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92601.

20 gm sample containing Mg. is treated with excess of 2M HCl. As a result 11.2 litre of H_(2) gas at NTP was evolved. Find % purity of Mg in sample

Answer»

0.6
0.3
0.75
0.72

Solution :`Mg+2HClrarrMgCl_(2)+H_(2)`
`n_(Mg)=n_(H_(2))=(11.2)/(22.4)=(1)/(2)`
`w_(Mg)=12gm`
`% "purity"=(12)/(20)xx100=60`
92602.

2.0 g of sample containing Na_(2)CO_(3) and NaHCO_(3), loses 0.248 g when heated to 300^(@) C, the temperature of which NaHCO_(3) decomposes to Na_(2)CO_(3) and H_(2)O. What is % of Na_(2)CO_(3) in mixture?

Answer»


ANSWER :0.66
92603.

2.0g of mixture of carconate , bicarbonate and chloride of sodium , on heating , produced 56 mL of CO_(2) at NTP . 1.6g of the same mixture required 25 mL of N HCl solution for neutralisation . Calculate the percentage ofNa_(2)CO_(3), NaHCO_(3) and NaCl in the mixture from the given data .

Answer»

Solution :On heating the given mixture , only `NaHCO_(3)` decomposes as
`2NaHCO_(3) to Na_(2)CO_(3) +H_(2)O + CO_(2)`
` :. ` EQ. of `NaHCO_(3)` = eq. of `CO_(2) = 56/11200 ""….(Eqn . 4ii)`
( 1 eq. of `CO_(2)` occupies 11200 mL at NTP )
` :. wt ofNaHCo_(3) = 56/1120 xx 84 = 0.42 g `
( eq. wt . of `NaHCO_(3) = 84`)
` :. % of NaHCO_(3) = (0.42)/2 xx 100 = 21 % `
Now , if x is the weight of NaCl in `1.6` g of the mixture
then wt. of `NaHCO_(3) = 0.336 " g " (i.e ., 21 % " of " 1.6 g)`
and wt. of `Na_(2)CO_(3) = 1.6 - 0.336 - x = (1.264 - x) g `
Since `Na_(2)CO_(3)` are NEUTRALISED by HCl solution as :
`Na_(2)CO_(3) +2HCL to 2NaCl +H_(2)O +CO_(2)`
`NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)`
m.e of `Na_(2)CO_(3)` + m.e of `NaHCO_(3)` = m.e of HCl
or eq. of `Na_(2)CO_(3) xx 1000" eq. of " NaHCO_(3) xx 100 ` = m.e of HCl...(Eqn.3)
`(1.264 -x)/53 xx1000 +(0.336)/84 xx 1000 = 1 xx 25 `
` x = 0.151 g `
` :. "" % of NaCl = (0.151)/(1.6) xx 100 = 9.42 % `
and% of `Na_(2)CO_(3) = 100 - (21 +9.42) = 69.58 % `
Thus , `{:{(Na_(2)CO_(3)=69.58%),(NaHCO_(3)=21.00%),(NaCl=9.42%):}` .
92604.

2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant (K_(f)) of benzene is "4.9 K kg mol"^(-1). What is the percentage association of the acid if it forms dimer in the solution?

Answer»

Solution :Mass of solute (benzoic acid), `w_(2)=2.0 G",Mass of solvent (benzene), "w_(1)=25.0g`
Observed `DeltaT_(f)=1.62K","K_(f)="4.9 K kg mol"^(-1)`
`therefore"Observed molar mass of benzoic acid"`
`M_(2)=(1000xxK_(f)xxw_(2))/(DeltaT_(f)xxw_(1))=("1000 g kg"^(-1)xx"4.9 K kg mol"^(-1)xx"2.0 g")/("1.62 K"xx"25.0 g")="242 g mol"^(-1)`
Calcualted molar mass of benzoic acid `(C_(6)H_(5)COOH)=72+5+12+32+1="122 g mol"^(-1)`
`"van't Hoff factor,i"=("Calculated mol. mass")/("Observed mol. mass")=(122)/(242)=0.504`
If `alpha` is the degree of association of benzoic acid, then we have
`{:(,2C_(6)H_(5)COOH,hArr,(C_(6)H_(5)COOH)_(2)),("Initial moles","1",,"0"),("After association",""1-alpha,,""alpha//2):}`
`therefore"Total number of moles after association "=1-alpha+(alpha)/(2)=1-(alpha)/(2) therefore i=(1-alpha//2)/(1)=0.504 or 1-(alpha)/(2)=0.504`
`alpha=(1-0.504)xx2=0.496xx2=0.992"or% AGE association = 99.2%".`
92605.

