94420 + Interview Questions in Chemistry in Class 12

1.	If 18g of glucose (C_(6)H_(12)O_(6)) is present in 1018 g of an aqueous solution of glucose, it is said to be
Answer» 1 MOLAL 1.1 molal 0.5 molal 0.1 molal Solution :18 g glucose `=18/180 ` mol =0.1 mol. As it is present in (1018-18) = 1000 g WATER, the solution is 0.1 molal.

Discussion

2.	If 193000 coulomb of electricity is passed through a metallic wire, how many electrons will flow through the wire ?
Answer» SOLUTION :1F = 96500C electricity contains 1 MOL or `6.023 xx 0^(23)` no. of ELECTRONS. `therefore ` 193000C = 2 F electricity contains 2 mol or `2xx6.023xx10^(23)` no. of electrons.

Discussion

3.	If 16 g of H_(2) and 56 g of N_(2) are present in a 2 liter vessel at STP then the total number of molecules in the vessel will be
Answer» ` 6.022 XX 10^(23)` `6.022 xx 10^(24)` ` 6.022 xx 10^(25)` `6.022 xx 10^(22)` ANSWER :B

Discussion

4.	If 1.71 g of sugar (molar mass=342) is dissoved in 500 cm^(3) of a solution at 300K. What will be its osmotic pressure?
Answer» Solution :MASS of solute=`W_(2)=1.71g"Molar mass of solute"=M_(2)=342` Volume of solution=500`CM^(3)=0.5L`Temperature=300K R=0.083 L bar `mol^(-1)K^(-1)` Formula: `pi=(W_(2)RT)/(M_(2)V)` `pi=(1.71xx0.083xx300)/(342xx0.5)` answer `pi=0.249` bar.

Discussion

5.	If 1.5 moles of oxygen combine with Al to form Al_2O_3 , the mass of Al in g (Atomic mass of Al = 27] used in the reaction is
Answer» 2.7 54 40.5 81 Solution :`underset(4 xx 27 = 108 g )(4Al) + underset("3 mol")(3O_2) to 2Al_2O_3` Amount of AL that combines with 3 moles of `O_2` =108 g Amount of Al that combines 1.5 moles of `O_2 = 54 g `

Discussion

6.	If 1.5 moles of oxygen combine with Al to form Al_(2)O_(3), the mass of Al in g (Atomic mass of Al = 27 ) used in the reaction is
Answer» 2.7 54 40.5 81 Solution :`4Al+3O_(2)rarr 2Al_(2)O_(3)` 3 moles of `O_(2)` combine with Al = 4 moles `=4xx27g=108g` `THEREFORE"1.5 moles of "O_(2)" will combine with Al = 54 G."`

Discussion

7.	If 15 gm of a solute in 100 gm of water makes a solution that freezes at -1.0 ^(@) C ,then 30 gm of the same solution the freezes at
Answer» `-0.5^(@) C` `-2.0^(@)C` `0^(@)C` `2.0^(@)C` Solution :As the amount of the solute is doubled , `DELTA T ` will ALSO be doubled.

Discussion

8.	If 12.6 of NaHCO_(3) is added to 20.0g of HCl solution, the residue solution is found to weigh 24.0g. What is the mass and volume of CO_(2) released at NTP in the reaction?
Answer» SOLUTION :8.6g `CO_(@)` relased, VOLUME at NTP `=(22.4)/(44)xx8.6=4.378`.

Discussion

9.	If 15 Faraday quantity of electricity is passed through Al_((l))^(3+) solution then how many gram of Al metal will be obtained ? (cell effeciency is 80%)(at wt. Al = 27 gm"mol"^(-1))
Answer» 135 gm 121.5 gm 108 gm 94.5 gm Solution :`Al^(+3) + 3e^(-) to Al THEREFORE CE=("At. Wt.")/(V)=(27)/(3)=9` `therefore ` 1 Faraday of electricity DEPOSITES 9 G of Al `therefore ` 15 Faraday will PRODUCE 135 g of Al. If cell efficiency is 80% then `(80xx135)/(100)` =108 g of Al.

