100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 1905 InterviewSolution

95201.	`R=(65 +- 1) ohm` , `l=(5+-0.1)mm` and `d=(10+-0.5)mm`.Find error in claculation of resistivity.A. 0.21B. 0.13C. 0.16D. 0.41
Answer» Correct Answer - B

Discussion

95202.	A organ pipe open on both ends in the `n^(th)` harmonic is in resonanance with a source of 1000 Hz The length of pipe is 16.6 cm and speed of sound in air is `332 m//s`.Find the vlue of n.A. 3B. 2C. 1D. 4
Answer» Correct Answer - C

Discussion

95203.	A particle doing SHM having amplitude 5 cm, mass 0.5 kg and angular frequency 5 `rad//s` is at 1cm from mean position.Find potential energy and kinetic energy.A. `KE=625xx10^(-4)J,PE=150xx10^(-3)J`B. `KE=150xx10^(-4)J,PE=625xx10^(-4)J`C. `KE=625xx10^(-4)J,PE=625xx10^(-4)J`D. `KE=150xx10^(-3)J,PE=150xx10^(-4)J`
Answer» Correct Answer - B

Discussion

95204.	The coefficient of volume expansion of glycerine is `49 xx 10^(-5)//^(@)C`. What is the fractional change in its density (approx.) for `30^(@)C` rise in temperature?A. 0.0155B. 0.0145C. 0.0255D. 0.0355
Answer» Correct Answer - B

Discussion

95205.	Name one special purpose fixative used in cytology laboratory. a) AAF fixative b) Carnoy’s fixative c) Formalin
Answer» b) Carnoy’s fixative Carnoy’s fixative is special purpose fixative used in cytology laboratory.

95205.

Name one special purpose fixative used in cytology laboratory. a) AAF fixative b) Carnoy’s fixative c) Formalin

Answer»

b) Carnoy’s fixative

Carnoy’s fixative is special purpose fixative used in cytology laboratory.

Discussion

95206.	The ideal fixative recommended in most of the laboratories for cytological specimens. It produces optimal nuclear details but some amount of cell shrinkage. Absolute (100%) ethanol produces a similar effect on cells. But is much more expensive. Identify the Routine fixative.
Answer» 95% Ethyl Alcohol (Ethanol)

Discussion

95207.	What do you mean by cytological fixatives and mention the properties of good cytological fixative.
Answer» It is critical to fix cytology specimens immediately after collection for proper preservation of the cellular components. It is important that no air-drying occurs prior to fixation. If a smear is already air-dried it should not be put in an alcohol fixative. Please note on the requisition if the slide (s) being submitted are fixed or air-dried. Properties of a good cytological fixative: • It should not excessively shrink or swell cells. • It should not distort or dissolve cellular components. • It should help preserve nuclear details. • It should improve optical differentiation and enhance staining properties of the tissues and cell components.

95207.

What do you mean by cytological fixatives and mention the properties of good cytological fixative.

Answer»

It is critical to fix cytology specimens immediately after collection for proper preservation of the cellular components. It is important that no air-drying occurs prior to fixation. If a smear is already air-dried it should not be put in an alcohol fixative. Please note on the requisition if the slide (s) being submitted are fixed or air-dried. Properties of a good cytological fixative:

• It should not excessively shrink or swell cells.

• It should not distort or dissolve cellular components.

• It should help preserve nuclear details.

• It should improve optical differentiation and enhance staining properties of the tissues and cell components.

