100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Unsteady state is also known as Transient state, the statement is(a) True(b) False(c) Applicable for some systems(d) None of the mentioned
Answer» Right option is (a) True For explanation: Unsteady state is also known as Transient state.

Discussion

2.	200 mL of 0.4M solution of `CH_(3)COONa` is mixed with 400 mL of 0.2 M solution of `CH_(3)COOH`. After complete mixing, 400 mL of 0.1 M NaCl is added to it. What is the pH of the resulting solution? `[K_(a)` of `CH_(3)COOH=10^(-5)]`A. 5.4B. 6C. 5D. 6.2
Answer» Correct Answer - C `pH=p_(Ka)+log(([CH_(3)COONa])/([CH_(3)COOH]))=5+log((200xx0.4)/(400xx0.2))=5`

Discussion

3.	Which one is a state variable ?(a) Internal Energy(b) Enthalpy(c) Entropy(d) All of the mentioned
Answer» Correct answer is (d) All of the mentioned The best I can explain: All a, b, c are state variables as these are only depended only on the state, not on the history.

Discussion

4.	Use the standard enthalpies of formation and calculation the enthalpy changes accompanying the following reaction: a. `CH_(4)(g)+2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l)` b. `4Al(s)+3O_(2)(g) rarr 2Al_(2)O_(3)(s)`
Answer» Given `Delta_(f)HCO_(2)(g)=-393.51 kJ mol^(-1)` `Delta_(f)HCH_(4)(g)=-74.81 kJ mol^(-1)` `Delta_(f)HH_(2)O(l)=-285.83 kJ mol^(-1)` `CH_(4)(g)+2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l)` `Delta_(R)H=Sigma(Delta_(f)H "Products")-Sigma(Delta_(f)H "Reactants")` `=(Delta_(f)H CO_(2)+2Delta_(f)HH_(2)O)-(Delta_(f)HCH_(4)+2Delta_(f)HO_(2))` `=(-393.51-2xx285.83)-(-74.81+2xx0)` `=-890.36 kJ mol^(-1)` b. Given `Delta_(f)H` of `Al_(2)O_(3)(s)=-1675.7 kJ mol^(-1)` `4 Al(s)+3O_(2)(g) rarr 2Al_(2)O_(3)(s)` `DeltaH_(R)=Sigma(Delta_(f)H "Products")-Sigma(Delta_(f)H "Reactants")` `=(2 Delta_(f)HAl_(2)O_(3))-(4 Delta_(f)HAl+3Delta_(f)HO_(2))` `=2xx-1675.7-(4xx0+3xx0)` `=-3351.4 kJ mol^(-1)`

Discussion

5.	Heat is negative when transferred to the _____________(a) System(b) Surrounding(c) Boundary(d) None of the mentioned
Answer» Correct choice is (a) System The best explanation: Heat is negative when transferred to the surrounding .

Discussion

6.	Consider the following set of reactions: `CHCl_(2)COOH+NaOH to CHCl_(2)COONa+H_(2)O " " DeltaH_(1)=-12830 Cal` `HCl+NaOH to NaCl to NaCl+H_(2)O " " DeltaH_(2)=-13680 Cal`. `NH_(4)OH+HCl to NH_(4)Cl+H_(2)O " " DeltaH_(3)=-12270 Cal`. Select the correct option(s) :A. Enthalpy of neutralization of `CHCl_(2)COOH` by `NH_(4)OH` is `-11420 Cal.`B. Enthalpy of dissociation of `NH_(4)OH=1410 Cal`C. Enthalpy of dissociatiion of `NH_(4)OH=1410 Cal `.D. Enthalpy change for the reaction `H_(2)O to H^(+)+OH^(-) is -13680 Cal `
Answer» (A),( C) `DeltaH=DeltaH_(1)+DeltaH_(3)-DeltaH_(2)` `=-(12830+12270-13680)` `=-11420 Cal` (B) `DeltaH_(2)-DeltaH_(1)=13680-12830` `=850 Cal` (C) `DeltaH_(2)-Delta_(3)=13680-12270` `=1410 Cal` (D) `H_(2)O to H^(+)+OH^(-)` `=+13680K`

Discussion

7.	In constant volume thermodynamic process work done is(a) Maximum(b) Minimum(c) Zero(d) None of the mentioned
Answer» Right answer is (c) Zero To explain: W = P(V2-V1).