Discussion

10.	If 12g of a sample is taken, then 6g of a sampledecays in 1 hr. Findthe amount of sample showing decay in next hour.
Answer» 3g 1g 2g 6g Solution :HALF-life is 1 HR and thus in each half-life, half of the sample DECAYS.

Discussion

11.	If 10gm of water4 is added to 150gm of oleum (104.5%), then the find solution:
Answer» Mass of `SO_(3)` left is 10gm Mass of `H_(2)SO_(4)` is 156.75gm No water will be left LABELLING of new solution is 102.25% Solution :100gm convert into maximum 104.5gm `H_(2)SO_(4)` 150gm convert into pure `(104.5)/(100)xx150=156.75gm H_(2)SO_(4)` and `w_(H_(2)O)` requried `=(4.5)/(100)xx150=6.75gm` we add water `gt6.75gm` THUS no `SO_(3)` REMAIN after 10gm water addition labelling of new solution `=100%` as no `SO_(3)` left in it.

Discussion

12.	If 10^(21) molecules are removed from 200 mg of CO_(2), then the number of moles of CO_(2), then the number of moles of CO_(2) left are
Answer» `2.88xx10^(-3)` `1.66xx10^(-3)` `4.54xx10^(-3)` `1.66xx10^(-2)` Solution :200 mg of `CO_(2)=0.200g=(0.2)/(44)mol` `=(0.2)/(44)xx6.023xx10^(23)` molecules After removing `10^(21)` molecules, molecules left `=(2.738-1)10^(21)=1.738xx10^(21)` molecules `=(1.738xx10^(21))/(6.022xx10^(23))mol=2.88xx10^(-3)mol.`

Discussion

13.	If 10^(21) molecules are removed from 100mg CO_(2). Then number of moles of CO_(2) left are:
Answer» `6.10xx10^(-4)` `2.8xx10^(-3)` `2.28xx10^(-3)` `1.36xx10^(-2)` Solution :Number of molecules in 100 mg `CO_(2)` `=("Mass ")/("Molar mass")xx6.023xx10^(23)` `=(0.1)/(44)xx6.023xx10^(23)` `=1.368xx10^(21)` Molecules REMAINING `=1.368xx10^(21)-10^(21)=0.368xx10^(21)` Number of moles remaining `=(0.368xx10^(21))/(6.023)xx10^(23)=6.1xx10^(-4)`

Discussion

14.	If 1000 ml of a gas A at 600 torr and 500 ml of gas B at 800 torr are placed in a 2L flask, the final pressure will be
Answer» 500 TORR 1000 torr 850 torr 200 torr Solution :Applying Boyle.s LAW, Pressure of gas A in 2L flask`= ( 1000 xx 600)/( 2000) = 300` torr Pressure of gas B in 2L flasks `= ( 500 xx 800)/( 2000) = 200 ` torr Total pressure `= 300 + 200 = 500 ` torr

Discussion

15.	If 100 ml of 1.0 M NaOH solution is diluted to 1.0 L, the resulting solution contains
Answer» 1.0 mol of NaOH 0.1 mol of NaOH 10.0 mol of NaOH 0.05 mol of NaOH Solution :`M_(1)V_(1)=M_(2)V_(2)` `1.0xx100=M_(2)xx1000` or `M_(2)=(1.0xx100)/(1000)=0.1` `:. 0.1` MOLE of NaOH is PRESENT in 1.0 L solution.

Discussion

16.	If 100 ml of 0.1 M CH_3COOH and 200 ml of 0.03 M NaOH solutions are mixed together, then the pH of resulting mixture will be given [pk_a (CH_3COOH) = 4.74 and log 1.5 = 0.18)]
Answer» 3.97 4.92 4.35 5.52 Answer :3

Discussion

17.	If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C, the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1))
Answer» `10.0` 100 1000 186 Solution :`M_(B)=(K_(F)xx1000xx w_(B))/(w_(A) XX Delta T_(f))` `=(1.86xx1000xx10.0)/(100xx1.86)=100`

