100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Abhijith owns an old car. He sells it and buys a new car. While Ajith purchases a few cars for selling at an assured profit. Do Abhijith and Ajith engage in business activities? Give reason.
Answer» The activity of Abhjith is not a business activity. But we can say Mr. Ajith engages in business activities. Business is a continuous activity, so there must be regularity in its dealings and its ultimate objective is earning profit. Abhijith’s transaction is casual. So it is not a business activity.

Discussion

2.	Below are given certain non-economic activities. Convert them into economic activities. Example: A nurse attending her ailing son. (Non-economic activity) A nurse attending patients in her hospital. (Economic activity) 1. A person working in his own garden. 2. A lady preparing food for her husband. 3. A man white-washing his own house. 4. A teacher teaching his son at home. 5. A chartered accountant preparing his own accounts.
Answer» 1. A person working in a school garden. 2. A lady preparing food in a restaurant. 3. A man white-washing the building of a trade centre. 4. A teacher teaching students in a school. 5. A chartered Accountant preparing accounts of a firm.

2.

Below are given certain non-economic activities. Convert them into economic activities. Example: A nurse attending her ailing son. (Non-economic activity) A nurse attending patients in her hospital. (Economic activity) 1. A person working in his own garden. 2. A lady preparing food for her husband. 3. A man white-washing his own house. 4. A teacher teaching his son at home. 5. A chartered accountant preparing his own accounts.

Answer»

1. A person working in a school garden.

2. A lady preparing food in a restaurant.

3. A man white-washing the building of a trade centre.

4. A teacher teaching students in a school.

5. A chartered Accountant preparing accounts of a firm.

Discussion

3.	……. removes hindrance of risk.
Answer» Insurance removes hindrance of risk.

Discussion

4.	……… removes hindrance of finance.
Answer» Banking removes hindrance of finance.

Discussion

5.	Classify the following activities into Economic and Non-Economic activities ……… 1. Running a stationery shop 2. Conducting a seminar3. Operating a Cine house 4. Watching television
Answer» 1. Running a stationary shop – Economic activity 2. Playing cricket – Non-economic activity 3. Operating a cine house – Economic activity 4. Watching television – Non-economic activity

5.

Classify the following activities into Economic and Non-Economic activities ……… 1. Running a stationery shop 2. Conducting a seminar3. Operating a Cine house 4. Watching television

Answer»

1. Running a stationary shop – Economic activity

2. Playing cricket – Non-economic activity

3. Operating a cine house – Economic activity

4. Watching television – Non-economic activity

Discussion

6.	…… is the process of selling goods to a foreign country.
Answer» Export trade

Discussion

7.	Buying and selling of goods and services between the two countries are called………
Answer» Foreign trade

Discussion

8.	Diary farming for milk is an example of ………. industry.
Answer» Diary farming for milk is an example of Genetic industry.

Discussion

9.	केन्द्रीकरण किसे कहते हैं ?
Answer» जब किसी भी इकाई में समस्त सत्ता संचालक के पास ही केन्द्रीत हो तो उन्हें केन्द्रीकरण कहते हैं ।

Discussion

10.	If the perimeter of an equilateral triangle is 36cm, calculate its area and height
Answer» Perimeter = 3a where a is the side of the equilateral triangle. 36cm = 3a side =12 cm area =√3÷4 a² = √3÷4×144. √336cm² area of triangle = 1÷2 base × height. 36√3= 6cm× height . height= 36√3÷6= 6√3= 1.732× 6. 10.4 cm

Discussion

11.	If the area of an equilateral triangle is 81√3cm2, find it's perimeter
Answer» We know that area is √3/4 a² area of the equilateral triangle = 81√3 √3/4 a² = 81√3 1/4 a² = 81 a² = 81(4) a² = 324 a = 18cm Then the perimeter of the equilateral triangle = 3×a = 3×18 =54cm.

11.

If the area of an equilateral triangle is 81√3cm2, find it's perimeter

Answer»

We know that area is √3/4 a²
area of the equilateral triangle = 81√3
√3/4 a² = 81√3
1/4 a² = 81
a² = 81(4)
a² = 324
a = 18cm

Then the perimeter of the equilateral triangle

= 3×a

= 3×18

=54cm.