Discussion

95208.	In a tyre of "Ferrari" car of Mr. Obama, a tube having a volume of 12.3 litres is filled with air at a pressure of 4 atm at 300 K.Due to travelling, the temperature of the tube and air inside it raised to 360 K and pressure reduced to 3.6 atm in 20 minutes. If the porosity (number of pores per unit area) of the tube material is `5xx10^3` pores/`cm^2` and each pore can transfer air from inside to outside of the tube with the rate of `6.023xx10^8` molecules per minute.Calculate the total surface area `(m^2)` of the tube.(R=0.082 Lt-atm/mole-K) Give your answer divide by 100.
Answer» Correct Answer - 50 Initial moles of air inside the tube `=(4xx12.3)/(0.082xx300)=2` moles of air after 20 minutes`=(3.6xx12.3)/(0.082xx360)=1.5` `:.` moles of air leaked in 20 minutes =2-1.5=0.5 `:.` molecules of air leaked in 20 minutes `=0.5 N_A` Let the total surface area of tube =`A cm^2` Total molecules escaped in 20 minutes`=(5xx10^6)A xx 6.023xx10^(8)xx20=0.5 N_A` `:. A=(0.5xx6.023xx10^23)/(5xx10^5xx6.023xx10^8xx20)=5xx10^7 cm^2 =5000 m^2`

Discussion

95209.	Match List : `{:("Column-I","Column-II"),((A)PV^(gamma)="Constant",(p)"Expansion of ideal gas in vacuum"),((B)DeltaT=O,(q)Z=1-a/(V_mRT)),((C )"For" H_2 and He at 0^@C,(r)"Adiabatic reversible process"),((D)"At low density of gas",(s)Z=1+(pb)/(RT)):}`
Answer» Correct Answer - A`to` r ; B`to` p ; C`to`S ; D`to`Q `{:((A)PV^(gamma)="Constant","Adibatic reversible process"),((B)DeltaT=O,"In free expansion"),(( C)"For" H_2 "and He"Zgt1,"repulsive forces diminate and" Z=1+(pd)/(RT)"at all pressure"),((D)"Low density"to"low pressure",):}` `Zlt1` `Z=1-a/(V_mRT)`

Discussion

95210.	Consider the following figure at 500 K.Assuming ideal gas behaviour, calculate the total pressure if the barriers are removed from the compartment.Assume that the volume of barriers is negligible.
Answer» Correct Answer - 2.85 atm Number of moles of A,B,C and D are given as : `n_A=(PV)/(RT)=(1xx1)/(RT),n_B=(1.5xx2)/(RT)=(3)/(RT)` `n_C=(2.5xx4)/(RT)=(10)/(RT),n_D=(2xx3)/(RT)=(6)/(RT)` `n=n_A+n_B+n_C+n_D=(1+3+10+6)/(RT)=20/(RT)` Total pressure is given as `P_("Total")=(nRT)/Vimplies P_("Total")=20/(RT)xx(RT)/7=2.85`

Discussion

95211.	For `I^(st)` order decomposition of `SO_2Cl_2(g)`, `SO_2Cl_2(g)toSO_2(g)+Cl_2(g)` a graph of log (a-x) `v//s` t is shown in figure answer the following using above information. What is the rate constant (in `sec^(-1)`) ?A. 0.2B. `4.6xx10^(-1)`C. `7.7xx10^(-3)`D. `1.15xx10^(-2)`
Answer» Correct Answer - C `{:(,SO_2Cl_2(g)to,SO_2(g)+,Cl_(2)(g)),(at " "t=0,a,0,0),(at " " t=t ,(a-x), x, x):}` `K=2.303/t "log" (a/(a-x))` `log (a-x)=[-K/2.303]t+log a` slope=`-K/2.303=-2/10` `K=2/10xx2.303/60sec^(-1)` `=7.7xx10^(-3) sec^(-1)`

Discussion

95212.	Assertion (A) : If the activation energy of a reaction is zero, temperature will have no effect on the rate constant. Reason (R ): Lower the activation energy, faster is the reaction.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - B According to arrhenius equation `K=Ae^(-Ea//RT)` when `E_a=0, K=A`

Discussion

95213.	Which of the following is incorrect statement?A. Stoichiometery of a reaction tells about the order of the elementary reactions.B. For a zero order reaction, rate and the rate constant are identical.C. A zero oder is controlled by factors other than concentration of reactions.D. A zero order reaction is an elementary reaction.
Answer» Correct Answer - D zero order reaction is always a complex reaction.