Discussion

18.	If 1.0 mole of I_(2) is introduced into 1.0 litre flask at 1000 K, at quilibrium (K_(e )=10^(-6)), which one is correct ?
Answer» `[I_(2)(g)] gt [1^(-1)(g)]` `[I_(2)(g)] LT [1^(-)(g)]` `[I_(2)(g)]=[I^(-)(g)]` `[I_(2)(g)]=(1)/(2)[I^(-)(g)]` Solution :`UNDERSET(I-x)(I_(2)) HARR underset(2x)(2I^(-))` `K_(E )=((2x)^(2))/((1-x))=10^(-6)` So In. shows that `(1-x)gt 2x therefore [I_(2)(g)] gt [I^(-)(g)]`

Discussion

19.	If 10 ml of 0.1 M aqueous solution of NaCl is divided into 1000 drops of equal volume. What will be the concentration one drop
Answer» 0.01 M 0.10 M 0.001 M 0.0001 M Solution :Molarity is INTRINSIC PROPERTY. HENCE it is INDEPENDENT of AMOUNT of solution.

Discussion

20.	If 10 mL of 0.1 M aqueous solution of NaCl is divided in to 1000 drops of equal volume, what will be the concentration of one drop ?
Answer» 0.01 M 0.10 M 0.001 M 0.0001 M ANSWER :B

Discussion

21.	If 1.0 kcal of heat is added to 1.2 L of O_(2) in a cylinder of constant pressure of 1 atm, the volume increases to 1.5 L. Calculate DeltaEandDeltaH of the process.
Answer» SOLUTION :`DeltaE=0.993"KCAL",DeltaH=1"kcal"`

Discussion

22.	If 10^(-4)dm^(3) of water is introduced into a 1.0dm^(3) flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established? (Given : Vapour pressure of H_(2)O at 300 K is 3170 Pa, R=8.314JK^(-1)mol^(-1))
Answer» `5.56xx10^(-3)` MOL `1.53xx10^(-2)` mol `4.46xx10^(-2)` mol `1.27xx10^(-3)` mol Solution :PV = NRT

Discussion

23.	If 1 mole of I_2 is introduced into 1.o litre flask at 1000 K, at equilibrium (K_c = 10^-6) which one is correct ?
Answer» [I_2(G)]GT`[I^-(g)]` [I_2(g)]LT`[I^-(g)]` [I_2(g)]=`[I^-(g)]` [I_2(g)]gt1/2`[I^-(g)]` ANSWER :A

Discussion

24.	If 1 mole H_(2) is reacted with 1 mole of the following compound. Which double bond will be hydrogenated ?
Answer» `C` `B` `a` `d` SOLUTION :N//A

Discussion

25.	If 1 mole of an ideal gas exapands isothermally at 37^(@)C from 15 L to 25L, the maximum work obtained is
Answer» 12.87 L atm 6.43 L atm 8.57 L atm 2.92 L atm Solution :`W_(rev)=-2.303nRT" log"(V_(2))/(V_(1))` `=-2.303xx1xx0.0821xx(273+37)XX"log"(25)/(15)` `=-2.303xx0.0821xx310xx"log"(5)/(3)` `=13J~~12.87L atm`

Discussion

26.	If 1 mL of a KMnO_(4) solution react with 0.140g Fe^(2+) and if 1 mL of KHV_(2)O_(4). H_(2C_(2)O_(4) solution react with o.1 mL of previous KMnO_(4) solution, how many millilitres of 0.20 M NaOH will react with 1 mL of previous KHC_(2)O_(4). H_(2)C_(2)O_(4) solution in which all the protons (H^(+)) are ionisable ?
Answer» 15/6 mL 13/16 `11/14` NONE of these Answer :A

Discussion

27.	If 1 ml of water contains 20 drops. Then no. of molecules in a drop of water is
Answer» `6.023xx10^(23)` molecules `1.376xx10^(26)` molecules `1.344xx10^(18)` molecules `4.346xx10^(20)` molecules Solution :22400 ML of water CONTAINS `=6.023xx10^(23)` molecules 1 ml of water contains `=(6.023xx10^(23))/22400` molecules `=20` drops `:.` 1 drop of water will contain `=(6.023xx10^(23))/(224xx2xx10^(3))` `=1.344xx10^(18)` molecules.