Discussion

8.	If `f:R to R ` satisfies f(x+y)=f(x)+f(y), for all x, y `in` R and f(1)=7, then `sum_(r=1)^(n)f( r)` isA. `(7n)/(2) `B. `(7(n+1))/(2)`C. `7n(n+1) `D. `(7n(n+1))/(2)`
Answer» D f(x+y)=f(x)+f(y) ltbgt Put x-1,y=1 `implies` f(2)=2`*`f(1)=2.7 Similarly f(3)=3.7 and so on `sum_(r=1)^(n)f(r)=7(1+2+3+......+n)=(7n(n+1))/(2)`

Discussion

9.	Which of the following statement(s) is / are correctA. The internal energy of an ideal gas may be increased by adding more molecules to it, at constant temperature.B. The molar enthalpy of any substance decrease on cooling , at constant pressure.C. A diabatic free expansion & isothermal free expansion are similar processs for an ideal gas .D. It is possible to have a process in which the entropy of an isolated system is decreased. .
Answer» Correct Answer - (A),(B),(C ),(D)

Discussion

10.	For a steady state closed system(a) U+25B3E > 0(b) U+25B3E < 0(c) U+25B3E = 0(d) None of the mentioned
Answer» Correct option is (c) U+25B3E = 0 Explanation: For a steady state closed system, U+25B3E = 0.

Discussion

11.	For adiabatic free expansion of a real gas, the correct relation are :A. W=0B. q=0C. `DeltaU=0`D. `DeltaT=0`
Answer» (A),(B),( C) For A diabatic free expansion `to` `q=0,W=0,DeltaU=0` `DeltaU=f(T,V)` for a real gas . hence `DeltaT ne0`

Discussion

12.	Arrange the following species in the increasing order of ionic radii. N 3- , F- , Na+ , Mg2+, O2- and Al3+. (c) H2O(g)  H(g) +OH (g) ; ∆aH°1= 502 kJmol-1 OH(g) H(g)+ O(g) ; ∆aH°2= 427kJmol-1 Why is ∆aH°1 and ∆aH°2 different for water?
Answer» All these ions have 10 electrons in their shell therefore these are isoelectronic speicies The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. therefore the order of ionic radii will be: Al³⁺< Mg²⁺< Na⁺< F⁻< O²⁻< N³⁻ (increasing order)

12.

Arrange the following species in the increasing order of ionic radii. N 3- , F- , Na+ , Mg2+, O2- and Al3+. (c) H2O(g)  H(g) +OH (g) ; ∆aH°1= 502 kJmol-1 OH(g) H(g)+ O(g) ; ∆aH°2= 427kJmol-1 Why is ∆aH°1 and ∆aH°2 different for water?

Answer»

All these ions have 10 electrons in their shell therefore these are isoelectronic speicies
The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. therefore the order of ionic radii will be:

Al³⁺< Mg²⁺< Na⁺< F⁻< O²⁻< N³⁻ (increasing order)

Discussion

13.	A hollow cubical box P is moving on a smooth horizontal surface in the x-y plane with constant acceleration of `vec(a) = 3 hat (i) + 4 hat (j) m//s^(2)`.A block Q of mass 2 kg is at rest inside the cubical box as shown in figure. If the coefficient of friction between the surface of the cube P and the block Q is 0.6. Then the force of friction between P and Q is : A. 5 NB. 8 NC. 12 ND. 10 N
Answer» Correct Answer - D `a = sqrt(3^(2)+4^(2))=5m//s` Friction force `f = ma = 2 xx 5 = 10 N`

Discussion

14.	A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is
Answer» The correct answer is: zero Net magnetic force on any current carrying loop in uniform magnetic field is zero. Explanation: Let's consider a small arc of the circular loop, the force is proportional to IdLxB. Using the right hand rule we see that the force is acting towards the centre. But a similarly elemental arc is exactly opposite to the considered arc. But the current flows in opposite direction. Hence the forces cancel each other.

