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92301.

30 g of urea (M = 60 g "mol"^(-1)) is dissolved in 846 g of water. Calculate the vapour pressureof water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.

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Solution :The solution MAY be taken as dilute solution.
Applying the RELATIONSHIP :
`(p_1^0 - p_1)/(p_1^0) = (w_2 XX M_1)/(M_2 xx w_1)`
` 1 - (p_1)/(p_1^0) = (w_2 xx M_1)/(M_2 xx w_1)`
Substituting the values of the above EQUATION, we have
` 1- (p_1)/(23.8) = (30 xx 18)/(60 xx 846)`
` 1 - (p_1)/(23.8) = 0.0106 `
`(p_1)/(23.8) = 1- 0.0106`
`(p_1)/(23.8) = 0.9894" or" p_1 = 0.9894 xx 23.8`
`p_1 = 23.55 mg Hg `
92302.

3.0 g of non-volatile solute when dissolved in 1 litre water, shows s osmotic pressure of 2 atmosphere at 300 K. Calculate the molecular mass of th solute. (R=0.082 L K^(-1)mol^(-1)).

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Solution :`W_(B)=3.0g, V=1L, pi=2 atm. T=300K, R=0.082" K atm K"^(-1)MOL^(-1), M_(B)=?`
`M_(B)=(W_(B)xxRxxT)/(pixxV)=((3.0g)XX(0.082"L atm K"^(-1)mol^(-1))xx(300K))/((2 atm)xx(1L))=20.0.116 g.`
92303.

3.0 g of H_(2) react with 29.0 g of O_(2) to form H_(2)O. (i) which is the limiting reagent? (ii) Calculate the maximum amount of H_(2)O that can be formed. (iii) Calculate the amount of the reactant left unreacted. Molecular mass of H_(2)=2.016.

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Solution :`UNDERSET(2xx2.016=4.032g)(2H_(2))+underset(32g)(O_(2))rarrunderset(2XX(1.016+16)=36.032)(2H_(2)O)`
`"3 g of "H_(2)" require "O_(2)=(32)/(4.032)xx3=23.8g`
Thus, `O_(2)(29g)` is present Hence, `H_(2)` is the limiting reactant.
`H_(2)O" FORMED "=(36.032)/(4.032)xx3g=26.8g`
`O_(2)" left unreacted "=29-23.8=5.2g`
92304.

30 g of CH_(3)MgBr reacts with excess of CH_(3)CH_(2)CH_(2)OH forming xg of a gas. What isthe value of x?

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Solution :`underset((=119g))(CH_(3)MGBR)+CH_(3)CH_(2)underset(OH)underset(|)(CH_(2))rarrunderset((=16g))(CH_(4)(g))+CH_(3)CH_(2)underset(OMgBr)underset(|)(CH_(2))`
`119g` of `CH_(3)MgBr` EVOLVE `CH_(4)=16 g`
30 g of `CH_(3)` MgBr evolve `CH_(4)=(30g)XX((16g))/((199g))=4g.`
`:.` Value of x=4
92305.

30 g. of acetic acid is dissolved in 1 dm^(3) of a solvent. The molality of the solution will be (Givent, density of solvent =1.25 g cm^(-3)

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`0.40`
`0.35`
`0.55`
`0.25`

SOLUTION :`m=W_(2)/(M_(2) xx W_(1) kg)=30/(60 xx 1 xx 1.25 kg )=0.4`
92306.

30 cc of (M)/(3) HCl, 20 cc of (M)/(2) HNO_(3) and 40 cc of (M)/(4) NaOH solutions are mixed and the volumes was made up to 1dm^(3). The pH of the resulting solution is

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2
1
3
8

Solution :Total MILLI equivalent of `h^(+) = 20 xx (1)/(2) + 30 xx (1)/(3) = 20`
Total milliequivalent of `OH^(-) = 40 xx (1)/(4) = 10`
Thus, milliequivalents of `H^(+)` LEFT `= 20 - 10 = 10`
Total volume of solution `= 1 dm^(3) = 1000 ml`
`:. [H^(+)] = (10)/(1000) = 10^(-2)`
`rArr PH = -log[H^(+)] = -log [10^(-2)] = 2`.
92307.