Discussion

28.	If 1 mL of water contains 20 drops, then number of molecules in a drop of water is
Answer» `6.023xx10^(23)` `1.376xx10^(26)` `1.6673xx10^(21)` `4.346xx10^(20)` Solution :1 ML of WATER = 1 g of water `(d_(H_(2)O)="1 g mL"^(-1))` `therefore" 20 drops of "H_(2)O=1 g` `therefore" 1 drop of "H_(2)O=(1)/(20)g=0.05g` `=(0.05)/(18)"mole"=(0.05)/(18)xx6.022xx10^(23)" molecules"` `=1.673xx10^(21)` molecules.

Discussion

29.	If 1 microgram of radium has disintegrated for 500 years, how many alpha particles will be emitted per second
Answer» `2.92 xx 10^(4) //s` `292 xx 10^(4)//s` `0.292 xx 10^(4)//s` `29.2 xx 10^(4)//s` SOLUTION :`2.92 xx 10^(4) ALPHA` -particles will be EMITTED per second

Discussion

30.	If 1 M CH_3COOHNa is added to 1M CH_3COOH:
Answer» PH of the SOLUTION increases pH decreases pH does not change None of these Answer :A

Discussion

31.	If 1 litre of N2 at 27^(@)C and 760 mm Hg contains N molecules, 4 litres of O_(2) under the same conditions of temperature and pressure, shall contain
Answer» N MOLECULES 2N molecules `N/4` molecules 4N molecules Answer :D

Discussion

32.	If 1 grain is equal to 64.8mg, how many moes of aspirin (mol. Wt. =169) are present in a 5 grain aspirin tablet?
Answer» SOLUTION :Mass of ASPIRIN in the tablet =`64.8xx5=324mg` =0.324g NUMBER of moles `=("Mass")/("Molar mass")=(0.324)/(169)` `=1.92xx10^(-3)`

Discussion

33.	If 1 faraday of electricity is passed through a solution of CuSO_4 the amount of copper deposited will be equal to its :
Answer» GRAM EQUIVALENT weight Gram MOLECULAR weight Atomic weight Electrochemical equivalent ANSWER :A

Discussion

34.	If 1/2 moles of oxygen combine with aluminium to form Al_(2)O_(3) then weight of Aluminium metal used in the reaction is (Al = 27 )
Answer» 27 g 18 g 54 g 40.5 g Answer :B

Discussion

35.	If 0.75 mole of an ideal gas expands isothermally at 27^(@)C from 15 litres to 25 litres, the maximum work obtained is
Answer» 8.40 J 9.34 J 10.86 J 10.43 JI Answer :B

Discussion

36.	If 0.561 g KOH is dissolved in water to give 200mL of solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is the pH ?
Answer» SOLUTION :`[KOH]=0.561/56 times1000/200M=0.050M` `As KOH toK^+ +OH^-,therefore[K^+][OH^-]=0.05M` `[H^+]=K_w//[OH^-]=10^-14//0.05=10^-14//(5 times10^-2)=2.0 times10^-13M` `pH=-LOG[H^+]=-log(2.0 times10^-13)=13-0.3010=12.699`

Discussion

37.	If 0.50 mol of CaCl_(2) is mixed with 0.20 mol of Na_(3)PO_(4), the maximum number of moles of Ca_(3)(PO_(4))_(2) which can be formed, is
Answer» `0.70` `0.50` `0.20` `0.10` Solution :`3 CaCl_(2)+2 Na_(3)PO_(4)to Ca_(3)(PO_(4))_(2)+6NaCl` `therefore` 2Moles of `Na_(3)PO_(4)=3` mole of `CaCl_(2) = 1` mole `Ca_(3)(PO_(4))_(2)` `therefore 0.2` mole of `Na_(3)PO_(4)=0.3` mole of `CaCl_(2)=0.1` mole of `Ca_(3)(PO_(4))_(2)`.