Discussion

12.	Water jet strikes a block of mass 2kg with velocity 10 m/s at 1 kg/sec rate. The acceleration of body will be:(1) 3 m/s2(2) 4 m/s2(3) 5 m/s2(4) 7 m/s2
Answer» Correct option is (3) 5 m/s² F = dp/dt = vdm/dt = 10 x 1 Acc. = F/M = 10/2 = 5 m/s²

12.

Water jet strikes a block of mass 2kg with velocity 10 m/s at 1 kg/sec rate. The acceleration of body will be:(1) 3 m/s2(2) 4 m/s2(3) 5 m/s2(4) 7 m/s2

Answer»

Correct option is (3) 5 m/s²

F = dp/dt = vdm/dt = 10 x 1

Acc. = F/M = 10/2 = 5 m/s²

Discussion

13.	Find radius of gyration of body having MOI 20 kgm2 and of mass 4 kg.
Answer» Radius of gyration K = \(\sqrt{I/M}\) where I = moment of inertia M = Mass K = \(\sqrt{20/4}\) K = √5 m

13.

Find radius of gyration of body having MOI 20 kgm2 and of mass 4 kg.

Answer»

Radius of gyration K = \(\sqrt{I/M}\)

where I = moment of inertia

M = Mass

K = \(\sqrt{20/4}\)

K = √5 m

Discussion

14.	A uniform conducting wires of length 12a and resistance R is wound up as a current carrying coil in the shape of,(i) any equilateral triangle of side a.(ii) a square of side a.The magnetic dipole moment of the coil in each case respectively are :(1) 4 Ia² and 3 Ia²(2) √3 Ia² and 3 Ia²(3) 3 Ia² and Ia²(4) 3 Ia² and 4 Ia²
Answer» Answer: (1). √3Ia² and 3Ia² M₁ = (\(\frac{\sqrt{3}}{4}a^2\))I x 4 = √3Ia² (no. of turns = 4) M₂= a²I × 3 = 3Ia² (no. of turns = 3)

14.

A uniform conducting wires of length 12a and resistance R is wound up as a current carrying coil in the shape of,(i) any equilateral triangle of side a.(ii) a square of side a.The magnetic dipole moment of the coil in each case respectively are :(1) 4 Ia² and 3 Ia²(2) √3 Ia² and 3 Ia²(3) 3 Ia² and Ia²(4) 3 Ia² and 4 Ia²

Answer»

Answer: (1). √3Ia² and 3Ia²

M₁ = (\(\frac{\sqrt{3}}{4}a^2\))I x 4 = √3Ia²

(no. of turns = 4)

M₂= a²I × 3 = 3Ia²

(no. of turns = 3)

Discussion

15.	Find the area enclosed the curve y=sin x and the X-axis between `x=0 and x=pi`.
Answer» Correct Answer - B Area`= int^(x_2)_(x_1) y dx = int ^pi_0 sin x dx =[cos x]^pi_0 = -cospi-(-cos0) =+1+1=2`

Discussion

16.	What is the relation between acidic nature of oxoacid and electronegativity of the central atom?
Answer» The acid strength of oxoacid decreases as the electronegativity f the central atom decreases.

Discussion

17.	Is formation of ozone from oxygen an endothermic or exothermic reaction?
Answer» Endothermic reaction.

Discussion

18.	Arrange `SO_3,SeO_3` and `TeO_3` in order of increasing acid strength.
Answer» Acid strength : `TeO_(3)ltSeO_(3)ltSO_(3)`.

Discussion

19.	Arrange the hydrides of group 16 elements in order of increasing boiling point.
Answer» Boiling point : `H_(2)SltH_(2)SeltH_(2)TeltH_(2)O`.

Discussion

20.	Arrange the hydrides of group 16 elements in increasing order of reducing power.
Answer» Reducing power : `H_(2)OltH_(2)SltH_(2)SeltH_(2)Te`

Discussion

21.	Arrange the hydrides of group 16 elements in order of increasing acid nature.
Answer» Acid Strength: `H_(2)OltH_(2)SltH_(2)SeltH_(2)Te`.