Discussion

95214.	The equilibrium constant `K_(P)` for the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `4.5 atm`. What would be the average molar mass (in `g//mol`) of an equilibrium mixture at a total pressure of `2 atm` of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` ?A. `69`B. `57.5`C. `80.5`D. `85.5`
Answer» Correct Answer - B `K_(P) = (4prop^(2))/(1 - prop^(2)) P` `P = 2 atm , K_(P) = (9)/(2)` `prop =` degree of dissociation of `N_(2)O_(4)(g)` `25prop^(2) = 9 , prop = (3)/(5) = 0.6` `prop = (M_(t) - M_(mix))/((n-1)M_(mix))` `M_(exp) = (92)/(1 + 0.6) = 57.5 g//mol`

Discussion

95215.	What will be the effect o the equilibrium constant on increasing temperature. If the reaction neither absorbs heat nor releases heat?A. Equilibrium constant will remain constant.B. Equilibrium constant will decreaseC. Equilibrium constant will increaseD. Can not be predicted.
Answer» Correct Answer - A `log((K_(2))/(K_(1)))=(DeltaH)/(Rxx2.303)[(1)/(T_(1)-(1)/(T_(2)))]` if `DeltaH=0` Then `K_(2)=K_(1)` Means no effect.

Discussion

95216.	The equilibrium constant `K_(P)` for the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `4.5 atm`. What would be the average molar mass (in `g//mol`) of an equilibrium mixture at a total pressure of `2 atm` of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` ?A. 69B. 57.5C. 80.5D. 85.5
Answer» Correct Answer - B `underset(P-Palpha)underset(P)(N_(2))O_(4)(g)hArrunderset(2Palpha)(2NO_(2)(g))` is 4.5 `P-Palpha+2Palpha=2` `P+Palpha=2` `4.5=(4P^(2)alpha^(2))/(P(1-alpha))` `4.5=(4Palpha^(2))/(1-alpha)` `4.5(1-alpha)=4Palpha^(2)` `P=(4.5(1-alpha))/(4alpha^(2))` `P(1+alpha)=2` `(4.5(1-alpha)(1+alpha))/(4alpha^(2))=2` `4.5(1-alpha^(2))=8alpha^(2)` `4.5-4.5alpha^(2)=8alpha^(2)` `4.5=12.5alpha^(2)` `alpha=sqrt((4.5)/(12.5))` `alpha=0.6` `alpha=(M-EMM)/(EMM(n-1))` `0.6EMM=92-EMM` `1.6EMM=92` `EMM=(92)/(1.6)` `=57.5`

Discussion

95217.	The conversion of ozne into oxygen is exotehrmin under what conditions is ozone is most stable? `2O_(3)(g)hArr3O_(2)(g)`A. At low pressure and low temperatureB. At high pressure and high temperatureC. At high pressure and low temperatureD. At low pressure and high temprature.
Answer» Correct Answer - B For exothermic reactio high temperature favour backward reaction and with increase in pressure reaction goes where number of moles are less

Discussion

95218.	Why in general does not proceed with a uniform rate throughout?
Answer» Because the rate of reaction at any time depends upon the concentration of the reactants at that time which keep on decreasing with time.

Discussion

95219.	In moist air, copper corrodes to produce a green layer on the surface. Give reason.
Answer» In moist air, copper reacts with oxygen `O_2` moisture `H_2O` and `CO_2` to produce basic copper corbonate. Which is green in colours. `2Cu+O_2+H_2O+CO_2toCuCO_3.Cu(OH)_2darr`

Discussion

95220.	While hatred against a member of the enemy race isjustifiable, especially during war time, what makes a human being rise above narrow prejudices?
Answer» News of war is fast becoming a way of life. The moment one picks up a newspaper, one is bombarded with news of wars between different countries, directly or indirectly. It is obvious that the countries at war are enemies and hatred is a part of this enmity. However, the success of humanity comes when we rise above this enmity and show our love towards the civilization as a whole. Dr Sadao did the same. He did whatever he could to save the life of a man whom he knew was a war prisoner. The instant he saw the injured man, he was filled with concern. Ignoring the fact that he was the enemy of his country and must have killed so many Japanese and may kill even more, if alive, he saved him.

