Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Consider the complex [co(NH_(3))_(5)CO_(3)]ClO_(4). The coordination numebr, oxidation number, number of d-electrons and number of unpaired d-electrons on the metal ion are , respectively.

Answer»

6,3,6,0
7,2,7,1
7,1,6,4
6,2,7,3

Solution :Coordination numbr of Co in
`[Co(NH_(3))_(5)CO_(3)]ClO_(4)=6`
Oxidation number of Co `[Co(NH_(3))_(5)CO_(3)]^(+)`
`implies x-2=+1"":.x=+3`
`Co^(3+)(Z=27)to3d^(6)"":.` Number of d-electrons =6

No unapired d- electron.
Discussion
2.

Consider the complex given below [(NH_(3))_(5)overset("II")Ru(N_(2))overset("II")Ru(NH_(3))_(5)]^(4+) and the statements regarding it . (i) N-N bond length is longer than that in free N_(2) molecule. (ii) It is diamagnetic having t_(2g)^(4)e_(g)^(2) configuration of each Ru^(2+) ion . (iii) Back bonding is involved in this The correct statement(s) is/are :

Answer»

Only (i)
Only(i) and (II)
Only (ii) and (iii)
all of these.

Solution : (ii) diamagnetic `RARR t_(2)^(6)eg^(0)`.
Discussion
3.

Consider the complex [Co(NH_(3))_(5)(CO_(3))]ClO_(4), the coordination number, oxidation number, number of unpaired d-electrons respectively are : If your answer is 4, 2, 1 write your answer as 0421.

Answer»


Solution :`[overset(+3)(CO)(Noverset(0)H_(3))_(5)(overset(-2)(CO_(3)))]Cloverset(-1)O_(4)`,
C.N.=6,`d^(6)` (spin paired complex)
Discussion
4.

Consider the complex [Co(NH_(3))_(4)CO_(3)]ClO_(4), in which coordination number, oxidation number and number of d-electrons on the metal are respectively

Answer»

6,3,6
6,2,7
5,3,6
5,3,7

Solution :`[overset(III)(Co)(NH_(3))_(4)CO_(3)]^(+)ClO_(4)^(-),CO_(3)^(-)` is working as BIDENTATE LIGAND, have COORDINATION number of `overset(III)(Co)=6`.
Discussion
5.

Consider the complex [CoH(N_(2))(PPh_(3))_(3)] havingtrigonal bipyramidal geometry in which cobalt is dsp^(3) hybridised. Select the correct statements regarding the given complex.

Answer»

IUPAC name of the COMPLEX is hydridodinitrogentris (triphenylphospine) cobalt (I).
It is diamagnetic in nature.
`N^(2)` occupies axial position.
N-N bond length is longer than free `N_(2)` molecule.

Solution :`Cc^(+)=[Ar]4s^(1)3d^(7)`
`PPh_(3)` and `N_(2) to S.F.L.`
`Co^(+)=[Ar]3d^(8)`

Trigonal bipyramidal geometry with `dsp^(3)`hybridisation.All electrons are paired so diamagnetic `(dsp^(3))`.
ACCORDING to BENT's rule, in a TBP geometry more electronegative SPECIES `(N_(2))` occupies axial position. (a) Dinitrogenhydridotris (triphenyphosphine) cobalt (I) is the correct IUPAC name.
Discussion
6.

Consider the chemical reaction N_(2) (g) + 3H_(2)(g) to 2NH_(3) (g). The rate of this reaction can be expressed in terms of time derivated of concentration of N_(2)(g) , H_(2) (g) or NH_(3) (g) . Identify the correct relationship amongst the rate expressions .

Answer»

RATE = -d[`N_(2)`]/dt = `-1//3` d [ `H_(2)`] / dt = 1/2 [`NH_(3)`] /dt
Rate = `-d[N_(2)`]/dt = `-3D[H_(2)]`/dt = 3d `[NH_(3)`]/dt
Rate = d`[N_(2)`] /dt = 1/3d[`H_(2)`]/dt = 1/2 d `[NH_(3)]`/dt
Rate = `-d[N_(2)]`/dt = `-d[H_(2)]`/dt = d`[NH_(3)]`/dt

Solution :Individual rates become equal when eachof these are divided by their respective stoichiometric COEFFICIENT .
Further the rate of the consumption of the reactant goes on decreasing and the rate of formation of the product goes on INCREASING with TIME hence -ve and +ve sign , respectively .
Discussion
7.

Consider the chemical reaction : N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g) The rate of this reaction can be expressed in terms of time derivative of concentration of N_(2)(g), H_(2)(g) "or" NH_(3)(g) . Identify the correct relationship amongst the rate expressions :

Answer»

RATE `= -(d[N_(2)])/(dt)=-1//3d[H_(2)]//dt=(1)/(2)d[NH_(3)]/dt`
Rate `= -d[N_(2)]//dt=-3d[H_(2)]//dt=(2d[NH_(3)])/(1)`
Rate `=d[N_(2)]//dt=1//3d[H_(2)]//dt=(1)/(2d[NH_(3)]dt)`
Rate `=-d[N_(2)]//dt=-d[H_(2)]//dt=d[NH_(3)]//dt`

SOLUTION :(A) Correct relationship is
Rate =` -(d[N_(2)])/(dt) = -(1)/(3) (d[H_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt)`
Discussion
8.