20 g of hydrogen is present in 5 litre vessel. The molar concentration of hydrogen is

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4
1
3
2

Solution :Molar CONCENTRATION `[H_(2)]=("Mole")/("V in litre")=(20//2)/(5)=2`.
92606.

20 g of hydrogen is present in 5 litre vessel, The molar concentration of hydrogen is :

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2
4
3
1

Answer :A
92607.

20 g of a monobasic acid furnishes 0.5 moles of H_3O^(+) ions in its aqueous solution. The value of 1 g eq. of the acid will be

Answer»

100 G
20 g
10 g
40 g

Answer :D
92608.

20 g of a substance in 2 litre of solution at 10^@C produces an osmotic pressure of 0.68 atm, the mol. wt. of solute is

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322
380
342
360

Answer :C
92609.

20 g of a sample of Ba(OH)_(2) is dissolved in 50 ml. of 0.1 N HCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20cc. Calculate the % Ba(OH)_(2) in the sample.

Answer»


ANSWER :0.0129
92610.

20 g of a non-volatile solute is added to 500 g of solvent. Freezing point of solvent = 5.48^(@)C and solution = 4.47^(@)C. K_(f)=1.93^(@)//m. Molecular mas of the solute is

Answer»

`72.2`
`76.4`
`73.2`
`70.6`

Solution :`Delta T_(f)=5.48-4.47=1.01`
`Delta T_(f)=K_(f)XX m`
`RARR 1.01=1.93xx(20)/(M)xx(1000)/(500)rArr M=76.4`.
92611.

20 g of a binary electrolyte (mol. Wt. = 100) are dissolved in 500 g of water. The freezing point of the solution is -0.74^@C, K = 1.86 k molality^-1. The degree of ionization of the electrolyte is:

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`50%`
`75%`
`100%`
Zero

Answer :D
92612.

2.0 g mixture of sodium carbonate and sodium bicarbonate on heating to constant mass gave 224mL of CO_(2), at NTP. The % mass of sodium bicarbonate in the mixture is:

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50
54
80
84

Answer :D
92613.

20% first order reaction is completed in 50 minute. Time required for the completion of 60% of the reaction is

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100
150
262
205

Answer :D
92614.

20 cm^(3)" so "SO_(2) diffuses through a porous plug in 60 sec. What volume of O_(2) will diffuse under similar conditions in 30 sec ?

Answer»

SOLUTION :`(V_(1))/(t_(1))xx(t_(2))/(V_(2))=SQRT((M_(2))/(M_(1)))`
`THEREFORE (20)/(60) xx (30)/(V) =sqrt((32)/(64))`
`therefore V=14.14 dm^(3)`
92615.

20 cc of a hydrocarbon mixed with 66 cc of oxygen were exploded in a eudiometer tube. The residual gases after cooling occupied 56 cc. On treatment with KOH solution, the volume decreased to 16 cc. Find the formula of the hydrocarbon.

Answer»

Solution :Experimental Values : Volume of hydrocarbon taken = 20 cc
Volume of oxygen added = 66 cc
Volume after explosion and cooling
`"i.e.,Volume of CO"_(2)" FORMED + unused oxygen = 56 cc"`
Volume after introducing KOH
`{:("i.e.,","Volume of unused oxygen = 16 cc"),(therefore"","Volume of CO"_(2)" formed "=56-16="40 cc"),("and","Volume of oxygen USED"=66-16="50 cc"):}`
Theoretical values : Let `C_(x)H_(y)` be the formula of the gaseous hydrocarbon. The oxidation equaiton will be
`{:(C_(x)H_(y),+,(x+y//4)O_(2),rarr,xCO_(2)+y//2H_(2)O"(NEGLIGIBLE volume)"),("1 vol.",,x+y//4"vol.",,"x vol."),("1 cc",,x+y//4" cc",,"x cc"),("20 cc",,20(x+y//4)" cc",,"20 x cc"):}`
Equating experimental and theoretical values of carbon dioxide formed and oxygen used, we GET
`{:(,20x=40,or,x=2,),(and,20(x+y//4)=50,or,20(2+y//4)=50,"("because x = 2")"),(,,or,y=2,):}`
Hence formula of hydrocarbon is `C_(2)H_(2)` (Acetylene).
92616.

20 CC of hydro carbon were exploded with excess of oxygen. After explosion and cooling a contracting of was noted on addition of KOH another contraction of 40 CC was noted. The molecular formula of hydrocarbon is

Answer»

`C_(2)H_(6)`
`C_(2)H_(4)`
`C_(2)H_(2)`
`CH_(4)`

ANSWER :C
92617.