Discussion

22.	State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.
Answer» Raoult’s law states that for a solution containing a non-volatile solute, the vapor pressure of the solution is directly proportional to mole fraction of the solvent. P_solution=K_HX_solution Consider a binary solution of two components A and B (both liquids). Let p total be the total vapor pressure of the solution due to both the components and let p_A and p_B be the partial pressures of component A and B, respectively. Then, p_A=x_A\(P_{A}^{°} \) p_B=x_B\(P_{B}^{°} \) The total vapor pressure of the solution of liquids A and B is then, by Dalton’s law of partial pressures, the sum of p_Aand p_B. p_Total=p_A+p_B Substituting for p_A and p_B, we get p_Total=x_A\(P_{A}^{°} \)+x_B\(P_{B}^{°} \) The characteristics of the solution which obeys Raoult’s law at all concentrations are as follows: (i) In a binary solution of components, A and B, the enthalpy of mixing ΔH_mix is zero, that is, in preparation of an ideal solution no thermal change is observed. (ii) In an ideal solution, the volume of mixing (ΔV_mix) is also zero, that is, the final volume of the solution is equal to the sum of volumes of components being mixed

22.

State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.

Answer»

Raoult’s law states that for a solution containing a non-volatile solute, the vapor pressure of the solution is directly proportional to mole fraction of the solvent.

P_solution=K_HX_solution

Consider a binary solution of two components A and B (both liquids). Let p total be the total vapor pressure of the solution due to both the components and let p_A and p_B be the partial pressures of component A and B, respectively. Then,

p_A=x_A\(P_{A}^{°} \)

p_B=x_B\(P_{B}^{°} \)

The total vapor pressure of the solution of liquids A and B is then, by Dalton’s law of partial pressures, the sum of p_Aand p_B.

p_Total=p_A+p_B

Substituting for p_A and p_B, we get

p_Total=x_A\(P_{A}^{°} \)+x_B\(P_{B}^{°} \)

The characteristics of the solution which obeys Raoult’s law at all concentrations are as follows:

(i) In a binary solution of components, A and B, the enthalpy of mixing ΔH_mix is zero, that is, in preparation of an ideal solution no thermal change is observed.

(ii) In an ideal solution, the volume of mixing (ΔV_mix) is also zero, that is, the final volume of the solution is equal to the sum of volumes of components being mixed

Discussion

23.	Arrange the hydrised of group 16 in order of increasing bond angle.
Answer» Bond angle : `H_(2)TeltH_(2)SeltH_(2)SltH_(2)O`.

Discussion

24.	Arrange the following in order of property indicated for each set: (a) H2O, H2S, H2Se, H2Te – increasing acidic character (b) HF, HCl, HBr, HI – decreasing bond enthalpy
Answer» (a) The order of increasing acidic strength is H₂O<H₂S<H₂Se <H₂Te. (b) The decreasing order of bond enthalpy is HF>HCl >HBr > HI

24.

Arrange the following in order of property indicated for each set: (a) H2O, H2S, H2Se, H2Te – increasing acidic character (b) HF, HCl, HBr, HI – decreasing bond enthalpy

Answer»

(a) The order of increasing acidic strength is H₂O<H₂S<H₂Se <H₂Te.

(b) The decreasing order of bond enthalpy is

HF>HCl >HBr > HI

Discussion

25.	Draw the structure of `XeF_4` molecule.A. B. C. D.
Answer» Correct Answer - C Lone pair- lone pair repulsion is maximum So to minimize the repulsion the lone pairs occupy the positions which are trans to each other in the given octahedral geometry.Therefore, (C ) option is correct.