14.

A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Answer»

The correct answer is: zero

Net magnetic force on any current carrying loop in uniform magnetic field is zero.

Explanation:

Let's consider a small arc of the circular loop, the force is proportional to IdLxB. Using the right hand rule we see that the force is acting towards the centre. But a similarly elemental arc is exactly opposite to the considered arc. But the current flows in opposite direction. Hence the forces cancel each other.

Discussion

15.	120 of an organic compound that contains only carbon and hydrogen gives 330g of CO2 and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are. (A) 25 and 75 (B) 40 and 60 (C) 60 and 40 (D) 75 and 25
Answer» Correct option is (D) 75 and 25 Given mass of organic compound = 120 mass of CO₂(g) = 330 g mass of H2O (l) = 270 g mass of carbon =\(n_{CO_2}\)x 12 = 330/44 x 12 = 90 g % of carbon = 90/120 x 100 = 75% mass of hydrogen = \(n_{H_2O}\) x 2 = 270/18 x 2 = 30g % of hydrogen = 30/120 x 100 = 25%

15.

120 of an organic compound that contains only carbon and hydrogen gives 330g of CO2 and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are. (A) 25 and 75 (B) 40 and 60 (C) 60 and 40 (D) 75 and 25

Answer»

Correct option is (D) 75 and 25

Given mass of organic compound = 120

mass of CO₂(g) = 330 g

mass of H2O (l) = 270 g

mass of carbon =\(n_{CO_2}\)x 12

= 330/44 x 12 = 90 g

% of carbon = 90/120 x 100 = 75%

mass of hydrogen = \(n_{H_2O}\) x 2

= 270/18 x 2 = 30g

% of hydrogen = 30/120 x 100 = 25%

Discussion

16.	A rectangular loop carrying current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and the plane of the loop is same of the left wire. If a steady current I is established in the wire as shown in the figure the loop will
Answer» The steady current I will move towards the wire Correct Answer is (C) Explanation: Wires placed close to each other carry current in the same direction and hence attract. Wires placed far apart carry current in opposite direction and hence repel each other. But here attraction is strong and repulsion is weak. So, loop moves towards the wire.

16.

A rectangular loop carrying current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and the plane of the loop is same of the left wire. If a steady current I is established in the wire as shown in the figure the loop will

Answer»

The steady current I will move towards the wire

Correct Answer is (C)

Explanation:

Wires placed close to each other carry current in the same direction and hence attract. Wires placed far apart carry current in opposite direction and hence repel each other. But here attraction is strong and repulsion is weak. So, loop moves towards the wire.

Discussion

17.	Name the different types of cables used in transmission.
Answer» The different types of cables used in transmission are: Unshielded Twisted Pair Cable (UTP cable) Shielded Twisted Pair (STP) cable Coaxial Cable Fiber Optic Cable.

17.

Name the different types of cables used in transmission.

Answer»

The different types of cables used in transmission are:

Unshielded Twisted Pair Cable (UTP cable)
Shielded Twisted Pair (STP) cable
Coaxial Cable
Fiber Optic Cable.

Discussion

18.	Mention the methods of opening files within C++ program.
Answer» The methods of opening file within C++ program Opening a file using constructor. Opening a file using member function open()

18.

Mention the methods of opening files within C++ program.

