94420 + Interview Questions in Chemistry in Class 12

1.	For a reaction, 2A +B to C + D ,-(d[A])/(dt)=k[A]^2[B]. The expression for (-d[B])/(dt) will be
Answer» `k[A]^2[B]` `1/2k[A]^2[B]` `k[A]^2[2B]` `k[2A]^2[B]` SOLUTION :`-1/2(d[A])/(dt)=(d[B])/(dt)`

Discussion

2.	For a reaction, 2A+BtoC+D,(d[A])/dt=K[A]^2[B]The expression for (d[B])/dtwill be:
Answer» `K[A]^2[B]` `(1/2)K[A]^2[B]` `K[A]^2[2B]` `K[2A]^2[B]` Answer :B

Discussion

3.	For a reaction 2A +B rarr C following information is known. Identify the options which is/are correct ? Information-1: When B is taken in very larger amount following graph was obtained. Information-2: When B = 1M and A = 2M are mixed and graph of (1)/([B]) us time is plotted adjoining graph is obtained.
Answer» order with RESPECT to A is 2 Order with respect to B is 2 Overall order is 2 Order with respect to ONE of the REACTANT will be negative. Answer :C::D

Discussion

4.	For a reaction 2A + B hArr Cwhere initial concentration of A=2M B=1M and C=0 the concentration of B at equilibrium is 0.5 M calculate the value of equilibrium constant for the reaction .
Answer» 0.5 2 1 1.5 Answer :C

Discussion

5.	For a reaction 2A+B rarr 3C, express the rate of reaction in terms of formation of the product ___________
Answer» `(1)/(2)(d[A])/(dt)` `(-1)/(3)(d[C])/(dt)` `(1)/(3)(d[C])/(dt)` `(-1)/(2)(d[B])/(dt)` Answer :C

Discussion

6.	For a reaction 2A+B hArr C+D, the partial pressures of A, B , C and D at equilibrium are 0.5, 0.8, 0.7 and 1.2 atmospheres respectively. The value of K_(p) for this reaction is
Answer» 0.24 2.4 0.42 4.2 Answer :D

Discussion

7.	For a radioactive substance with half-life period 500 years, the time for complete decay of 100 milligram of it would be
Answer» 1000 years `100 xx 500` years 500 years INFINITE time Solution :The time required for COMPLETE DECAY (I ORDER) is always infinite.

Discussion

8.	For a reaction 2A+ B ⇌ C + D , the partial pressure of A, B, C and D at equiliberium are 0.5, 0.8 0.7 and 1.2 atmosphere respectively. The value of k_p for this reaction is :
Answer» 4.2 2.4 0.42 0.24 Answer :A

Discussion

9.	For a reaction (1)/(2) A to 2B , rate of disapperance of 'A' is related to the rate of appearance of B by the expression
Answer» `- (d[A])/(dt) = (1)/(4) (d[B])/(dt)` `- (d[A])/(dt) = (d[B])/(dt)` `- (d[A])/(dt) = 4 (d[B])/(dt)` `- (d[A])/(dt) = (1)/(2) (d[B])/(dt)` SOLUTION :`(1)/(2) A to 2B , -2 (d[A])/(dt) = (1)/(2) (d[B])/(dt) , -(d[A])/(dt) = (1)/(4) (d[B])/(dt)`.

Discussion

10.	For a reaction 1//2 A rarr 2B rate of disappearance of A is related to rate of appearance of B by the expression
Answer» `(-d[A])/(dt)=4(d[B])/(dt)` `(-d[A])/(dt) = (1)/(4)(d[B])/(dt)` `(-d[A])/(dt)=1/2(d[B])/(dt)` `(-d[A])/(dt)=(d[B])/(dt)` Solution :For the given CHEMICAL EQUATION, we have `(1)/(V_A)(d[A])/(dt) = (1)/(V_B)(d[B])/(dt)` i.e `-(1)/(1//2)(d[A])/(dt) = (1)/(2)(d[B])/(dt)` HENCE, `-(d[A])/(dt) = 1/4 (d[B])/(dt)`