Discussion

26.	When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P-V curve, V-T curve,P-T curve etc. For example, P-V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V-T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed. Now, answer the following questions : Two moles of an ideal gas is changed from its initial state (16 atm, 6L) to final state (4 atm, 15L) in such a way that this change can be represented by a straight line in P-V curve.The maximum temperature attained by the gas during the above change is :(Take : `R=1/12L atm K^(-1) "mol"^(-1)`)A. 324 KB. 648 KC. 1296 KD. 972 K
Answer» Correct Answer - B Equation of staight line `(y-y_1)=((y_2-y_1)/(x_2-x_1))(x-x_1) " " :. (P-16)=((4-16)/(15-6))(V-6)` `:. 3P+4V=72 implies T_("max")=((PV)_(max))/(nR)` For `(PV)_(max)((PV)_(max)).3P=72/2` and `4V=72/2 :. P=12,V=9 " " T_(max)=(12xx9)/(1xx(1//12))=648 K`

Discussion

27.	When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P-V curve, V-T curve,P-T curve etc. For example, P-V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V-T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed. Now, answer the following questions : Which of the following graphs is not possible for a fixed amount of ideal gas upon moving from initial state A to final state B :A. B. C. D. None of these
Answer» Correct Answer - D (A)This graph is possible if temperature of gas is kept constant and pressure is increased from A to B (B)This graph is possible if temperature of gas is increased continuously during the process A to B. ( C)This graph is possible if during the process : `P prop 1/v^2` `because VT="constant" " " :. " " V((PV)/(nR))="Constant" or PV^2="constant"`

Discussion

28.	When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P-V curve, V-T curve,P-T curve etc. For example, P-V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V-T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed. Now, answer the following questions : The figure shows the effect of pressure on the compressibility factor, Z of a gas : A. The curve AE can be explained by PV=RTB. The curve AF and CD can be explained by PV=RT+PbC. The curve AB can be explained by PV=RT-`a/V`D. All the three curves AF,AE and AD shows the real behaviour
Answer» Correct Answer - D (A)For AE, Z = constant `implies PV=RT` (B)At higher pressure PV=RT+Pb ( C)At lower pressure `PV=RT-a/V`

Discussion

29.	When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P-V curve, V-T curve,P-T curve etc. For example, P-V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V-T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed. Now, answer the following questions : Which of the following statements is correct regarding a fixed amount of ideal gas undergoing the following process : A. Root mean square (RMS) speed of gas molecules increase during the process `AtoB`B. Density of the gas increases during the process `AtoB`C. Such a graph is not possibleD. If `P_B=4P_A` then `V_A=4V_B` (where `P_A,V_A, P_B` & `V_B` represent pressure and volume values at states A and B).
Answer» Correct Answer - B (A)Temperature decreases during process A to B, so RMS speed decreases (`because V_(RMS)prop sqrtT`) (B)`d_(gas)=(PM)/(RT)`upon moving from A to B, P increases and temperature decreases.So, density of gas increases. ( C)This graph is possible if during the process :`P prop 1/(V^(1//2))` `because PT="Constant" therefore P((PV)/(nR))="Constant" :. P^2V="Constant" or PV^(1//2)=` Constant (D)If `P_B=4P_A`, then `V_A=16V_B`(according to `P^2V` =Constant for process)

Discussion

30.	The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at TK.standard vapour pressure of A and B are `P_A^@ and P_B^@` respectively at TK.We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at `(3T)/2`K.At the temperature greater than TK, both A and B exist in only gaseous form.Total vapour pressure of the system at TK is given as `P=P_A^@X_A+P_B^@X_B` where `X_A and X_B` are the mole fraction of A and B in a liquid mixture. In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as `AoversetAtoB` In container (II) At the given temperature `(3T)/2`,A and B are ideal in nature and non mixing in nature.A small pin hole is made into container.We can determine the initial rate of effusion of both gases in vaccum by the expression `r=(kp)/sqrtm` Where P= pressure difference between system and surrounding K=positibe constant M=Molecular weight of the gas. Vapour is collected and passed into a container of volume 8.21 lit, maintainer at 100 K and after 5 min number of mole of `B=8/3`.Then calculate pressure develop into the container after two half lives.A. 2 atmB. 4 atmC. 1 atmD. 0.5 atm
Answer» Correct Answer - A `{:(,AtoB),(t=0,8/3" "4/3),(t=5"min",4/3" "8/3):}` Now after two half life `n_A=2/3` and `n_B=10/3` `P_T V=(n_A+n_B)RT` `P_Txx8.21=12/3xx0.082xx100 " "implies P_T=4 "atm"`