Discussion

38.	If 0.5 mole of BaCl_(2) is mixed with 0.2 mole of Na_(3)PO_(4) the maximum number of mole of Ba_(3)(PO_(4))_(2) that can be formed is
Answer» `0.7` `0.5` `0.30` `0.1` ANSWER :B

Discussion

39.	If 0.5 mole of BaCl_2 is mixed with 0.2 mole of Na_3PO_4the maximum number of mole ofBa_3(PO_4) that can be formed is :
Answer» 0.7 0.5 0.3 0.1 Answer :D

Discussion

40.	If 0.5 mol of BaCl_(2) is mixed with 0.2 mole of Na_(3)PO_(4), the maximum number of moles of Ba_(3)(PO_(4))_(2) that can be formed is
Answer» 0.7 0.5 0.3 0.1 Solution :`3BaCl_(2)+2Na_(3)PO_(4) rarr Ba_(3)(PO_(4))_(2)+6NaCl` `"3 moles of "BaCl_(2)" react with 2 moles of "Na_(3)PO_(4).Na_(3)PO_(4)" will be the LIMITING reactant."`

Discussion

41.	If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes the mass of silver deposited on cathode, is (eq. wt. of silver nitrate=108) :
Answer» 2.3523 g 3.3575 g 5.3578 g 6.3575 g Solution :(b) Current (I) =0.5 amp. TIME (t)=100 MM `=100xx60 sec.=6000 s`. EQ. MASS of Ag (E ) =108 `W=(EIt)/(96500)=(108xx0.5xx6000)/(96500)=3.3575 g`.

Discussion

42.	If 0.5 g of a solute (molar mass 100 g mol^(-1)) in 25 g of solvent elevates the boiling point by 1 K, the molar boiling point constant of the solvent is
Answer» 2 8 5 `0.5` Solution :`K_(B)=(DELTA T_(b)xx m_(B)xx w_(A))/(w_(B)xx1000)` `K_(b)=(1xx100xx25)/(0.5xx1000)=5.00`

Discussion

43.	If 0.4 gm NaOH is present in 1 litre solution, then its pH will be
Answer» 2 10 11 12 Solution :`[NAOH] =(0.4)/(40) = 0.01 M, [OH^(-)] = 10^(-2)M` `[H^(+)] = 10^(-12), pH = -log[H^(+)] = 12`.

Discussion

44.	If 0.228 g of silver salt of dibasic acid gabe a residue of 0.162 g of silver on ignition then molecular weight of the acid is
Answer» 70 80 90 100 Solution :Mass of silver SALT taken `=0.228 gm` Mass of silver LEFT `= 0.162gm` Basicity of ACID `= 2` Step 1 : To calculate the EQUIVALENT mass of the silver salt (E ). `(" Eq.mass of silver salt")/(" Eq. mass of silver ") = (" Mass of acid taken ")/(" Mass of silver left ")` `= E/ 108 = (0.0228)/(0.162)` `= E = (0.228)/(0.162) xx 108 = 152` (Eq. mass of silver salt) Step 2. to calculate the eq. mass of acid . Eq. mass of acid = Eq. mass of silver salt - Eq. mass of Ag + Basicity `= 152- 108 +1 = 152 - 107 = 45` (Eq. mass of acid) Step 3 : To determine the molecular mass of acid. Mol. mass of the acid = Eq. mass of acid `xx` basicity `= 45 xx 2 = 90`.

Discussion

45.	If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of AgNO_(3) yields 0.50 g of AgCl, the equivalent mass of the metal is (Ag=108, Cl = 35.5)
Answer» 21.9 20.04 40.08 43.80 Solution :`(E+"EQ. mass of" CL^(-))/("Eq. mass of AG + Eq. mass of" Cl^(-))=(E+35.5)/(108+35.5)` `=(0.2)/(0.5) implies E = 21.90`

Discussion

46.	If 0.2g of a gas 'X' occupies a volume of 440 ml and if 0.1 g of CO_(2) gas occupies a volume of 320 ml at the same temperature and pressure , X could be
Answer» `O_(2)` NO `C_(4) H_(10)` `SO_(2)` Solution :At same T and p, `( V_(1))/( V_(2))= ( n_(1))/(n_(2)) = ( W_(1)//M_(1))/(W_(2)//M_(2))` `( 440 ML )/( 320ml ) = ( 0.2 //M_(1))/( 0.1 // 44 )` `:. M _(1) = 66g mol^(-1)` Hence gas X could be `SO_(2)`