30 cc of M/3 HCl, 20 cc of M/2 HNO_(3) and 40 cc of M/4 NaOH solutions are mixed and the volume wasmade up of 1 dm^(3) . The pH of the resulting solution is

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2
1
3
8

SOLUTION :Total milliequivalents of `H^(+)`
`=20xx1/2+30xx1/3=20`
Total milliequivalents of `OH^(-)=40xx1/4=10`
Thus, milliequivalents of `OH^(-)=40xx1/4=10`
Thus, milliequivalents of `H^(+)`left `=20-10=10`
Total VOLUME of solution `=1dm^(3)=1000` ml
`:.[H^(+)]=10/1000=10^(-2)`
`impliespH=-LOG[H^(+)]=-log[10^(-2)]=2`
92308.

3^(0)aminescontain

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NITRILE GROUP
IMINO group
NITRO group
aminogroup

ANSWER :A
92309.

3 xx 10^(3) kg of acetic acid is added to 500 cm^(3) of water. If 25% acetic acid is dissociated in water, what will be DeltaT_(f)K_(f)=1.86 K//m, d of H_(2)O = 1g cm^(-3).

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ANSWER :0.231 K
92310.

3 Which is the common oxidation state of the first transition series of elements

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`+2`
`+6`
`+8`
 `+4`

ANSWER :A
92311.

3-Phenylpropene on reaction with HBr gives (as a major produt)

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`C_(6)H_(5)CH_(2)CH(BR)CH_(3)`
`C_(6)H_(5)CH(BrCH_(2)CH_(3)`
`C_(6)H_(5)CH_(2)CH_(2)CH_(2)Br`
`C_(6)H_(5)CH(Br)CH=CH_(2)`

Solution :According to Markownikoff.s rule, the negative part of the ATTACKING reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.
`C_(6)H_(5)CH_(2)CH=CH_(2)+HBr to C_(6)H_(5)CH_(2)-underset("Br ")underset("| ")("C ")H=CH_(3)`
92312.

3-Pentanol is produced by which of the following reactions? 1. CH_(3)CH_(2)Br overset(Mg)underset("Ether")to overset(CH_(3)CH_(2)OH)to overset(H^(oplus)underset(H_(2)O)to 2.H-=CH overset(NaNH_(2))to overset(CH_(3)CH_(2)CHO)to overset(H^(oplus))underset(H_(2)O)to 3.(CH_(3)CH_(2))_(2)CO overset(LiAlH_(4))underset("Ether")to overset(H^(oplus))underset(H_(2)O)to

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(i) only
(i) & (II) only
(i) & (III) only
(ii) & (iii) only

ANSWER :C
92313.

3-Phenylpropene on reaction with HBr gives (as a major product)

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`C_6H_5CH_2CH(Br)CH_3`
`C_6H_5CH(Br)CH_2CH_3`
`C_6H_5CH_2CH_2CH_2Br`
`C_6H_5CH(Br)CH=CH_2`

Solution :According to Markownikoff's rule , the negative part of the UNSYMMETRICAL REAGENT adds to LESS hydrogenated (more substituted ) carbon ATOM of the double bond.
`C_6H_5CH=CH-CH_3+HBr to oversetoverset(Br)(|)(C_6H_5CHCH_2CH_3)`
92314.

When 3-phenylpropene reacts with HBr in the presence of peroxide, the major product formed is

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`C_(6) H_(5)CH_(2) CH_)(2) CH_(2) Br`
`C_(6) H_(5) CH_(2) CH(Br) CH_(3)`
`C_(6)H_(5) CH(Br) CH_(2) CH_(3)`
`C_(6)H_(5)CH (Br) CH= CH_(2)`

Answer :B
92315.

3- pentanol

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PRIMARY alcohol
Secondary alcohol
Tertiary alcohol
None of above

Answer :B
92316.

3-pentanol is :

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`3^@` ALCOHOL
`2^@` alcohol
`1^@` alcohol
phenol

Answer :B
92317.

3-ntrio 2-methyl butane -2-ol is the condensation products of

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ethnal and NITROMETHANE
ACETONE and nitromethane
acetone and nitroethane
methanal and 1-nitropane

ANSWER :C
92318.

Write chemical reactions of affect the 3-Nitrobromobenzene to 3-nitrobenzoic acid transformations.

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SOLUTION :
92319.

3-Nitrobromobenzene to 3-nitrobenzoic acid .

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SOLUTION :
92320.