Discussion

31.	The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at TK.standard vapour pressure of A and B are `P_A^@ and P_B^@` respectively at TK.We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at `(3T)/2`K.At the temperature greater than TK, both A and B exist in only gaseous form.Total vapour pressure of the system at TK is given as `P=P_A^@X_A+P_B^@X_B` where `X_A and X_B` are the mole fraction of A and B in a liquid mixture. In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as `AoversetAtoB` In container (II) At the given temperature `(3T)/2`,A and B are ideal in nature and non mixing in nature.A small pin hole is made into container.We can determine the initial rate of effusion of both gases in vaccum by the expression `r=(kp)/sqrtm` Where P= pressure difference between system and surrounding K=positibe constant M=Molecular weight of the gas. If vapour is collected in container of volume 8.21 lit maintained at 75 K then ratio of initial rate of effusion of Gas A and B is given as :A. `2:1`B. `1:2`C. `4:1`D. `1:4`
Answer» Correct Answer - A `r_A/r_B=P_A/P_Bsqrt(M_B/M_A)implies r_A/r_B=n_A/n_Bsqrt(M_B/M_A)=((8//3))/((4//3))sqrt(1/1)` `r_A/r_B=2`

Discussion

32.	The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at TK.standard vapour pressure of A and B are `P_A^@ and P_B^@` respectively at TK.We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at `(3T)/2`K.At the temperature greater than TK, both A and B exist in only gaseous form.Total vapour pressure of the system at TK is given as `P=P_A^@X_A+P_B^@X_B` where `X_A and X_B` are the mole fraction of A and B in a liquid mixture. In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as `AoversetAtoB` In container (II) At the given temperature `(3T)/2`,A and B are ideal in nature and non mixing in nature.A small pin hole is made into container.We can determine the initial rate of effusion of both gases in vaccum by the expression `r=(kp)/sqrtm` Where P= pressure difference between system and surrounding K=positibe constant M=Molecular weight of the gas. If partial vapour pressure of A is twice that of partial vapour pressure of B and total V.P. of equal to 2 at T=50 K and V=8.21 lit, then number of mole of A and B in vapour phase is :A. `8/3 , 4/3`B. 3,1C. 2,2D. 1,3
Answer» Correct Answer - A `P_T=P_A+P_B=3P_B` So, `P_B=2/3` and `P_A=2-2/3=4/3` Now `P_A/P_B=n_1/n_2=2` So, `n_1=2n_2` `P_T V=n_T R_T` `2xx8.21=n_Txx0.082xx50` `n_T=4=n_1+n_2 " " So, n_2=4/3 " " n_1=8/3`

Discussion

33.	With reference of the above graph, choose the correct alternativesA. `P_BgtP_A`B. `P_AgtP_B`C. Pressure first increases then decreasesD. Pressure first decreases then increases
Answer» Correct Answer - B,C At const. temperature `V alpha1//p`

Discussion

34.	A gaseous organic compound has a density of `2.5 kg//m^3` at 2 atm and at `273^@C`.The molecular formula of the compound can be :A. `C_3H_4O_2`B. `C_4H_4`C. `C_3H_4O`D. `C_4H_8`
Answer» Correct Answer - C,D `d=(PM)/(RT) " " :. 2.5=(2xxM_("gas"))/(0.082xx546) " " :. M_("gas")=56` `:.` Both ( C) and ( D) options are correct

Discussion

35.	In the calcum fluaride structure, the coordination bumber of the cations and anions are respectively ,A. ` 6` and ` 6`B. ` 8` and `4`C. ` 4 ` and `4`D. ` 4` and `8`
Answer» Correct Answer - B In fluorite-type structure, ` CN=8 : 4` (factula statement ).

Discussion

36.	Which of the following statements is correct ?A. There is regular increase in negative value of electron gain enthalpy with increasing atomic number from left to right in the periodB. Electropositive nature of elements decrease with increasing atomic number from left to right in the periodC. Ionisation energies of elements decreases from left to right in the periodD. Effective nuclear charge of elements decreases from left to right in the period
Answer» Correct Answer - B ionisation potential and effective nuclear charge increase in a period from left to right . In a period the values of electron gain enthalpy (negative) do not increase regularly.