20% aqueous FeCl_(3) solution has density of 1.1 gm/mL then molar concentration of this solution is ……….

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`0.028`
`1.357`
`0.163`
`1.47`

Solution :`20% FeCl_(3)` solution MEANS 20 gm `FeCl_(3)` is present in 100 gm of solution.
`therefore` VOLUME of 100 gm solution `= (100)/(1.1)=90.91` mL
Moles of 20 gm `FeCl_(3)` solution `= (20)/(162)=0.1234` mol
So, molar concentration solution
`= (0.1234xx1000)/(90.91)=1.357 M`
92618.

2 SO_3 rarr 2SO_2 + O_2 is at equilibrium. The SO_2concentration is 0.6 M. Initial concentration of SO_3 is 1M. The equilibrium constant is:

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2.7
1.36
0.34
0.675

Answer :D
92619.

2-propylbenzene on air oxidation and followed by decomposition by dilute acid gives

Answer»

PHENOL and propanal
phenol and propanone
phenol and propanol
phenol and propionic acid

Answer :B
92620.

2-propanol when heated with copper at 570 k yields:

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`CH_3CH=CH_2`
`CH_3CHO`
`CH_3COCH_3`
`CH_2=C=CH`

ANSWER :C
92621.

2-Propanol reacts with sulphuric acid at 170^(@)C, gives an alkene which reacts with bromine to give dibromoalkane. Alcoholic potassium hydroxide dehydrogenates the dibromo product to X. The structure of X is :

Answer»

`CH_(2)=CH-CH_(3)` <BR>`CH_(3)-UNDERSET(Br)underset("|")(CH)=CH_(2)Br`
`CH_(3)-C-=CH`
`underset(Br)underset("|")(CH_(2))-C-=CH`

Solution :`CH_(3)underset(OH)underset("|")(CH)CH_(3)underset(170^(@)C)overset(H_(2)SO_(4))rarrCH_(3)CH=CH_(2)overset(Br_(2))rarrCH_(3)underset(Br)underset("| ")(CH)-underset(Br)underset("|")(CH_(2))overset("alc. KOH")rarrunderset((X))(CH_(3)C-=CH)`
92622.

"2-propanol "+NaBr overset("Reflux")rarr X. What is X ?

Answer»

2-bromopropane
PROPANE
Propene
Propanone

Solution :`CH_(3)-underset(OH)underset(|)CH-CH_(3)+NaBr underset(Delta)OVERSET("conc." H_(2)SO_(4))to underset("2-bromo propane")(CH_(3)-underset(Br)underset(|)CH-CH_(3)+H_(2)O`
92623.

2-Phenylethanol may be prepared by the reactionof phenylmagnesium bromide with

Answer»

`HCHO`
`CH_(4)CHO`
`CH_(3)COCH_(3)`

Solution :2-Phenylethanol, `OVERSET(1)(C)H_(2)OH overset(2)(C)H_(2)C_(6)H_(5)`, is a `1^(@)` alcohol which can be PREPARED from `C_(6)H_(5)MgBr` by treating with ethylene oxide (note that HCHO will introduced only one carbon atom, i.e., it will give `C_(6)H_(5)CH_(2)OH` annd not `C_(6)H_(5)CH_(3)CH_(2)OH`).
92624.

2-Phenylcycloprop-2-en-1-one is allowed to react with phenylmagnesium bromide and the reaction mixture is hydrolysed with perchloric acid.The product formed is

Answer»

`PH-undersetunderset(O)(||)C-undersetunderset(OH)(|)oversetoverset(Ph)(|)C-CHO`


SOLUTION :
92625.

2-Phenylbutan-2-ol is best prepared by which of the following combinations?

Answer»

`C_(6)H_(5)COCH_(3)+ C_(2)H_(5)MgBr`
`C_(6)H_(5)COCH_(3)+ C_(6)H_(5)MgBr`
`C_(6)H_(5)COC_(2)H_(5)+ CH_(3)MgBr`
All of these

Solution :All of these
92626.

2-Phenyl propene on acidic hydration gives ,

Answer»

2-phenyl-2-propanol
2-phenyl-1-propanol
3-phenyl-1-propanol
1-phenyl-2-propanol

Solution :Acidic hydration of 2-phenyl propene follows electrophilic REACTION mechanism FORMING an intermediate `3^@` carbocation (more STABLE ), there by forming 2-phenyl-2-propanol.
92627.