Discussion

95221.	Which one of the following is a reversible reaction? (a) Ripening of a banana (b) Rusting of iron (c) Tarnishing of silver (d) Transport of oxygen by Hemoglobin in our body
Answer» (d) Transport of oxygen by Hemoglobin in our body All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.

95221.

Which one of the following is a reversible reaction? (a) Ripening of a banana (b) Rusting of iron (c) Tarnishing of silver (d) Transport of oxygen by Hemoglobin in our body

Answer»

(d) Transport of oxygen by Hemoglobin in our body

All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.

Discussion

95222.	_____ repeats itself. A) children B) language C) everybody D) stories E) history
Answer» Correct option is E) history

Discussion

95223.	Which two factors displacement requires ? A. magnitude B. direction C. Both A and B D. none of the above
Answer» C. Both A and B

Discussion

95224.	Give the hazard motion of a honey bee. A. two-dimensional. B. three-dimensional C. linear D. one-dimensional
Answer» B. three-dimensional

Discussion

95225.	Which type of motion is “the pendulum of a wall clock moves at regular intervals” ? A. rectilinear B. Periodic C. Circular D. none of the above
Answer» Periodic. Periodic motion is motion that repeats itself at regular intervals of time. Every body executing circular motion can be said to be executing periodic motion

95225.

Which type of motion is “the pendulum of a wall clock moves at regular intervals” ? A. rectilinear B. Periodic C. Circular D. none of the above

Answer»

Periodic.

Periodic motion is motion that repeats itself at regular intervals of time. Every body executing circular motion can be said to be executing periodic motion

Discussion

95226.	Which type of motion is “The hands of a clock” ? A. Circular B. rectilinear C. Periodic D. none of the above
Answer» Circular. A Here, all objects rotate in circular motion

95226.

Which type of motion is “The hands of a clock” ? A. Circular B. rectilinear C. Periodic D. none of the above

Answer»

Circular.

A Here, all objects rotate in circular motion

Discussion

95227.	Which type of motion repeats itself at regular intervals of time? A. Circular motion B. Periodic motion C. Rectilinear motion D. none of the above
Answer» Periodic motion. Periodic motion is motion that repeats itself at regular intervals of time. Everybody executing circular motion can be said to be executing periodic motion.

95227.

Which type of motion repeats itself at regular intervals of time? A. Circular motion B. Periodic motion C. Rectilinear motion D. none of the above

Answer»

Periodic motion.

Periodic motion is motion that repeats itself at regular intervals of time. Everybody executing circular motion can be said to be executing periodic motion.

Discussion

95228.	`e//m` ratio in case of anode ray experiment is different for different gases. The ion of gases formed after the ejection of electron are different of gas is different.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A Specific charge depends on mass of ion, which is different for different gases.

Discussion

95229.	Statement-1 : If an electron is located within the range of `0.1 Å` then the uncertainly in velocity is approximately `6xx10^6 m//s` Statement-2 : Trajectory (path of motion) of above electron can be defined. `[h=6.6xx10^(-34), m_e=9.1xx10^(-31) kg]`A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - C `mDeltav Deltax=h/(4pi)implies Deltav=(6.62xx10^(-34))/(9.1xx10^(-31)xx4xx3.14xx10^(-11))=6xx10^6 m//s` as uncertainly in velocity is very high so we cannot define the trajectory of an electron.