Discussion

47.	If 0.2 mol of O_(2) vapours can effuse from an opening in a heated vessel in 20 second how long will it take 8 gm H_(2) (g) to effuse under same conditions.
Answer» `600 sec` `100 sec` `400 sec` `200 sec` Solution :`(r_(O_(2)))/(r_(H_(2))) = SQRT((M_(H_(2)))/(M_(O_(2))))` `((DELTA n)/(t))_(o_(2))/(((Delta n)/(t))_(H_(2))) = sqrt((2)/(32))` `((0.2)/(20))/((4)/(t)) = sqrt((1)/(16))` `t = (400)/(4) = 100 sec`

Discussion

48.	If 0.2 mol of H_(2(g)) and 2.0 mol of S_((s)) are mixed in a 1dm^(3) vessel at 90^(@)C, the partial pressure of H_(2)S_((g)) formed according to the reaction H_(2(g))+S_((s))iffH_(2)S,K_(p)=6.8xx10^(-2) would be
Answer» <P>0.19 ATM 0.38 atm 0.6 atm 0.072 atm Solution :Suppose x moles of `H_(2)` have reacted, then at eqm., `[H_(2)]=(0.2-x),[H_(2)S]=x` `p_(H_(2))=(0.2-x)/(0.2-x+x)=(x)/(0.2)xxP` `K_(p)=(p_(H_(2)S))/(p_(H_(2)))i.e.,6.8xx10^(-2)=(x)/((0.2-x))` or, `0.068(0.2-x)=xorx=0.0127mol` Pressure of 0.0127 mol of `H_(2)S` at 363 K in 1 L vessel, `P=(NRT)/(V)=(0.0127xx0.0821xx363)/(1)=0.38atm`

Discussion

49.	If 0.2 g of fine animal charcoal is mixed with half litre of acetic acid (1 M) solution and shaken for 30 minutes
Answer» CONCENTRATION of the solution REMAINS same concentration of the solution increases concentration of the solution decreases none of these. Solution :Because acetic ACID gets adsorbed on charcoal.

Discussion

50.	If 0.2 ampere can deposit 0.1978 g of copper in 50 minutes, how much of copper will be deposited by 600 coulombs?
Answer» 19.78 g 1.978 g 0.1978 g 197.8 g Answer :C

Discussion

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If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes the mass of silver deposited on cathode, is (eq. wt. of silver nitrate=108) :

If 0.5 g of a solute (molar mass 100 g mol^(-1)) in 25 g of solvent elevates the boiling point by 1 K, the molar boiling point constant of the solvent is

If 0.4 gm NaOH is present in 1 litre solution, then its pH will be

If 0.228 g of silver salt of dibasic acid gabe a residue of 0.162 g of silver on ignition then molecular weight of the acid is

If 0.20 g chloride of a certain metal, when dissolved in water﻿ and treated with excess of AgNO_(3) yields 0.50 g of AgCl, the﻿ equivalent mass of the metal is (Ag=108, Cl = 35.5)﻿

If 0.2g of a gas 'X' occupies a volume of 440 ml and if 0.1 g of CO_(2) gas occupies a volume of 320 ml at the same temperature and pressure , X could be

If 0.2 mol of O_(2) vapours can effuse from an opening in a heated vessel in 20 second how long will it take 8 gm H_(2) (g) to effuse under same conditions.

If 0.2 mol of H_(2(g)) and 2.0 mol of S_((s)) are mixed in a 1dm^(3) vessel at 90^(@)C, the partial pressure of H_(2)S_((g)) formed according to the reaction H_(2(g))+S_((s))iffH_(2)S,K_(p)=6.8xx10^(-2) would be

If 0.2 g of fine animal charcoal is mixed with half litre of acetic acid (1 M) solution and shaken for 30 minutes

If 0.2 ampere can deposit 0.1978 g of copper in 50 minutes, how much of copper will be deposited by 600 coulombs?

If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of AgNO_(3) yields 0.50 g of AgCl, the equivalent mass of the metal is (Ag=108, Cl = 35.5)