3 moles of ideal gas X (C_(p,m) = (5)/(2)) and 2 molesof idealgas Y(C_(p,m) = (7)/(2)R) are takenin vessa andcompressed reversibly and adiabitically, duringthisprocess temperatureof gaseousmixture increasedfrom300 K to 400 K. Calculate change in internal energy (DeltaU) in cal of gaseous mixture (Given R = 2 cal//mol.K)

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Solution :`DELTAU =DeltaU_(1) + DeltaU_(2)`
`n_(1)C_(V_(1)m)DELTAT + n_(2)C_(V_(1)m)DeltaT`
`=[3 XX(3)/(2)R xx 100 ]+[2 xx (5R)/(2)xx100]`
`= 450 R + 500 R`
`= 950 R`
`= 1900 cal`
92321.

3 moles of ethanol react with one mole of phosphorus tribromide to form 3 moles of bromaethane and one mole of X. Which of the following is X?

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`H_(3)PO_(4)`
`H_(3)PO_(2)`
`HPO_(3)`
`H_(3)PO_(3)`

Answer :D
92322.

3 moles of CO_(2) gas expands isothermally against external pressure of 1 bar. Volume increases from 10 L to 30 L respectively. The system is in thermal contact of surroundings at temperature 15^(@)C. Entropy change in isotehrmal process is: DeltaS=2.303nR" log "((V_(2))/(V_(1))). Q. Select the correct relation:

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`DeltaS_("system")GT0,DeltaS_(surr).=0`
`DeltaS_(surr)lt0,DeltaS_("system")gt0`
`DeltaS_("system")=0,DeltaS_(surr).=0`
`DeltaS_(surr)gt0,DeltaS_("system")lt0`

Answer :B
92323.

3 moles of CO_(2) gas expands isothermally against external pressure of 1 bar. Volume increases from 10 L to 30 L respectively. The system is in thermal contact of surroundings at temperature 15^(@)C. Entropy change in isotehrmal process is: DeltaS=2.303nR" log "((V_(2))/(V_(1))). Q. If CO_(2) behaves like an ideal gas, then entropy change of system (DeltaS_("system")) will be:

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`+27.4JK^(-1)`
`9.1JK^(-1)`
`-27.4JK^(-1)`
`-9.1JK^(-1)`

ANSWER :A
92324.

3 moles of an ideal gas is compressed from 30 dm^3 to20 dm^3 against a constant pressure of 3.039 xx 10^5Nm^-2. The work done in calories is ___________ . (1 J = 0.239 cal)

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`+30.39 CAL`
`+72.63 cal`
`+726.3 cal`
`+303.9 cal`

ANSWER :C
92325.

3 moles of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction :A(g)+4B(g)hArr 2C(g)+3D(g)When equilibrium is reached, there is 1 mole of C. The equilibrium extent of the reaction is

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`1//4`
`1//3`
`1//2`
1

Answer :C
92326.

3 mole of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction 3A_((g))+4B_((g))hArr2C_((g))+3D_((g))

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`1//4`
`1//3`
`1//2`
1

Solution :`3A+4BhArr2C+3D`
`impliesK_(1)=([C]^(2)[D]^(3))/([A]^(3)[B]^(4))""...(i)`
When moles of `C=1` the equation becomes
`3/2A+2BhArrC+3/2D`
`impliesK_(2)=([C][D]^(3//2))/([A]^(3//2)[B]^(2))""...(ii)`
On comparing Eq. (i) with Eq. (ii), we GET
`K_(2)^(2)=K_(1)^(2)or K_(2)=(K_(1))^(1//2)`
HENCE, equlibrium extent`=1/2`
92327.

3 mof of mixture of NaHCO_(3) and Na_(2)CO_(3) is strongly heated at 200°C, when 1 mol of CO2 is obtained. What is the weight of residue? What is molar percentage of NaHCO_(3) and Na_(2)CO_(3) in the mixture?

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SOLUTIONRESIDUE : 2 MOLE,
92328.

3-Methylpent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

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TWO
four
six
zero.

Solution :.
With two chiral carbon atoms in the PRODUCT, total number of STEREOISOMERS are `=2^(n)=2^(2)=4`.
92329.