Discussion

95230.	If hydrogen atoms (in the ground state) are passed through an homogeneous magnetic field, the beam is split into two parts.This interaction with the magnetic field shows that the atoms must have magnetic moment.However, the moment cannot be due to the orbital angular momentum since l=0. Hence one must assume existance of intrinsic angular momentum, which as the experiment shows, has only two permitted orientations. Spin of the electron produces angular momentum equal to `s=sqrt(s(s+1))h/(2pi)` where s=+`1/2`. Total spin of an atom =+`n/2 or n/2` where n is the number of unpaired electron. The substance which contains species with unpaired electrons in their orbitals behave as paramagnetic substances.The paramagnetism is expected in terms of magnetic moment. The magnetic moment of an atom `mu=sqrt(s(s+1))(eh)/(2pimc)=sqrt(n/2(n/2+1))(eh)/(2pimc)s=n/2` `rArrmu_s=sqrt(n(n+2))B.M.` n=number of unpaired electrons 1. B.M.(Bohr magneton)`=(eh)/(2pimc)` If magnetic moment is zero the substance is diamagnetic. Which of the following is a paramagnetic substanceA. `Mg^(2+)`B. `Cu^(+)`C. `Mn^(+7)`D. `Ti^(+2)`
Answer» Correct Answer - D There is two unpaired electron in `Ti^(2+)`

Discussion

95231.	Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Answer» It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process.

Discussion

95232.	The French physical Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength `lambda` of a material particle, its linear momentum p and planck constant h. `lambda=h/p=h/(mv)` The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than fighter particles.The wave associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they (I)have lower velocities (II)have no electrical and magnetic fields and (III)are not emitted by the particle under consideration The experiment confirmation of the de Broglies relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffrated by a nickel crystal.As diffraction is a characteristic property of waves hence the beam of electron behaves as a wave, as proposed by de broglie. De-Broglie wavelength of an electron travelling with speed equal to 1% of the speed of lightA. 400 pmB. 120 pmC. 242 pmD. 375 pm
Answer» Correct Answer - C `lambda=h/(mv)=(6.6xx10^(-34))/(9.1xx10^(-31)xx3xx10^6)=242` pm

Discussion

95233.	Why is it necessary to remove CO (carbon monoxide) when ammonia is obtained by Haber"s process ?
Answer» CO acts as poison catalyst for Habers process therefore, it will lower the activity of the Fe catalyst.

Discussion

95234.	Instead of principle quantum number (n), azimuthal quantum number (L) &magnetic quantum number m, a set of new quantum number s, t & u was introduce with similar logic but different value as defined below `s=1,2,3,...........oo` all positive integral values. `t=(s^(2)-1^(1)),(s^(2)-2^(2)),(s^(2)-3^(3))`......... No negative value `u=((t+1))/2"to"+((t+1))/2`(including zero , if any ) in integral step . Each orbital can have maximum four electrons . (s+t) rule is defined ,similar to `(n+l)` rule. The number of subshells in which the third shell is sub divided equal toA. 1B. 3C. 5D. 7
Answer» Correct Answer - B S=1 t=0 `u=-1/2,1/2` No of subshell=1 , No of orbital=2 No of electrons =8 S=2 t=0,3 `u=-1/2,1/2`(for t=0) u=-2,-1,0,+1,+2(for t=3) No of subshell=2, No of orbital=7 No of electrons=28 S=3 t=0,5,8 `u=-1/2,1/2`(for t=0), u= -3,-2,-1,0,1,2,3 (for t=5) `u= -9/2, -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2, 9/2`(for t=8) No of subshell=3, No of orbital=19 No of electrons =76 (1,0),(2,0),(3,0),(4,0),(2,3)....Energy order of (S,t) For Atomic No. 24 (S+t) rule. There is no electron in S=2,t=3