Discussion

37.	The increasing order of acidic nature of `Li_2O, BeO,B_2O_3,CuO` is :A. `Li_2Olt BeOltCuOltB_2O_3`B. `BeOltCuOltB_2O_3ltLi_2O`C. `Li_2Olt CuOltBeOltB_2O_3`D. `B_2O_3lt CuOltBeOltLi_2O`
Answer» Correct Answer - C On moving left to right in periodic table, acidic nature of oxide increases.So, metallic oxides are basic, non-metallic oxides are acidic and BeO is amphoteric.

Discussion

38.	Which of the following statements is wrong for the transition elements?A. Transition elements are placed from `3^(rd)` to `6^(th)` periodB. Last electron enters in (n-1) d orbitalC. Exhibits variable valencyD. General electronic configuration is (n-1) `d^(1-10)ns^(0-2)`
Answer» Correct Answer - A Transition elements starts from `4^(th)` period as they have valence shell configuration `(n-1)d^(1-10)ns^(0-2)`Thus for n=3, there is no d-orbital available.

Discussion

39.	If the same element is forming oxides in different oxidation states then :a)that oxide will be neutral in nature in which element will be in its highest oxidation state.b)that oxide will be highest acidc in nature in which element will be in its highest oxidation state.c)that oxide will be amphoteric in nature in which element will be in its highest oxidation stated)that oxide will be highly basic in nature in which element will be in its highest oxidation state.A. That oxide will be neutral in nature in which element will be in its highest oxidation state.B. That oxide will be highest acidic in nature in which element will be in its highest oxidation stateC. That oxide will be amphoteric in nature in which element will be in its highest oxidation stateD. That oxide will be highly basic in nature in which element will be in its highest oxidation state.
Answer» Correct Answer - B Oxidation state `prop` electronegativity Electronegativity increases with increase in oxidation state, so the difference in electronegativity decreases (between element and oxygen) and acidic character increases.

Discussion

40.	[Co(C2O4)3]3- is a : (1) Low spin complex (2) Paramagnetic (3) High spin (4) sp3d2 hybridized
Answer» Correct option: (1) Low spin complex Explanation: CO⁺³ has d²sp³ hybridisation so all electron are paired therefore it is Low spin complex and diamagnetic.

40.

[Co(C2O4)3]3- is a : (1) Low spin complex (2) Paramagnetic (3) High spin (4) sp3d2 hybridized

Answer»

Correct option: (1) Low spin complex

Explanation:

CO⁺³ has d²sp³ hybridisation so all electron are paired therefore it is Low spin complex and diamagnetic.

Discussion

41.	At 527 ºC temperature the activation energy is 54.7 KJ/mole. The value of Arrhenius factor is 4 × 1010. The rate constant will be (1) 12.28 × 1011 (2) 14.58 × 1013 (3) 12.28 × 1017 (4) 14.58 × 10–13
Answer» Correct option: (2) 14.58 × 10¹³

Discussion

42.	Choose incorrect statement :-A. `1^("st")` ionisation potential `Mg lt S lt P lt F`B. Atomic radii `Cl lt Mg lt P lt Ca`C. Basic nature `Al_(2)O_(3) lt MgO lt Na_(2)O lt K_(2)O`D. Acidic nature `NH_(3) lt PH_(3) lt AsH_(3)`
Answer» Correct Answer - B `"Size" prop "Shell" prop (1)/(Z^(+))`

Discussion

43.	An element X occurs in short period having configuration `ns^(2)np^(1)`. The formula and nature of its oxide is:A. `XO_(3)`, basicB. `XO_(3),` acidicC. `X_()O_(3)`, amphotericD. `X_(2)O_(3)`, basic
Answer» Correct Answer - C `Al_(2)O_(3)rarr"Amphoteric oxide"`