2-phenyl prop-1-ene on oxidation by ozone and followed by hydrolysis in the presence of zinc gives

Answer»

MIXTURE of ACETOPHENONE and ethanal
mixture of BENZALDEHYDE and ethanal
mixture of acetophenone and methanal
mixture of benzaldehyde and methanal

Answer :C
92628.

2-phenyl ethyl chloride undergoes alkaline hydrolysis to give 2-phenyl ethanol with

Answer»

COMPLETE INVERSION of configuration
racemization PULS some inversion.
RETENTION of configuration
complete racemization

Answer :B
92629.

2-Phenyl-2-hexanol can be prepared by Grignard synthesis The pair of compounds giving the desired product is

Answer»




NONE of these

SOLUTION :
92630.

2-Phenyl butane on oxidation by KOH+KMnO_(4) gives

Answer»

BENZOIC ACID
1-phenyl ETHANOIC acid
2-methyl BUTANOIC acid
3-methyl butanoic acid

Answer :A
92631.

2-Pentanone can be differentiated from 3-Pentanone by_____.

Answer»

SOLUTION :IODOFORM TEST.
92632.

2-Phentlethanol may be prepared by the reaction of phenylmagnesium chloride with

Answer»

HCHO
`CH_3CHO`
`CH_3COCH_3`

SOLUTION :
92633.

2-pentanone and 3-pentanone can be distinguished by one of the following:

Answer»

TOLLEN's reagent
Fehling's solution
Schiff's test
Iodoform test

Answer :D
92634.

2-pentanone and 3-pentanone are predominantly

Answer»

Positional isomers
Functional isomers
Metamers
RING CHAIN isomers

Answer :C
92635.

2-pentanone and 3- pentanone are

Answer»

CHAIN ISOMERS
position isomers
metamers
FUNCTIONAL isomers

Answer :C
92636.

2-Pentanone and 3-methylbutan-2-one are

Answer»

Optical ISOMERS
GEOMETRICAL isomers
Chain isomers
Tautomers

ANSWER :C
92637.

2% of aqueous urea and 6% of aqueous 'X' are isotonic. If 'X' is a molecular solid, calculate its molar mass.

Answer»

Solution : For isotonic solutions, osmotic PRESSURES are same, `pi_1=pi_2`
Hence, concentrations of solutions are also same,`C_1=C_2`
Since MOLES are equal , `W_1/M_1=W_2/M_2` (or) `2/60=6/M_2`
Molar MASS of solute X=`M_2="180 g mol"^(-1)`
92638.

2-nitro 2 -methylpropaneon reductiongives

Answer»

`(CH_(3))_(3)N`
`(CH_(3))_(2)NH`
`(CH_(3))_(3)CNH_(2)`
`(CH_(3))_(2)CHCH_(2)NH_(2)`

Solution :`CH_(3) - underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(C) - NO_(2) + 6[H] overset(FE+ HCl) toCH_(3) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - NH_(2)+2H_(2)O`
92639.

2- Nitrbutanceis heated with dilH_(2)SO_(4) gives

Answer»

`CH_(3) - overset(OH) overset(|)(CH) -CH_(2) - CH`
`CH_(3) - overset(O)overset(||)(C) - CH_(2) - CH_(2) - CH_(3)`
`CH_(3) - overset(O)overset(||)(C) - CH_(2) - CH_(3)`
`CH_(3) - overset(OSO_(3)H)overset(|)(CH) - CH_(2) - CH_(3)`

ANSWER :C
92640.

2 N HCl solution will have same molar conc. as a

Answer»

4.0 N `H_(2)SO_(4)`
`0.5 N H_(2)SO_(4)`
`1 N H_(2)SO_(4)`
`2 N H_(2)SO_(4)`

Solution :From the relation
`("Normality")/("Molarity") = ("Molecular mass")/("Equivalent mass") =n`
For 2 N HCl
Molarity `=("Normality " xx " Equivalent WEIGHT")/("Molecular weight")`
Molarity `=(2 xx 36.5)/36.5=2`
For `4N H_(2)SO_(4)`
Molarity `=("Normality " xx "Equilvalent weight")/("Molecular weight")`
Molarity `= (4 xx 49)/98 =2`
HENCE, `4 N H_(2)SO_(4)` and 2N HCl solution will have same molar consentration.
92641.

2 moles of the same gas are enclosed in two containers A and B at 27^(@)C and at pressure 2 and 3 atms respectively. The rms velocity .

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will be same in both the container 'A' and 'B'
will be more in container 'B'
will be more in container 'A'
can't be SAID as VOLUME of the container is not MENTIONED.

Answer :A
92642.