Discussion

11.	fora raction, I^(-)+OCI^(-)toIO^(-)+Cl^(-)in anaqueousmedium , therateof reactionis givenby (d[IO]^(-))/(dt)=k([I^(-)][OCI^(-)])/([OH^(-)])theoverallorderof reactionis
Answer» `-1` `0` `1` `2` Solution :`(d[IO]^(-))/(dt)=K([I^(-)]^(1)[OCI^(-)]^(1))/([OH^(-)]^(1))` ORDEROF reaction`=1+1-1=1`

Discussion

12.	For a radioactive decay the value ofk = 2.7xx10^(-3)s^(-1) and initial concentration is 160 moles /L. After 100s the concentration of radioactive elementis :
Answer» 76 mol /L 122 mol /L 50 mol /L 80 mol /L Solution :(B) `K = (2.303)/(t) "log" ([A]_(0))/([A])` `2.7xx10^(-3) = (2.303)/(100) "log" (160)/([A])` log` (160)/([A]) = 0.1172 ` `(160)/([A])` = 1.310 `:. [A] = (160)/(1.310)= 122 .13~~ 122 ` mol /L

Discussion

13.	For a pure liquid (Specific gravity 2) having molar mass 50gm/mol. Ratio of molarity to molality is:
Answer» 20kg/LITRE `(1)/(2)kg//litre` 200kg/litre 2000 kg/ `m^(3)` Solution :For pure LIQ. TAKE 1 litre volume `d=(2gm)/(ml)=("Mass of liquid")/(1000ml)rArr` Mass of liquid =200gm=2KG `(M)/(m)=(n_("liq"))/(V("lit"))xx("Mass of solvent"(kg))/(n_(("liq")))=(2)/(1)kg//litre` `=(2kg)/(10^(-3)m^(3))=2000kg//m^(2)`

Discussion

14.	For a primitive cubic crystal with a = 3 xx 10^10 m, what are the smallest diffraction angles 6 for (a) (1 0 0) and (b) (1 1 1) planes for lambda = 1.50 xx 10^(-10) m?
Answer» ANSWER :(a) `15.38^(@)`, (B) `25.66^(@)`

Discussion

15.	For a process A + B rarr product, the rate of reaction is second order with respect to A and zero order with respect to B. When 1 mole each of A and B are taken in 1 litre vessel the initial rate is 1xx10^(-2) mol/L-sec. The rate of reaction when 50% of the reactant have been converted to product would be :
Answer» `1XX10^(-2)` mol/L-sec `2.5XX10^(-3)` mol/L-sec `5xx10^(-2)` mol/L-sec `5xx10^(-3)` mol/L-sec Answer :D

Discussion

16.	For a primitive cubic crystal witha = 3 xx 10^(3)m,what is the smallest diffraction angle theta, for (1 1 0) plane for lambda = 1.50 xx 10^(-10) m?
Answer» SOLUTION :We have, `nlambda = 2dsin theta` and `d = a/SQRT(H^(2) + k^(2) + l^(2))` `therefore n lambda = (2a)/sqrt(h^(2) +k^(2) + l^(2)) sin theta` For first-degree reflection, n=1 `1 xx 1.50 xx 10^(-10) = (2 xx 3 xx 10^(10))/sqrt(1^(2) + 1^(2) +0^(2)) sin theta` `sin theta = (sqrt(2) xx 1.50 xx 10^(-10))/(6 xx 10^(10))` `therefore theta = 20.70^(@)`.

Discussion

17.	For a precious stone, 'carat' is used for specifying its mass. If 1 carat=3.168grains (a unit of mass) and 1 gram=15.4 grains find the total mass in kilogram of the ring that contains 0.5 carat diamond and 7 gram gold.
Answer» ANSWER :`7.1xx10^(-3)KG`

Discussion

18.	For a phase change H_2O(l)⇌H_2O(s) 0^@C,1 bar
Answer» `TRIANGLEG`=0 `TRIANGLES`=0 `TRIANGLEH`=0 `TRIANGLEU`=0 Answer :A

Discussion

19.	For a particular value of azimuthal quantum number (l), the total number of magnetic quantum number (m) is given by:
Answer» `L=(m+1)/2` `l=(2m+1)/2` `m=(2l-1)/2` `l=(m-1)/2` ANSWER :D

Discussion

20.	For a particular reversible reaction at temperature T,DeltaH and DeltaS were found to be +ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when
Answer» `T_(e) GTT` `T GT T_(e)` `T_(e)` is 5 times T `T=T_(e)` SOLUTION :`DeltaG=DeltaH-TDeltaS` At equilibrium `DeltaG=0` For a REACTION to be spontaneous `DeltaG` should be negative, so T should be greater than `T_(e)`.