Answer»

The methods of opening file within C++ program

Opening a file using constructor.
Opening a file using member function open()

Discussion

19.	Explain the features of SQL.
Answer» The features of SQL: SQL is an ANSI and ISO standard computer language for creating and manipulating databases. SQL allows the user to create, update, delete, and retrieve data from a database. SQL is very simple and easy to learn.

19.

Explain the features of SQL.

Answer»

The features of SQL:

SQL is an ANSI and ISO standard computer language for creating and manipulating databases.
SQL allows the user to create, update, delete, and retrieve data from a database.
SQL is very simple and easy to learn.

Discussion

20.	What is buffer solution?
Answer» A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %) or when solution is diluted is called buffer solution.

Discussion

21.	Define buffer action.
Answer» The capacity of a solution to resist alteration in its pH value is known as buffer capacity and the mechanism of buffer solution is called buffer action.

Discussion

22.	How many types of buffer solution?
Answer» Types of buffer solutions : (A) Simple buffer solution (B) Mixed buffer solution

22.

How many types of buffer solution?

Answer»

Types of buffer solutions :

(A) Simple buffer solution

(B) Mixed buffer solution

Discussion

23.	what is a liner equation
Answer» A linear equation is an algebraic equation of the form y=mx+b. involving only a constant and a first-order (linear) term, where m is the slope and b is the y-intercept. Occasionally, the above is called a "linear equation of two variables," where y and x are the variables.

Discussion

24.	Figure shows a hemisphere of charge Q and radius R and a sphere of charge 2Q and radius R. The total potential energy of hemisphere is `U_(H)` and that of the sphere is `U_(s)` Then. A. `2U_(H) = U_(s)`B. `2U_(H) lt U_(s)`C. `2U_(H) gt U_(s)`D. `U_(H) = U_(s)`
Answer» Correct Answer - B `U_(s)=2U_(h)` + interaction energy

Discussion

25.	The galvanometer shown in the figure reads 3A, while the ideal voltmeter reads 24 volt. The value of `R = 7 Omega`. The galvanometer resistance is : A. `10 Omega`B. `5 Omega`C. `1 Omega`D. `20 Omega`
Answer» Correct Answer - C `V = i(G+R)` or `24 = 3 sqrt(G+7)` or `G = 1 Omega`

Discussion

26.	A 400 ml sample of 1M NaOH is left in a hot plate ovemight, the following morning solution is 1.6M. Then volume of water evaporated is:A. 150mlB. 250mlC. 200mlD. 100ml
Answer» `M_(1)V_(1)=M_(2)V_(2)` `1Mxx400=1.6xxV_(2)` `v_(2)=(400)/(1.6)=250mL` `DeltaV=400-250=150ml`

Discussion

27.	3/4x+7/2/5x-4=5/4
Answer» \(\frac{3}{4x}+\frac{7}{2}\times5x-4=\frac{5}{4}\) \(\Rightarrow \frac{3}{4x}+\frac{35x}{2}=\frac{5}{4}+4\) \(\Rightarrow \frac{3\,+\,70x^2}{4x}=\frac{5\,+\,16}{4}\) \(\Rightarrow \frac{73x^2}{4x}=\frac{21}{4}\) \(\Rightarrow\) \(73x^{2-1}=21\) \(\Rightarrow 73x=21\) \(\Rightarrow x=\frac{21}{73}\)

27.

3/4x+7/2/5x-4=5/4

Answer»

\(\frac{3}{4x}+\frac{7}{2}\times5x-4=\frac{5}{4}\)

\(\Rightarrow \frac{3}{4x}+\frac{35x}{2}=\frac{5}{4}+4\)

\(\Rightarrow \frac{3\,+\,70x^2}{4x}=\frac{5\,+\,16}{4}\)

\(\Rightarrow \frac{73x^2}{4x}=\frac{21}{4}\)

\(\Rightarrow\) \(73x^{2-1}=21\)

\(\Rightarrow 73x=21\)

\(\Rightarrow x=\frac{21}{73}\)

Discussion

28.	Statement-1 : The `5^(th)` period of periodic table contains 18 elements not 32. Statement-2 : n=5,l=0,1,2,3. The order in which the energy of available orbitals 4d, 5s, 5p increases is `5slt4dlt5p` and the total number of orbitals available are 9 and thus 18 electrons can be accomodated.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 The number of elements in a period is equal to the number of electrons that can be accomadated in available orbitals that are being filled for the given n.