2 "mole" of PCl_(5) were heated in a closed vessel of 2 litre capacity. At equilibrium 40% of PCl_(5) dissociated into PCl_(3) and Cl_(2). The value of the equilibrium constant is:

Answer»

`0.266`
`0.53`
`2.66`
`5.3`

Solution :`underset(2)(PCl_(5))hArrunderset(0)(PCl_(3))+underset(0)(Cl_(2))`
`(2xx60)/(100)(2xx40)/(100)(2xx40)/(100)`
Volume of CONTAINER =2 litre.
`K_(c)=((2xx40)/(100xx2)xx(2xx40)/(100xx2))/((2xx60)/(100xx2))=0.266.`
92643.

2 moles of Ne gas and 5 moles of He gas, both samples having average velocity 7xx10^2 m//s, are mixed.Find the average translational kinetic energy per mole of the given gas mixture (in joules).Report your answer as 'X' where X=(Average translational KE in joules)xx0.002.The reported answer should be upto nearest integer.

Answer»


SOLUTION :Average Translational kinetic energy (per mole of gas)`=3/2RT=(3M)/2xx(RT)/M=(3piM)/16xx(8RT)/(piM)`
`=(3piM)/16xx(sqrt((8RT)/(piM)))^2=(3piM)/16xx(v_(AV))^2`
`:.` For one mole of gas MIXTURE,
Average translational kinetic energy `=(3piM_("mix"))/16xx(v_(av))^2`
`=3/16xx22/7xx[((2xx20+5xx4)//7)/1000]xx(7xx10^2)^2[because M_(mix) " is in" kg//"mole"]`
=2475 Joules.
Reported ANSWER=`2475xx0.002=4.75~~5`
92644.

2 moles of N_(2) O_(4 (g)) is kept in a closed container at 298 K and under 1 atm pressure . It is heated to 596 K when 20% by mass of N_(2) O_(4 (g)) decomposes to NO_(2) . The resulting pressure is

Answer»

2.4 atm
1.2 atm
4.8 atm
2.8 atm

SOLUTION :`{:(, 2 N_(2) O_(4) to 4 NO_(2)), (" Initially " ,2 ""0) , ("At 596 K" , 2p ""4p), ("Finally" , 2-2p ""4p):}`
As given , `2p = (20 xx 2)/(100)`
2p = 0.4
`THEREFORE p = 0.2` atm
Resulting pressure (P) = 2- 2p + 4p
P = 2 +2 p
P = 2 + 2 (0.2) = 2.4 atm
92645.

2 moles of helium gas expanded isothermally and irreversibly at 27^(@)C from volume 1 dm^(3) to 1 m^(3) at constant pressure of 100 kPa. Calculate the work done.

Answer»

99900 kJ
99900 J
34464.65 kJ
34464.65 J

Solution :`W=-pDeltaV`
Given, `p=100" kPa"=10^(5)Pa`,
`V_(1)=1dm^(3)=10^(-3)m^(3),V_(2)=1m^(3)`
`W=10^(5)xx(1-10^(-3))J=99900J`
92646.

2 moles of He gas (r=5/3) are initially at a temperature of 27^(@)C and occupy a volume of 20L. The gas is first expanded at constant pressure until the volume is doubled, then undergoes reversible adiabatic change until the temperature returns to its initial value. Which curve is correct for Pvs V?

Answer»




SOLUTION :NA
92647.

2 moles of H_(2)SO_(4) is kept in a beaker. Find (i) Moles of H-atoms (ii) Moles of S-atoms (iii) Moles of O-atoms (iv) Number of O-atoms

Answer»


Answer :`(##ALN_NC_CHM_MC_E01_003_A01##)`
(i) 4 (ii) 2(iii) 8 (iv) `8N_(4)`
92648.

2 moles of H_2 at NTP occupy a volume of

Answer»

11.2 litre
44.8 litre
2 litre
22.4 litre

Answer :B
92649.

2 moles of ann ideal gas at 27^(@)C are expanded reversibly from 2L to 20 L. find entropy change. (R=2 cal/mol K)

Answer»

0
4
9.2
92

Solution :`DeltaS=2.303"nR LOG"(V_(2))/(V_(1))`
`=2.303xx2xx2xx"log"(20)/(2)=9.2`
92650.

2 moles of an ideal monoatomic gas undergo a reversible process for which PV^(2)= constant. The gas sample is made to expand from initial volume of 1 litre of final volume of 3 litre starting from initial temperature of 300 K. find the value of DeltaS_("sys") for the above process. Report your answer as Y where DeltaS_("sys")=-YRIn3

Answer»


ANSWER :1