Discussion

95235.	Instead of principle quantum number (n), azimuthal quantum number (L) &magnetic quqntum number m, a set of new quantum number s, t & u was introduce with similar logic but different value as defind below `s=1,2,3,...........oo` all positive integral values. `t=(s^(2)-1^(1)),(s^(2)-2^(2)),(s^(2)-3^(3))`......... No negative value `u=((t+1))/2"to"+((t+1))/2`(including zero , if any ) in integaral step . Each orbital can have maximum four electrons . (s+t) rule is defined ,similar to (n+l) rule. Number of electrons that can be accommodated in s=2 and s=3 shell.A. 14, 38B. 28, 76C. 8, 28D. None of these
Answer» Correct Answer - B S=1 t=0 `u=-1/2,1/2` No of subshell=1 , No of orbital=2 No of electrons =8 S=2 t=0,3 `u=-1/2,1/2`(for t=0) u=-2,-1,0,+1,+2(for t=3) No of subshell=2, No of orbital=7 No of electrons=28 S=3 t=0,5,8 `u=-1/2,1/2`(for t=0), u= -3,-2,-1,0,1,2,3 (for t=5) `u= -9/2, -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2, 9/2`(for t=8) No of subshell=3, No of orbital=19 No of electrons =76 (1,0),(2,0),(3,0),(4,0),(2,3)....Energy order of (S,t) For Atomic No. 24 (S+t) rule. There is no electron in S=2,t=3

Discussion

95236.	Instead of principle quantum number (n), azimuthal quantum number (L) &magnetic quqntum number m, a set of new quantum number s, t & u was introduce with similar logic but different value as defind below `s=1,2,3,...........oo` all positive integral values. `t=(s^(2)-1^(1)),(s^(2)-2^(2)),(s^(2)-3^(3))`......... No negative value `u=((t+1))/2"to"+((t+1))/2`(including zero , if any ) in integaral step . Each orbital can have maximum four electrons . (s+t) rule is defined ,similar to (n+l) rule. Number of electronc for which s=2,t= 3 for an element with atomic number 24.A. 8B. 4C. 0D. None of these
Answer» Correct Answer - C S=1 t=0 `u=-1/2,1/2` No of subshell=1 , No of orbital=2 No of electrons =8 S=2 t=0,3 `u=-1/2,1/2`(for t=0) u=-2,-1,0,+1,+2(for t=3) No of subshell=2, No of orbital=7 No of electrons=28 S=3 t=0,5,8 `u=-1/2,1/2`(for t=0), u= -3,-2,-1,0,1,2,3 (for t=5) `u= -9/2, -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2, 9/2`(for t=8) No of subshell=3, No of orbital=19 No of electrons =76 (1,0),(2,0),(3,0),(4,0),(2,3)....Energy order of (S,t) For Atomic No. 24 (S+t) rule. There is no electron in S=2,t=3

Discussion

95237.	which of the following order/s is/are cprrect fpr the properties mentione in the breaket ?A. `-overset(o+)NH_(3)gt-NO_(2)[-I]`B. `-overset(ɵ)(O)gt-CH_(2)^(ɵ)[+I]`C. `-oversetoverset(O)(\|\|)C-O-oversetoverset(O)(\|\|)C-Rgt-oversetoverset(O)(\|\|)C-OR[-M]`D. `overset(..)(-NH_(2))gt-overset(..)underset(..)(S)H[+M]`
Answer» Correct Answer - A::C::D From order of `+I, -I, +M` & `-M`.

Discussion

95238.	In which of the following one benzene ring is attched to `+ m` and another is attached to `- m` group ?A. B. C. D.
Answer» Correct Answer - A::C::D

Discussion

95239.	Phenols are weakly avidic
Answer» Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form.

Discussion

95240.	Given below are two statements. Statement I : Phenols are weakly acidic. Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option: (A) Both Statement I and Statement II are correct. (B) Both Statement I and Statement II are incorrect. (C) Statement I is correct but Statement II is incorrect. (D) Statement I is incorrect but Statement II is correct
Answer» (C) Statement I is correct but Statement II is incorrect. Phenol are weakly acidic. Phenol is more acidic than alcohol & H₂O statement (I) is correct. (II) is incorrect.