Discussion

29.	Which of the following is correct order of wavelength for radiation?A. Infra red rays `gt` Red color rays `gt` Ultraviolet rays `gt` Cosmic raysB. Cosmic rays `gt` Ultraviolet rays `gt` Red color rays `gt` Infra red raysC. Ultraviolet rays `gt` Cosmic rays `gt` Red color rays `gt` infra red raysD. Cosmic rays `gt` Red color rays `gt` Ultraviolet rays `gt` Infra red rays
Answer» Correct Answer - Theory based

Discussion

30.	Select the incorrect statement(s).A. `IE_1` of nitrogen atom is less than `IE_1` of oxygen atomB. Electron gain enthalpy of oxygen is less negative than seleniumC. Electronegativity on pauling scale is 2.8 times the electronegativity on Mulliken scaleD. `Cr^(6+)` is smaller than `Cr^(3+)`
Answer» Correct Answer - A,C (A)IE (I) of N is more than O due to stable half filled electronic configuration of valence shell in N. (B)Electron gain enthalpy of O (-141 `"kJmol"^(-1)`) is less than sulphur and selenium due to its exceptionally small atomic size. (C )Electronegativity on Mulliken scale is 2.8 larger than electronegativity on Pauling scale. (D)The ionic radius decreases as more electrons are ionized off. `Cr^(6+)=44"pm",Cr^(3+)=61.5"pm"`

Discussion

31.	Which of the following statement(s) in (are) true ?A. Ionisation energy `prop 1/("Screening effect")`B. The first ionisation energies of Be and Mg are more than ionisation energies of B and Al respectivelyC. Atomic and ionic radii of Niobium and Tantalum are almost sameD. Metallic and covalent radii of potassium are 2.3 Å and 2.03 Å respectively
Answer» Correct Answer - A,B,C,D (A)As screening effect increases, effective nuclear charge decreases and thus valence shell electron is loosely bound.Hence ionization energy decreases. (B)Be and Mg and have `ns^2` configuration (stable configuration) ( C)Due to lanthanide contraction. (D)`r_("metallic")gtr_("covalent")`(covalent bond formation involves the overlapping of orbitals)

Discussion

32.	If the concentration of `Mg^(2+)` ions in sea water is 1200 ppm. How many moles of NaOH are required to precipitate all `Mg^(2+)` ions into `Mg(OH)_(2)(S)` present in 1 litre solution.
Answer» `"ppm"=(W_(Mg^(+2)))/(W_("water"))xx10^(6)` `10^(6)ml` solution contain `W_(Mg^(+2))=1200`ppm `[Mg^(+2)]=(1200)/(24)xx(10^(3))/(10^(6))` `[Mg^(+2)]=50xx10^(-3)=5xx10^(-2)=n_(Mg^(+2))` in 1 litre solution `n_(NaOH)` used =`2xxn_(Mg^(+2))` `=2xx5xx10^(-2)` `=10^(-1)xx10=1`

Discussion

33.	Which of the following represent the correct order of electron affinities?A. FgtClgtBrgtIB. CltNltClltFC. NltCltOltFD. CltSigtPgtN
Answer» Correct Answer - C,D Generally the electron affinity values of `3^(rd)` period elements are higher than the elements of 2nd period of the same group because of less interelectronic repulsions.Atomic size, nuclear charge and electrons configuration also affect the electron affinity. Electron gain enthalpes in KJ `mol^(-1)` are: `C=-121,Si=-134,P=-74,N~~O,O=-142,F=-328`