Discussion

21.	For a particular reversible reaction at temperature T, DeltaHandDeltaS were found to be both +ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when
Answer» `T_(e)GTT` `TgtT_(e)` `T_(e)` is 5 TIMES T `T=T_(e)` SOLUTION :`DeltaG` should be NEGATIVE. `DeltaG=DeltaH-T_(e)DeltaS=0`

Discussion

22.	For a particular reversible reaction at temperature T, DeltaH and DeltaS were found to be both +ve. If is the temperature T_(e) at equilibrium, the reaction would be spontaneous when
Answer» `T=T_(E)` `T_(e) GT T` `T gt T_(e)` `T_(e)` is 5 TIME T Solution :`DeltaG=DeltaH-TDeltaS""[DeltaH=+ve, DeltaS=+ve]` `DeltaG=+ve-T_(e)(+ve)DeltaG=+ve-T_(e)(+ve)` if`T gt T_(e)` then `DeltaG=-ve` (SPONTANEOUS).

Discussion

23.	For a particular reversible reaction at temperature T. Delta H and Delta S were found to be both +ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when
Answer» `T_(E)` is five times T `T = T_(e)` `T_(e) GT T` `T gt T_(e)` Solution :`Delta G = Delta H - T Delta S` At EQUILIBRIUM, `Delta G = 0` `0 = Delta H - T Delta S` or `T_(e) Delta S = Delta H` `T_(e) = (Delta H)/(Delta S)` Since it is an endothermic reaction, its is favoured by high temperature so that `T gt T_(e)`.

Discussion

24.	For aparticular reaction DeltaH^(@) =-76.6 KJ and DeltaS^(@) JK^(-1). This reaction is :
Answer» SPONTANEOUS at the temperatures non- spontaneous at all temperatures Spontaneous at TEMPERATURE below `66^(@)C` Spontaneous at temperature above `66^(@)C` SOLUTION :N//A

Discussion

25.	For a particular reaction DeltaH^(@)=-38.3kJ and DeltaS^(@)=-113JK^(-1). this reaction is
Answer» spotaneous at all TEMPERATURE non-spotaneous at all temperature spontaneous at a temperature below 338 K spontaneous at a temperature above 339 K Answer :C

Discussion

26.	For a p - block element, its 3d, 3s, 3p and 4s orbitals are completely filled and the differentiating electron goes to the 4p orbital. The element should have its atomic number in the range
Answer» 13-18 21-26 31-36 49-54 Solution :`31-36 IFF` GA to KR.

Discussion

27.	For a particular reaction (A rightarrow 6) the rate constant is 0.693 min^(-1) If the initial concentration of the reactant, A is 1 M, then the rate of reaction after 1 minute will be
Answer» 0.35 M `MIN^(-1)` 0.14 M `min^(-1)` 0.693 M `min^(-1)` 0.30 M `min^(-1)` ANSWER :A

Discussion

28.	For a order reaction,show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer» Solution :90% reaction is complete so,CALCULATION of t (90%): Here ,90 % reaction is complete `therefore` REMAINING concentration is 10% If the initial concentration =`[R]_(0)` So, concentration after completion of 90% Reaction=`([R]_(0)xx10)/(100)=0.1 [R]_(0)=[R]_(90)` So,`t_(90)=(2.303)/(K)` log `([R]_(0))/([R]_(t))` `=(2.303)/(k)` log `(1)/(0.1)`{But `(1)/(0.1)`=10.0} `therefore t_(90)=(2.303)/(k)(1)=(2.303)/(k)` 99% reaction is complete,so , calculation of t(99%) : 99% reaction is complete so,1% reactant is remaining so,If the initial concentration =`[R]_(t)(99)` `[R]_(t)=1% of [R]_(0)=(1)/(100)[R]_(0)=0.01[R]_(0)` `t_((99%))=(2.303)/(k)` log `([R]_(0))/(0.01[R]_(0))=(2.303)/(k)xx2` `therefore (t_(99%))/(t_(90%))=((2.303)/(k))xx2xx((k)/(2.303))=(2)/(1)` So,if the first order reaction then time required for 99% completion is twice then time required for the completion of 90% of reaction.