95240.

Given below are two statements. Statement I : Phenols are weakly acidic. Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option: (A) Both Statement I and Statement II are correct. (B) Both Statement I and Statement II are incorrect. (C) Statement I is correct but Statement II is incorrect. (D) Statement I is incorrect but Statement II is correct

Answer»

(C) Statement I is correct but Statement II is incorrect.

Phenol are weakly acidic. Phenol is more acidic than alcohol & H₂O statement (I) is correct. (II) is incorrect.

Discussion

95241.	Lithium forms bcc crystals. Calculate the atomic radius of lithium if the length of the side of a unit cell of lithium is 351 pm.
Answer» Radius of unit cell of lithium = 351 pm Then, (351 pm)³ = (351 x 10^-10 cm)³ = 8.432 x 10^-23 cm³ approx

95241.

Lithium forms bcc crystals. Calculate the atomic radius of lithium if the length of the side of a unit cell of lithium is 351 pm.

Answer»

Radius of unit cell of lithium = 351 pm

Then, (351 pm)³ = (351 x 10^-10 cm)³

= 8.432 x 10^-23 cm³ approx

Discussion

95242.	A gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of 1 atm from a volume of 2 litre to a volume of 6 litre. In doing so, it absorbs 800 J heat from surroundings. Determine increase in internal energy of process.
Answer» Since,work is done against constant pressure and thus, irreversible. Given, ΔV = (6 – 2) = 4 litre; P = 1 atm ∴ W = – 1 × 4 litre-atm = – 4 × 1.01325 × 10² J = 405.3 J Now from first law of thermodynamics q = ΔU – W 800 = ΔU + 405.3 ∴ ΔU = 394.7 Joule

95242.

A gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of 1 atm from a volume of 2 litre to a volume of 6 litre. In doing so, it absorbs 800 J heat from surroundings. Determine increase in internal energy of process.

Answer»

Since,work is done against constant pressure and thus, irreversible.

Given, ΔV = (6 – 2) = 4 litre; P = 1 atm

∴ W = – 1 × 4 litre-atm = – 4 × 1.01325 × 10² J

= 405.3 J

Now from first law of thermodynamics

q = ΔU – W

800 = ΔU + 405.3

∴ ΔU = 394.7 Joule

Discussion

95243.	The osmotic pressure of solution containing `34.2 g` of cane sugar (molar mass = 342 g `mol^(-1)` ) in 1 L of solution at `20^(@)C` is (Given `R = 0.082` L atm `K^(-1) mol^(-1)` )A. 2.40 atmB. 3.6 atmC. 24 atmD. 0.0024 atm
Answer» Correct Answer - A Data: `W_(2) = 34.2 g, M_(2) = 342 gmol^(-1),V=1 L` `T = 20 xx 273 = 293 K` and `R=0.082 L atm K^(-1), pi=?` Calculation: To calculate `pi` `pi= n_(2)RT` `pi = W_(2)/M_(2) xx (RT)/V` `=(34.2 g xx 0.082 L. atm K^(-1) mol^(-1) xx 293 K)/(342 g mol^(-1) xx 1L)` `=(34.2 xx 82 xx 293 xx 10^(-3))/(342)` atm `pi = 2.40` atm

Discussion

95244.	The pressure of a gas is 100 kPa. If it is compressed form 1m3 to 10 dm 3, find the work done. (a) 990 J (b) 9990 J (c) 9900 J (d) 99000J
Answer» The Correct option is (d) 99000J