3-methylpentan-3-ol will be prepared from

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ETHYL formate and methyl magnesium bromide
ethyl ethanoate and ethyl magnesium bromide
ethyl propanoate and methyl magnesium bromide
ethyl formate and ethyl magnesium bromide

SOLUTION :`{:(CH_(3)-C=O+ C_(2)H_(5)MgBrto),(""| """ethyl magnesium"),(""OC_(2)H_(5) """bromide"),("""ethyl ethanoate"):}`
`{:(""CH_(2)H_(5)""CH_(2)CH_(3)),(""|""|),(CH_(3)-C-OHgBrunderset(-MG(OC_(2)H_(5))Br)toCH_(3)C=O),(""|),(""OC_(2)H_(5)):}`
`{:(""CH_(2)CH_(3)),(""|),(overset(C_(2)H_(5)MgBr)toCH-C-OHgBr),(""|),(""CH_(2)-CH_(3)):}`
`{:(""CH_(2)CH_(3)),(""|),(overset(H.OH)toCH_(3)-C-OH),(""|),(""CH_(2)CH_(3)),("""3-methylpentan-3-ol"):}`
92330.

3-Methylpent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible steroisomers for the product is:

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TWO
four
six
zero.

Answer :B
92331.

3-menthyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possiblestereoisomersfor the product is

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six
zero
two
four

Answer :D
92332.

3-methylbut-1-eneon reaction with HBr gives (as major product) ?

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`(CH_(3))_(2)CHCH_(2)CH_(2)BR`
`(CH_(3))_(2)CBrCH_(2)CH_(3)`
`(CH_(3))_(2)CHCH Br CH_(3)`
`(CH_(3))_(3)C CH_(2)Br`

SOLUTION :
92333.

3 - methyl cyclohexene on oxidation will give

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ANSWER :D
92334.

3-Methyl butane-2-ol on heating with HI gives

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2-iodo-3-methylbutane
2-iodo-2-methylbutane
1-iodo-3-methylbutane
1-iodo-2-methylbutane

Answer :A::B::D
92335.

3-methyl but-1-ene on HBO reaction gives

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3-methyl butan-2-ol
2-methyl butan-2-ol
3-methyl butan-1-ol
2-methyl butan-1-ol

Answer :C
92336.

3-methyl-2-butanol

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SOLUTION :
92337.

3-methyl-2 butanol on treatment with HClgives predominantly

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2, 2-dimethylpentane
2-chloro-3-methyl BUTANE
2-chloro-2-methyl butane
NONE of these.

ANSWER :C
92338.

3-methyl-2-butanol on treatment with HCI gives predominantly :

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2-chloro-2-methylbutane
2-chloro-3-methylbutane
2,2-dimethylpentane
None of the above

Answer :A
92339.

3-Hydroxy butanal is formed when (X) reacts with (Y) in dilute (Z) solution. What are(X),(Y) and(Z)?

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`{:(ULX,ulY,ULZ),(CH_3CHO,(CH_3)_2CO,NAOH):}`
`{:(ulX,ulY,ulZ),(CH_3CHO,CH_3CHO,NaCl):}`
`{:(ulX,ulY,ulZ),((CH_3)_2CO,(CH_3)_2CO,HCl):}`
`{:(ulX,ulY,ulZ),(CH_3CHO,CH_3CHO,NaOH):}`

ANSWER :D
92340.

3- hydroxy 2- methyl pentanal is formed when Xreact with Y in dilute Z solution. What are X,Y,Z

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`{:(""X,""Y,""Z),(CH_(3) - CHO,CH_(3) - CHO,NaOH):}`
`{:(""X,""Y,""Z),(C_(2)H_(5) - CHO,CH_(3) - CHO,NaOH):}`
`{:(""X,""Y,""Z),(C_(2)H_(5) - CHO,C_(2)H_(5) - CHO,NaOH):}`
`{:(""X,""Y,""Z),(C_(2)H_(5) - CHO,C_(2)H_(5) - CHO,NaCl):}`

ANSWER :C
92341.

3 gram of non-volatile solute in a 1000 cm^(3) of water shows an osmotic pressure of 2 bar at 300K. Calculate the molar mass of the solute (R=0.0853" L bar "K^(-1)mol^(-1)).

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SOLUTION :`37.35" G "MOL^(-1)`.
92342.

3 gram of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the fitrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

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`18 MG`
`36 mg`
`42 mg`
`54 mg`

ANSWER :A
92343.