Discussion

29.	For a non-ideal solution showing positive deviations, DeltaV_("mixing") is ………………..and DeltaH_("mixing") is ………… .
Answer» SOLUTION :POSITIVE, positive

Discussion

30.	For a non-ideal solution with a-ve deviation
Answer» `Delta H_(MIX)=-ve` `Delta V_(mix)=-ve` `Delta S_(mix)=-ve` `Delta G_(mix)=-ve` Solution :Explanation : (a) For -ve deviation `Delta H_(mix)` = -ve due to greater FORCES of attraction after mixing. STATEMENT (a) is correct. `Delta V_(mix)` = -ve because volume of resulting solution decreases as PARTICLES come close together. Statement (b) is correct `Delta G_(mix)=-ve` because `Delta G=Delta H-T Delta S=(-)Delta H-T(+Delta S)=-ve` Hence, choice (d) is correct. `Delta S_(mix) =+ve`, Entropy ALWAYS increases on mixing. Statement (c )is incorrect.

Discussion

31.	For a non-ideal gas, the compressibility factor (Z) is defined as: Z=(pV_(m))/(RT),V_(m)= Molar volume Compressibility of an unknown gas at 600K and 1.0 atm was found to be 1.2 Also this gas was found to effuse 1.58 times slower than the puremethane gas under identical condition. Answer the following three questions based on the above mentioned information and the information provided in an individual question. Molar volume of the gas in the given experimental condition is
Answer» `40.8L` `39.2L` `58.8L` `27.2L` ANSWER :C

Discussion

32.	For a non-ideal solution showing positive deviation from Raoult's law, Delta H_("mixing") mixing is ............ and delta V_("mixing") is ...............
Answer» SOLUTION :`+ve , - ve`

Discussion

33.	For a non-ideal gas, the compressibility factor (Z) is defined as: Z=(pV_(m))/(RT),V_(m)= Molar volume Compressibility of an unknown gas at 600K and 1.0 atm was found to be 1.2 Also this gas was found to effuse 1.58 times slower than the puremethane gas under identical condition. Answer the following three questions based on the above mentioned information and the information provided in an individual question. The value of the Virial coefficient 'B' in the Virial equation is, (Ignore the higher terms from equation during calculation) Virial equation : Z=1+B/(V_(m))+C/(V_(m)^(2))+D/(V_(m)^(3))+........., V_(m) is the molar volume
Answer» `8.16 L MOL^(-1)` `7.84Lmol^(-1)` `11.76 mol^(-1)` `5.44Lmol^(-1)` ANSWER :C

Discussion

34.	For a non-ideal gas, the compressibility factor (Z) is defined as: Z=(pV_(m))/(RT),V_(m)= Molar volume Compressibility of an unknown gas at 600K and 1.0 atm was found to be 1.2 Also this gas was found to effuse 1.58 times slower than the puremethane gas under identical condition. Answer the following three questions based on the above mentioned information and the information provided in an individual question. Density of the gas in the above mentioned experimental condition is
Answer» `0.98gL^(-1)` `0.68gL^(-1)` `1.02gL^(-1)` `1.47gL^(-1)` ANSWER :B

Discussion

35.	For a neutral amino acid (X), isoelectric point is 5.8. Now its solubility at this point in water is
Answer» maximum MINIMUM ZERO UNPREDICTABLE ANSWER :B

Discussion

36.	For a non-electrolyte solution, the Van't Hoff factor is equal to
Answer» Zero 1 2 Between 0 and 1 Answer :B

Discussion

37.	For a neutral amino acid (X), isoelectric point is 5.8 . Now is solubility at this piont in water is
Answer» MAXIMUM MINIMUM zero unpredicatable Answer :A

Discussion

38.	For a (N)/(10) solution of KMnO_4, its molarity will be:
Answer» 2M `(M)/(50)` `(M)/(20)` `(M)/(40)` SOLUTION :EQ. wt of `KMnO_4 =("Mol. wt")/(5)` `:. " Molarity " =(M)/(50) `