Discussion

34.	If the quantum numbers n,l,m and s were defined as: R=shell number =1,2,3,4,. . . . In integral steps. l=Type of subshell =0,1,2,3,. . . To n in integral steps. m=Number of orbitals corresponding to any subshell =-(l+1)to+(l+1), in integral steps, incuding zero. s= Spin quantum number `=-(1)/(2)` or `+(1)/(2)` The l-values correspond to the subshells as actual representations, like l=0 (s-subshell). l=1(p-subshell),l=2(d-subshell),and so on. In the modern long form of periodic table, the 2nd period should (Assume that (n+l)rule is perfectly obeyed).A. 8 elementsB. 12 elementsC. 16elementsD. 18elements
Answer» For n=3 `m=-(l+1)"to" +(l+1)` `=0m=-1,0,+1=3` orbital =1 m=02,01,0,+1,+2=5 orbital =2 m=-3,-2,-1,0,+1,+2,+3=7 orbital =3 m=-4,-3,-2,-1,0,+1+2,+1,+4=9 orbital ltbr. Total orbitals =3+5+7+9=24 orbitals `s=-(1)/(2) "or" +(1)/(2)` (i.e. two electrons) Total electrons =24xx2=48 electrons (ii) lr 2nd period Given: n=2,l=0,1,2 l=0 l=0s in n=1,l=0,1 l=0,1 l=1 p l=2 d Now `1s^(2)1p^(6)2s^(6)2p^(6)=16` electron (as electron will not enter in 2d)

Discussion

35.	If the quantum numbers n,l,m and s were defined as: R=shell number =1,2,3,4,. . . . In integral steps. l=Type of subshell =0,1,2,3,. . . To n in integral steps. m=Number of orbitals corresponding to any subshell =-(l+1)to+(l+1), in integral steps, incuding zero. s= Spin quantum number `=-(1)/(2)` or `+(1)/(2)` The l-values correspond to the subshells as actual representations, like l=0 (s-subshell). l=1(p-subshell),l=2(d-subshell),and so on. The maximum number of electrons is `3^(rd)` shell should be:A. 18B. 48C. 24D. 32
Answer» For n=3 `m=-(l+1)"to" +(l+1)` `=0m=-1,0,+1=3` orbital =1 m=02,01,0,+1,+2=5 orbital =2 m=-3,-2,-1,0,+1,+2,+3=7 orbital =3 m=-4,-3,-2,-1,0,+1+2,+1,+4=9 orbital ltbr. Total orbitals =3+5+7+9=24 orbitals `s=-(1)/(2) "or" +(1)/(2)` (i.e. two electrons) Total electrons =24xx2=48 electrons (ii) lr 2nd period Given: n=2,l=0,1,2 l=0 l=0s in n=1,l=0,1 l=0,1 l=1 p l=2 d Now `1s^(2)1p^(6)2s^(6)2p^(6)=16` electron (as electron will not enter in 2d)

Discussion

36.	Consider three flasks in diagram below. Assuming that connecting tube has negligible volume and all three falsks are at same temperature. If only 1 and 2 stopcocks are opened then select correct option(s).A. `P_(HCl)=(1)/(2)` atmB. `P_(NH_(3))=0` atmC. `P_(NH_(3))+P_(HCl)=(1)/(2)` atmD. `P_(He)=5` atm
Answer» `n_(HCl)=(PV)/(RT)=(2xx1)/(RT)` `n_(NH_(3))=(PV)/(RT)=(1xx1)/(RT)` `NH_(3)+HClrarrNH_(4)Cl(s)` moles `(1)/(RT) (2)/(RT)` `0 (1)/(RT)` `n_(HCl)` remain after reaction =`(1)/(RT)` For `HCl(g):-P_("new")(V+V)=nRT` `P_("new")xx2=(1)/(RT)xxRT` `(P_(HCl))_("final")=P_("new")=(1)/(2)atm=P_("total")` `P_(NH_(3))=0(LR)`