Consider the chemical reaction N_(2)(g)+3H_(2)(g) rarr 2NH_(3)(g) . The rate of this reaction can be expressed in terms of time derivative of concentration of N_(2) (g) , H_(2)(g) or NH_(3)(g) . Indentifythe correct relationship amongst the rate expressions :

Answer»

RATE `= - d[N_(2)]//DT=-1//3d[H_(2)]//dt=1//2d[NH_(3)]//dt`
Rate `=-d[N_(2)]//dt=-3d[H_(2)]//dt=2d[NH_(3)]//dt`
Rate `=d[N_(2)]//dt=1//3d[H_(2)]//dt=1//2d[NH_(3)]//dt`
Rate = -`d[N_(2)]//dt=-d[H_(2)]//dt=d[NH_(3)]//dt`

Solution :(A) For the REACTION :
`N_(2) +3H_(2) rarr 2NH_(3)`
Rate =` -(d[N_(2)])/(dt) =-(1)/(3) (d[H_(2)])/(dt) = (1)/(2) (d[NH_(2)])/(dt)`
Discussion
9.

Consider the chemial reaction N_2(g) + 3H_2(g)to 2NH_3(g). The rate of this reaction can be expressed in term of time derivatives of concentration of N_2(g), H_2(g) or NH_3(g). Identify the correct realtionship amongst the rate expressions:

Answer»

Rate `= -(d[N_2])/(DT) = 1/3(d[N_2])/(dt) = 1/2(d[NH_3])/(dt)`
Rate `= (d[N_2])/(dt) = -3(d[H_2])/(dt) = 2(d[NH_3])/(dt)`
Rate `=(d[N_2])/(dt) = 1/3(d[N_2])/(dt) = 1/2(d[NH_3])/(dt)`
Rate `=(d[N_2])/(dt) = -(d[H_2])/(dt) = (d[NH_3])/(dt)`

Solution :For chemical reaction
`N_2(g)+ 3H_2(g) to 2NH_2(g) `
The rate of reaction is EXPRESSED as FOLLOWING
Rate `=-(d[N_2])/(dt) = -1/3(d[H_2])/(dt) = + 1/2(d[NH_2])/(dt)`
In the progress of reactionsd, concentration of `N_2` and `H_2` are decreased, so negative sign is REPRESENTED in expression, while concentration of `NH_2` are decreased, so negative sign is represented in expression, while concentration of `NH_3` is increased, so POSITIVE sign is represented inexpression
Discussion
10.

Consider the change in oxidation state ofBromine corresponding to different emf values as shown in thediagram below: BrO_(4)^(-) overset(1.82V)(rarr)BrO_(3)^(-)overset(1.5V)(rarr) HBrO overset(1.595V)(rarr) Br_(2) overset(1.0652V)(rarr)Br^(-) Then the species undergoing disproportional is

Answer»

`Br_(2)`<BR>`BrO_(4)^(-)`
`BrO_(4)^(3)`
HBrO

Answer :D
Discussion
11.

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below : BrO_(4)^(-) overset(1.82 V)(rarr)BrO_3^(-) overset(1.5V)(rarr) HBrO overset(1.595V)(rarr)Br_2 overset(1.0652V)(rarr)Br^(-) Then the species undergoing disproportional is .................. .

Answer»

`Br_2`
`BrO_4^-`
`BrO_3^-`
`HBrO`

Solution :
`(E_("CELL"))_A = -1.82 + 1.5 = -0.32 V`
`(E_("cell"))_B = -1.5 + 1.595 = +0.095 V`
`(E_("cell"))_C = 1.595 + 1.0652 = -0.529 V`.
`:.` The species undergoing disproportional is `HBrO`
Discussion
12.

Consider the change in oxidation state of bromide corresponding to different emf values as shown in the diagram below- BrO_(4)^(-) overset(1.82 V)to BrO_(3)^(-) overset(1.5V)to HBrO overset(1.595 V) to Br_(2) overset(1.0652V) to Br^(-) Then the species undergoing disproportionation is -

Answer»

`Br_(2)`<BR>`BrO_(4)^(-)`
`BrO_(3)^(-)`
HBrO

Answer :D
Discussion
13.

Consider the cell Zn|Zn^(2+) (aq) (1.0M) ||Cu^(2+) (aq) (1.0M) |Cu The standard reduction potentials are +0.35 V for 2e+ Cu^(2+) (aq) to Cu and -0.763V for 2e+Zn^(2+) (aq) to Zn Is the cell reaction spontaneous or not?