Discussion

44.	Select correct order of `IP_(2)` :-A. `V lt Mn lt Cr lt Fe`B. `V lt Me lt Fe lt Cr`C. `Cr lt V lt Fe lt Mn`D. `Cr lt Mn lt V lt Fe`
Answer» Correct Answer - B `IP prop Z^(+)`

Discussion

45.	Consider the following cell reaction. `2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq)+2H_(2)O(l),` `E^(@)=1.67V` At `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` isA. `1.27V`B. `1.77V`C. `1.87V`D. `1.57V`
Answer» Correct Answer - D `[H^(+)]=10^(-3),Q=([Fe^(2+)])/([H^(+)]^(4)P_(O_(2))),n=4` `E=1.67-(0.06)/(4)log.((10^(-3))^(2))/((10^(-3))^(4)(10^(-1)))` `=1.67-(0.06)/(4)xx7` `=1.57V`

Discussion

46.	`Cu(s)\|Cu^(+2)(aq,10^(-3))\|\|Ag^(+)(10^(-5)M)\|Ag(s)` if `E_(Cu^(+2)//Cu=0.34V` ltBrgt `E_(Ag^(+)//Ag=+0.80V` `E_("cell")` will be :-A. `0.46 V`B. `0.46-(RT)/(2F)ln 10^(7)`C. `0.46+(RT)/(2F) ln 10^(7)`D. `0.46+(RT)/(2F)ln 10^(2)`
Answer» Correct Answer - B `E_("cell")=0.85-0.34-(RT)/(2F)ln.((10^(-3)))/((10^(-5))^(2))` `=0.46-(RT)/(2F)ln10^(7)`

Discussion

47.	What will be the percent fraction of edge length not covered by atoms when a metal crystallises in bcc lattice?A. `11.4%`B. `10.4%`C. `13.4%`D. `12.4%`
Answer» Correct Answer - C `"Fraction"=(a-2R)/(a),R=(sqrt3a)/(4)` `=(a(1-(2sqrt3)/(4)))/(a)=1-(sqrt3)/(2)=0.134` `%=13.4`

Discussion

48.	Following three planes `(P_(1), P_(2), P_(3))` in an fcc unit cell are shown in the figure below. Consider the following statements and choose the correct option/options that follow: A. `P_1` contains no three dimensional voidsB. `P_2` contains only octahedral voidsC. `P_3` contains both octahedral and tertahedral voidsD. All are true
Answer» Correct Answer - ABCD In an FCC, Octahedral voids lies on the edges & in centre of unit cell & tetrahedral voids on the body diagonal

Discussion

49.	Write the balanced chemical equation of the reaction between Sodium hydroxide and Hydrochloric acid
Answer» Balanced chemical equation of the reaction between Sodium hydroxide and Hydrochloric acid is given by: Acid + base → water + salt NaOH(aq)+HCl(aq)→NaCl(aq)+H₂O(l)

49.

Write the balanced chemical equation of the reaction between Sodium hydroxide and Hydrochloric acid

Answer»

Balanced chemical equation of the reaction between Sodium hydroxide and Hydrochloric acid is given by:

Acid + base → water + salt

NaOH(aq)+HCl(aq)→NaCl(aq)+H₂O(l)

Discussion

50.	A solution of a substance 'X' is used for whitewashing.(i) Name the substance 'X' and write its formula.(ii) Write the reaction of the substance 'X' named in (i) above with water.
Answer» (i) The substance 'X' used for whitewashing is quick lime (calcium oxide). Its formula is CaO. (ii) When quick lime is mixed with water, the following reaction takes place: CaO(s) + H₂O(l) ------> Ca(OH)₂(aq) + Heat (Quick lime calcium oxide) water calcium hydroxide(slaked lime)

Discussion

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Consider the following cell reaction. `2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq)+2H_(2)O(l),` `E^(@)=1.67V` At `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` isA. `1.27V`B. `1.77V`C. `1.87V`D. `1.57V`

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Write the balanced chemical equation of the reaction between Sodium hydroxide and Hydrochloric acid

A solution of a substance 'X' is used for whitewashing.(i) Name the substance 'X' and write its formula.(ii) Write the reaction of the substance 'X' named in (i) above with water.