Discussion

37.	A flask of `4.48 L` capacity contains a mixture of `N_(2)` and `H_(2)` at `0^(@)C` and `1 atm` pressure. If thed mixture is made to react to form `NH_(3)` gas at the same temperature, the pressure in the flask reduces to `0.75 atm`. Select statement(s) :A. Initially total moves of gases in the mixture is `0.2 mol`.B. Initially total moves of gases in the mixture is `0.4 mol`.C. The partial pressure of `NH_(3)` gas in the final mixture is `0.33 atm`.D. The partial pressure of `NH_(3)` gas in the final mixture is `0.25 atm`.
Answer» Correct Answer - B::C `\|{:(,,N_(2),+,3H_(2)rarr2NH_(3)),("Initally",a atm, b atm,0,),("Finally",a-x,,b-3x,2x):}\|` `:. a + b = 1` and `a + b - 2x = 0.75` `:. P_(NH_(3)) = 2x = 0.25 atm`

Discussion

38.	Which of the following pair(s) represent(s) the isoelectronic species ?A. `S^(2-)` & `Sc^(3+)`B. `SO_(2)` & `NO_3^(-)`C. `N_(2)` & `CN^(-)`D. `NH_3` & `H_3O^+`
Answer» Correct Answer - A,B,C,D Species having same number of electrons are called isoelectronic specis.

Discussion

39.	Which of the following is/are aromatic?A. B. C. D.
Answer» (B) `6pie^(-)` with complete conjugation (A romatic)

Discussion

40.	The chemical formula of Baking soda is(A) NaHCO3 (B) Al2O3(C) P2O5 (D) N2O
Answer» The chemical formula of Baking soda is NaHCO₃

Discussion

41.	Two flask A and B of equal volume are taken. Flask a contains `(H_(2)(g)` at `27^(@)C` and 1 atm pressure. Flask B contain `N_(2)(g)` at `27^(@)C` and 2 atm pressure. Then select incorrect statements.A. Average kinetic energy per molecule is same for bothB. Number of molecuels in both compartment are same.C. Mass of `H_(2)` is more than `N_(2)`.D. `(U_("rms"))_(H_(2))gt(U_("rms"))_(N_(2)`
Answer» (A) `KE_("avg")=(3)/(2)KT` (only depend on temperature) (B) `(n_(H_(2)))/(n_(N_(2)))=(1xxV)/(RT)xx(RT)/(2xxV)=(1)/(2)rArrn_(H_(2))=(n_(N_(2)))/(2)` (C ) `(W_(H_(2)))/(2)=(W_(N_(2)))/(28xx2)rArrW_(H_(2))=(W_(H_(2)))/(28) W_(H_(2))ltW_(N_(2))` (D) `(U_(rms)H_(2))/(U_(rms)N_(2))=sqrt(M_(N_(2))/(M_(H_(2))))=sqrt(14)gt1` `(U_(rms))_(H_(2))gt(U_(rms))N_(2)`

Discussion

42.	Which of the following is amphoteric oxide—(A) Fe2O3 (B) Al2O3(C) P2O5 (D) N2O
Answer» Al₂O₃ is amphoteric oxide.

Discussion

43.	Which of the following is/are more acidic then A. B. C. D.
Answer» (A) Cl is more E.N. than I (B) F is more E.N. than I ( C) Br is more E.N. than I (D) `-I` is more if distance is less.

Discussion

44.	The incorrect statement is(A) The first ionization enthalpy of K is less than that of Na and Li(B) Xe does not have the lowest first ionization enthalpy in its group(C) The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.(D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.
Answer» Correct option is (D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30. Ionization enthalpy order : Li > Na > K He > Ne > Ar > Kr > Xe > Rn Sr > Rb Zn > Ga

44.