Discussion

39.	For a monoatomic gas, kinetic energy is equal to E, its relation with rms velocity is
Answer» `C=sqrt((2E)/M)` `C=sqrt((3E)/(2M))` `C=sqrt((E)/(2M))` `C=sqrt((E)/(3M))` ANSWER :A

Discussion

40.	For a molecule to be optically active, it should
Answer» contain at LEAST two `sp^(2)` hybridized CARBON atoms not be super imposable on its mirror image have TETRAHEDRAL geometry. super-imposable on its mirror image. ANSWER :B

Discussion

41.	For a mixture of two volatile , completely miscible liquids A and B , with P_(A)^(@)=500 " torr and " P_(B)^(@)=800 torr , what is the composition of last droplet of liquid remaining in equilibriumwith vapour ? Provided the initial ideal solution has a composition of x_(A) = 0.6 and x_(B)=0.4
Answer» `x_A=0.6,x_B=0.4` `x_A=0.5,x_B=0.5` `x_A=0.7,x_B=0.3` `x_A=0.3,x_B=0.7` ANSWER :C

Discussion

42.	For a lyophilic colloid in column 1, select correct combination
Answer» (II)(IV)(P) (III)(i)(P) (II)(i)(Q) (II)(iii)(Q) Solution :NA

Discussion

43.	For a liquid the vapour pressure is given by: log_(10)P=(-400)/(T)+10 Vapour pressure of the lqiuid is 10^(x) mm Hg. The value of x will be:
Answer» ANSWER :9

Discussion

44.	For a linear plt of log (x/m) versus log p in a Frundlich adsorption isoterms, which of the following statements is correct ? (K and n are constants)
Answer» Both K and 1/n appear in the slope term 1/n appears as the INTERCEPT Only 1/n appears as the slope log (1/n) appears as the intercept SOLUTION :For the Freundlich ADSORPTION isotherm EQUATION is `log ""((x)/(m))=log k +1/n log p` Comparing this equatin with `y=mx+c` Slope `(m) =1/m,` intercept (c ) = log k

Discussion

45.	For a linear plot of log(x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)
Answer» 1/n APPEARS as the intercept Only 1/n appears as the SLOPE. log(1/n) appears as the intercept. Both K and 1/n APPEAR in the slope term. Answer :B

Discussion

46.	For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct/ (k and n are constants)
Answer» LOG (1/n) appears as the intercept Both k and 1/n appear in the SLOPE term 1/n appears as the slope Only 1/n appears as the slope Solution :`(x)/(m)=KP^(1//n)` `log""(x)/(m)=(1)/(n)logP+logK therefore"Slopeis "(1)/(n)`

Discussion

47.	For a I order reaction A toBthe reaction rate at reactant concentration 0.01 M is found to be 2.5 xx 10^(-5) Ms^(-1). The half life period of the reaction is
Answer» 30s 300 s 220 s 347 s Solution :rate `=k[A]^1` `k = (2.0xx 10^(-5))/0.01 = 2.0 XX 10^(-3)s^(-1)` `:.t_(1//2)= 0.693/k = 0.693/(2XX 10^(-3))= 347s`

Discussion

48.	For a hypothetical reaction AtoL the rate expression is rate =-(dC_(A))/(dt)
Answer» NEGATIVE SIGN represents that rate is negative Negative sign pertains to the decrease in the CONCENTRATIONS of reactant Negative sign indicates the ATTRACTIVE forces between reactants. All of the above are correct Answer :B

Discussion

49.	For a hypothetical reaction, A+B to C+D, the rate =k[A]^(-1//2)[B]^(3//2). On doubling the concentration of A and B the rate will be
Answer» `4` TIMES `2` times `3` times none of the above Answer :B

Discussion

50.	For a hypothetical H like atom which follows Bohr's model, some spectral lines were observed as shown.If it is known that line 'E' belongs to the visible region, then the lines possibly belonging to ultra violet region will be (n_1 is not necessarily ground state ) [Assume for this atom, no spectral series shows overlap with other series in the emmission spectrum]
Answer» B and D D only C only A only Solution :In the GIVEN figure if line 'E' is in visible region then line belonging to ULTRAVIOLET region will have more energy then 'E' i.e. line A

Discussion

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