Discussion

95245.	Calculate the work done during compression of 2 mol of an ideal gas from a volume of `1m^(3)` to `10 dm^(3)` 300 K against a pressure of 100 KPa .A. 99 kJB. `-99` kJC. `114.9` kJD. `-114.9` kJ
Answer» Correct Answer - A n=2 moles, `V_(1) = 1m^(3) = 10^(3)dm^(3), V_(2) = 10 dm^(3), P_(ex) = 100 kPa` For isothermal Irreversible process `W = -Pex (V_(2)-V_(1))` `=-100 kPa (10 dm^(3) -1000 dm^(3))` `=-100 (-900) kPa xx dm^(3) = 99000 J` = 99 kJ

Discussion

95246.	The overall reaction taking place at anode during electrolysis of fused sodium chloride using suitable electrode isA. Oxidation of ChlorineionsB. Reduction of sodium ionsC. Reductions of chlorineD. Oxidation of sodium atoms
Answer» Correct Answer - A Electrolysis of fused NaCl `NaCl to Na^(+) + Cl^(-)` At cathode `(-Ve)` `2Na + 2e^(-) to 2Na` (Reduction) At anode `(+Ve)`: `2Cl^(-) to Cl_(2) + 2e^(-)` (Oxidation)

Discussion

95247.	`0.5F` of electricity is passed through `500mL` of copper sulphate solution. The amount of copper which can be deposited will beA. ` 63. 5G`B. ` 31. 75 g`C. ` 15. 8g`D. Unpredictable
Answer» Correct Answer - C `Cu^(2+) +2e rarr Cu` ` 2F=1 mol of Cu = 63. 5 g ` of `Cu` ` 0.5 F= (63.5)/2 xx 0 =15 . 8 g` .

Discussion

95248.	The process of passing of a precipitate into colloidal solution on adding an electrolyte is calledA. DialysisB. PeptizationC. ElectrophoresisD. Electro-osmosis
Answer» Correct Answer - B Peptization is the process of passing of a precipitate into colloidal solution on the addition of electrolyte.

Discussion

95249.	The modern atomic mass scale is based on :-A. `C^(12)`B. `o^(16)`C. `H^(1)`D. `C^(13)`
Answer» Correct Answer - A

Discussion

95250.	The statement that is not correct for the periodic classification of elements is `:`A. the properties of elements are the periodic functions of their atomic numbers.B. non`-` metallic elements are lesser in number than metallic elementsC. the first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic numberD. for transition elements the `d-` subshells are filled with electrons monotonically with increase in atomic number.
Answer» Correct Answer - 4 The `d-` sub shells are not filled with electrons monotonically with increase in atomic number. There are some exceptions like `Cr,Cu ` etc.

Discussion

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The osmotic pressure of solution containing `34.2 g` of cane sugar (molar mass = 342 g `mol^(-1)` ) in 1 L of solution at `20^(@)C` is (Given `R = 0.082` L atm `K^(-1) mol^(-1)` )A. 2.40 atmB. 3.6 atmC. 24 atmD. 0.0024 atm

The pressure of a gas is 100 kPa. If it is compressed form 1m3 to 10 dm 3, find the work done. (a) 990 J (b) 9990 J (c) 9900 J (d) 99000J

Calculate the work done during compression of 2 mol of an ideal gas from a volume of `1m^(3)` to `10 dm^(3)` 300 K against a pressure of 100 KPa .A. 99 kJB. `-99` kJC. `114.9` kJD. `-114.9` kJ

The overall reaction taking place at anode during electrolysis of fused sodium chloride using suitable electrode isA. Oxidation of ChlorineionsB. Reduction of sodium ionsC. Reductions of chlorineD. Oxidation of sodium atoms

`0.5F` of electricity is passed through `500mL` of copper sulphate solution. The amount of copper which can be deposited will beA. ` 63. 5G`B. ` 31. 75 g`C. ` 15. 8g`D. Unpredictable

The process of passing of a precipitate into colloidal solution on adding an electrolyte is calledA. DialysisB. PeptizationC. ElectrophoresisD. Electro-osmosis

The modern atomic mass scale is based on :-A. `C^(12)`B. `o^(16)`C. `H^(1)`D. `C^(13)`