3 gm of copper metal in cathode is deposited on electrode when 3 ampere current is pass for 2 hours through aqueous solution of CuSO_(4). Then what is the efficiency of current ? (Atomic weight of Cu=63.5 gm/mol)

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0.33
`48.7%`
`42.2%`
`54.4%`

Solution :Coulomb=Ampere`xx`Second
`=3xx7200`
`=(3xx7200)/(96500)=(3xx72)/(965)F`. . . (i)
MOLE=`("weight")/("MOLECULAR weight")=(3gm)/(63.5" gm mole"^(-1))`
`Cu_((aq))^(2+)+2e^(-) to Cu_((S))`
So, 2F gives 1 mole Cu . . . (Cathodic reaction)
So, `(3xx72)/(965)` F Gives
Theoretical mole of Cu
`=(3xx72)/(965)F((1" mole")/(2F))=(3xx36)/(965)`
So, efficiency
`=("PRACTICAL value")/("theoretical value")=(3)/(63.5)xx(965)/(3xx36)=0.4221`
`=42.21%`
92344.

3g of urea is dissolved in 45g of H_(2)O. The relative lowering in vapour pressure is

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0.05
0.04
0.02
0.01

Solution :Relative lowering of VAPOUR pressure,
`(p_(A)^(@)-p_(A))/(p_(A)^(@))=x_(B)`
where `x_(B)` is the mole fraction of solute (UREA)
`x_("urea")=(n_("urea"))/(n_("urea")+n_(H_(2)O))~~(n_("urea"))/(n_(H_(2)O))` (for dilute solutions)
`n_("urea")=(3)/(60)=0.05`
`n_(H_(2)O)=(45)/(15)=2.5`
`n_("urea")=(0.05)/(2.5)=0.02`
92345.

3 g of facetic acid is added to 250 mLof 0.1 HCl and the solutions made up to 500mL.To 20 mLof this solutions 5mLof 5M NaOHis added . The pH of the solutions is b/wn& n+ 1value of 'n' is :[Given : pKa of acetic acid = 4.75 molar mass of acetic acid =60 g/mol ] Neglect any changes in volumes.

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ANSWER :2
92346.

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid absorbed (per gram of charcoal) is

Answer»

18 mg
36 mg
42 mg
54 mg

Solution :INITIAL m moles of `CH_(3)COOH=0.06xx50`
FINAL m moles of `CH_(3)COOH=0.043xx50`
Hence, MASS of `CH_(3)COOH` adsorbed per gram of
charcoal`=((0.06-0.042)xx50xx10^(-3)xx60xx10^(3))/(3)=18mg`
92347.

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was fitered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid adsorbed (per gram of charcoal) is:

Answer»

18 mg
36 mg
42 mg
54 mg

Solution :Moles of acetic acid initially present
`= (0.06 XX 50)/(1000) = 3 xx 10 ^(-3)`
Moles of acetic acid after ADSORPTION
`= (0.042 xx 50)/(1000) = 2.1 xx 10 ^(-3)`
Moles of acetic acid adsorbed
`= 3.0 xx 10 ^(-3) -2.1 xx 10 ^(-3)`
`= 0.9 xx 10 ^(-3)`
Mass of acetic acid
`=0.9 xx 10 ^(-3) xx 60 = 54 xx 10 ^(-3) g`
Amount of acetic acid adsorbed PER gram of charcoal
`= (54 xx 10 ^(-3))/(3) = 18 xx 10 ^(-3) g`
`or =18` mg
92348.

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amont of acetic acid absorbed (pr gram of charcoal) is

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42 MG
54 mg
18 mg
36 mg

Solution :Milleq of acetic acid present initially `=0.06xx60=3`
millieq of acetic acid left after adsorption
`=0.042xx50=2.1`
`THEREFORE` millieq of acetic acid adsorbed by 3 g of charcoal
`=3-2.1=0.9meq=0.9xx60 mg=54mg`
`therefore` Amount of acetic acid adsorbed per GRAM of charcoal `=54/3=18` mg
92349.

3 g of activated charcoal was added to 50 ml . Of acetic acid solution (0.06 N ) in a flask . After an hour it was filtrate was found to be 0.042 N . The amount of acetic acid adsorbed ( per gram of charcoal )is

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18 mg
36 mg
42 mg
54 mg

Solution :m.e of `CH_(3)COOH` adsorbed = `50(0.6 -0.042)`
92350.

3 g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of CO_(2) and 5.4 g of H_(2)O. The data illustrates the law of:

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CONSERVATION of mass
multiple PROPORTIONS
CONSTANT proportions
RECIPROCAL proportions

Answer :A