The incorrect statement is(A) The first ionization enthalpy of K is less than that of Na and Li(B) Xe does not have the lowest first ionization enthalpy in its group(C) The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.(D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Answer»

Correct option is (D) The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Ionization enthalpy order :

Li > Na > K

He > Ne > Ar > Kr > Xe > Rn

Sr > Rb

Zn > Ga

Discussion

45.	A gaseous mixture of CO and `CO_(2)` having total volume 150ml with excess of red hot charcoal to cause following reaction: `CO_(2)(g)+C(s)rarr2CO(g)` The volume increases to 250ml. Identify correct statement(s)A. Original mixture contain `(100)/(3)%` of CO.B. Original mixture contain 150 ml of `CO_(2)`.C. Original mixture contain 100ml of `CO_(2)`.D. Original mixture contain 50ml of CO,.
Answer» `CO_(2)+C(s)rarr2CO(g)` Vml excess `0ml` remain 2V ml `DeltaV=2V-V=V ml=250-150=100` `V_CO_(2))=100ml,V_(CO)=50ml` `%"of" CO=(V_(CO))/(V_("total"))xx100=(50)/(150)xx100` `=(100)/(3)%`

Discussion

46.	The ionization energy of `He^(+)` is x times that of H. The ionization energy of `Li^(2+)` is y times that of H. find `\|y-x\|`
Answer» Correct Answer - 5 `x=4, y=9 I.E.=13.6z^(2)`

Discussion

47.	At STP the density of a gas `X` is three times that of gas `Y` while molecule mass of gas Y is twice that of X. The ratio of pressures of X and Y will be:A. `B.C.D.
Answer» Correct Answer - 6 `P=(dRT)/M`

Discussion

48.	In an experiment, 50 mL of 0.1 M solution off a metallic salt `M(NO_(3))` reacted exactly with 25 mL of 0.1 M solution of `NaSO_(3)` in the reaction `SO_(3)^(2-)` is oxidized to `SO_(4)^(2-)`. If in the original salt oxidation number fo the metal was 3, what would be the new oxidation number of metal?
Answer» Correct Answer - 2 Meq of `Na_(2)SO_(3)=` meq of salt `25xx0.1xx2=50xx0.1xxn` factor

Discussion

49.	What is the number of ways of choosing 4 cards from a pack In how many of these (i) four cards are of the same suit? (ii) four card belong to four different suits?
Answer» Given 52 cards → select 4 cards. (i) 4 cards are of same smit. ∴ total number of selection = [(13C₄ + 13C₄ + 13C₄ + 13C₄)/4] = 4[13C₄] = 4\|13!/(4!x 9!)\| = (4 x 13 x 12 x 11 x 10)/(4 x 3 x 2) = 2860 (ii) 4 cards are of different suits Total number of selection = 13C₁ × 13C₁× 13C₁ × 13C₁ = (134)⁴ = (13)⁴ = 28.561

49.

What is the number of ways of choosing 4 cards from a pack In how many of these (i) four cards are of the same suit? (ii) four card belong to four different suits?

Answer»

Given 52 cards → select 4 cards.

(i) 4 cards are of same smit. ∴ total number of selection = [(13C₄ + 13C₄ + 13C₄ + 13C₄)/4]

= 4[13C₄] = 4|13!/(4!x 9!)| = (4 x 13 x 12 x 11 x 10)/(4 x 3 x 2) = 2860

(ii) 4 cards are of different suits

Total number of selection

= 13C₁ × 13C₁× 13C₁ × 13C₁

= (134)⁴ = (13)⁴ = 28.561

Discussion

50.	At STP the density of a gas `X` is three times that of gas `Y` while molecule mass of gas Y is twice that of X. The ratio of pressures of X and Y will be:
Answer» Correct Answer - 6 `P=(dRT)/(M)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

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What is the number of ways of choosing 4 cards from a pack In how many of these (i) four cards are of the same suit? (ii) four card belong to four different suits?

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