=(75.5xx1000)/(56xx540)=2.50M`.

Discussion

50.	Molarity of a solution does not change with temparature .
Answer»

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Molecular formula C4H_10O has ........ isomeric ethers ,

Molecular formula C_(3)H_(6)O_(2) on acid hydrolysisgives A and B. The compound 'A' reduce Tollens reagent. The possible structure of C_(3)H_(6)O_(2) will be

Molecular formula C_(3)H_(9)N canshown. (1) chain isomerism (2) positionisomerism (3) functionalisomerism (4) metamerism .

Molecular formula C_4H_10O exhibits

Molecular formula C_(3)H_(6)O_(2) have two structures A & B. Structure A releases CO_(2) gas with NaHCO_(3) but B does not. Compound B is fruity smelling liquid. Write the structures & IUPAC name of A and B.

Molecular formula C_(3)H_(6)O represents

Molecular formula C_(2)H_(7)N represents

Molecularformula C_(2)H_(7)N showswhich typesof isomerism ?

Molecular formula C_3H_3O shows

Molecular formula C_2H_6O shows

Molecular formula C_2H_6O represents

Molecular formula C_(2)H_(4)O_(2) shows

Molecular formula C_(2)H_(4)O_(2) represents

Molecular formula 'A' (C_94)H_(8)O) reacts with CH_(3)MgI gives 3- methyl 2-butanol . The structure of A is

Molecular formula and empirical formula are related as :

Molecular formual C_(2)H_(7)Nshownwhichtypeofisomerism ?

Molecular forces that bind polymer chains are __________.

Molecuels ions and their magnetic properties are given below {:("Molecular/ion","Magnetic property"),(C_(6)H_(6),"Antiferromagnetic"),(CrO_(2),"Ferrimagnetic"),(MnO,"Ferromagnetic"),(Fe_(3)O_(4),"Paramagnetic"),(Fe^(3+),"Diamagnetic"):} The correctly matched pairs in the above is

Molecular folmula of salvarsan is

Molecular crystals exist in:

Molecular association is highes in

Molecular attraction and size of the moleculesin a gas are negligible at :

Molecuar formula of chloroxylenol is

Molecualar formula of prontosil is

Molecualarformula C_(4)H_(11)N

Molecluar foumula of BHA is

Mole fracton of helium in a saturated solution at 20^(@)C is1.2xx10^(-6). Find out the pressure of helium above the solution. Given that at 20^(@)C, Henry's Law Constat (k_(H))=144.97 K bar.

Mole of Chlorobenzene and bromo benzene is 0.1 and 0.2 respectively. The vapour pressure of chlorobenzene and bromo benzen is 0.350 bar and 0.500 bar respectively. So find out total vapour pressure of solution prepared by mixing of chlorobenzene and bromo benzene.

Mole fration of a solute in 2.0 molal aqueous solutions is :

Mole fraction of x M aqueous urea solution is : [Given : density of solution is d gm//ml ]

Molefractionof urea in itsaqueous solution is 0.25 .Whatis themassof ureapresentin 200 gureasolution? (Molecularmass of ureais 60 g mol^(-1))

Mole fraction of nitrogen in air is

Mole fraction of K_(2)CO_(3) in a mixture of K_(2)CO_(3) and KHCO_(3) is 0.5. What will be the volume of 0.1 N HCl required to neutralize 1.252 g of the mixture?

Mole fraction of glucose in 18% (wt./wt.) solution of glucose is

Mole fraction of a solute in 2.5 molal aqueous solution is:

Molarity of H_(2)SO_(4) is 0.8 M and its density os 1.06 g//cm^(3). What will be the concentration in terms of molality and molar fraction?

Molasses contains :

Molarity of the liquid HCl if density of the solution is 1.17 g/cc is

Molarity of pure water is

Molarity of H_(2)SO_(4) is 0.8 and its density is "1.06 g/cm"^(3). What will be its concentration in terms of molality and mole fraction ?

Molarity of aqueous glucose (C_(6)H_(12)O)^(6) will be, if mole fraction of glucose is 0.4.

Molarity of 4% NaOH sollution is

Molarity of a solution prepared by dissolving 75.5 g ofpure KOH in 540 ml solution is

Molarity of a solution does not change with temparature .