Answer»

Solution :Further `E_(cell)=E_(Ag^+,Ag)^@-E_(Cd^(2+),Cd)`
`={E_(Ag^+,Ag)^@+0.0591/1 LOG [Ag^+]}- {E_(Cd^(2+),Cd)^@+0.0591/2log[Cd^(2+)]}`
`={E_(Ag^+,Ag)^@-E_(Cd^(2+),Cd)^@}+0.0591/1 log""([Ag^+])/([Cd^(2+)]^(1/2))`
`=E_(cell)^@+0.0591 log""([Ag^+])/([Cd^(2+)]^(1/2))`
SINCE `[Cd^(2+)]` is in the denominator, DECREASE in `[Cd^(2+)]` will make `E_(cell)` more POSITIVE.
Discussion
14.

Consider the cell Zn|Zn^(2+) (aq) (1.0M) ||Cu^(2+) (aq) (1.0M) |Cu The standard reduction potentials are +0.35 V for 2e+ Cu^(2+) (aq) to Cu and -0.763V for 2e+Zn^(2+) (aq) to Zn Calculatethe emf of the cell.

Answer»

SOLUTION :In a GALVANIC CELLS , the anode is regarded as the negative ELECTRODE, The electrode `Cd|Cd^(2+)` is negative being the anode where oxidation occurs.
Discussion
15.

Consider the cell Zn|Zn^(2+) (aq) (1.0M) ||Cu^(2+) (aq) (1.0M) |Cu The standard reduction potentials are +0.35 V for 2e+ Cu^(2+) (aq) to Cu and -0.763V for 2e+Zn^(2+) (aq) to Zn Write down the cell reaction.

Answer»

SOLUTION :CELL REACTIONS are as FOLLOWS ,
RHS ELECTRODE `Cu^(2+) +2e=Cu` (reduction)
LHS electrode `Zn=Zn^(2+)+2e` (oxidation)
`therefore E_(cell)^@=E_(Ag^+,Ag)^@-E_(Cd^(2+),Cd)^@`
`=0.80-(-0.40)`
`=1.20` volts
Discussion
16.

Consider the cell Pt|H_(2(g,1atm))|H_((aq.1M))^(+)||Fe_((aq))^(3+),Fe_((aq))^(2+)||Pt_((s)) Given that E_(Fe^(3+)|Fe^(2+))^(o)=0.771V the ratio of conc. Of Fe_((aq))^(2+) to Fe_((aq))^(3+) is, when the cell potential is 0.830V

Answer»

0.101
0.924
0.12
None of these

Solution :For oxidation-reduction in half-cell:
The half-cell REACTION is `Fe_((aq))^(3+)+etoFe_((aq))^(2+)`
`Fe_(FE^(3+)//Fe^(2+))=E_(Fe^(3+)//Fe^(2+))^(o)-(0.059)/(1)log(([Fe^(2+)])/([Fe^(3+)]))`
`0.83=0.771-(0.0591)/(1)log(([Fe^(2+)])/([Fe^(3+)]))`
`IMPLIES([Fe^(2+)])/([Fe^(+3)])=0.10039~0.1004`
Discussion
17.

Consider the cell, Pt|H_(2(g))|H_((aq))^(+)"||"I_((aq))^(-)|I_(2(s)). If the standard cell potential is 0.54 V then the standard potential for cathode half reaction will be

Answer»

0 V
`0.54` V
`+0.54` V
`1.08`V

Answer :C::D
Discussion
18.

Consider the cell reaction Zn+Cu^(2+)(aq)iff Cu + Zn^(2+) (aq). Reaction quotient is Q = ([Zn^(2+)])/([Cu^(2+)])E^(0) of the cell is 1.10V. Now, E_(cell) will be 1.159V when:

Answer»

`([Cu^(2+)])/([ZN^(2+)]) = 0.01`
`([Zn^(2+)])/([Cu^(2+)]) = 0.01`
`([Zn^(2+)])/([Cu^(2+)]) = 0.1`
`([Zn^(2+)])/([Cu^(2+)]) = 1`

ANSWER :B
Discussion
19.

Consider the cellPt|Cl_(2(g))|HCl_((aq))|Br_(2(l))|Pt. If concentration of HClis increased, the cell potential will

Answer»

INCREASE
DECREASE
REMAIN the same
become maximum

Answer :A::C
Discussion
20.

Consider the cell: Cu|Cu^(2+)||Cl^(-)|Cl_(2),Pt Write the reaction that occur at anode and cathode.

Answer»

SOLUTION :ANODE: `CutoCu^(2+)+2e^(-)` (Oxidation): Cathode: `Cl_(2)+2e^(-)to2Cl^(-)`(REDUCTION).
Discussion
21.

Consider the cell Ag|AgBr(s)|Cr^(-)||Cl^(-)|AgCl(s)|Ag at 25^(@)C the solubility product constants of AgBr & AgCl are respectively 5xx10^(-13)&1xx10^(-10). For what ratio of the concentrations of Br^(-) & Cl^(-) ions would the e.m.f. of the cell be zero?

Answer»


Solution :If CELL is taken to be conc, cell `E_(cell)^(@)=0`
Anode: `AgtoAg_(a)^(+)+e^(-)`
cathode: `underline(Ag_(c)^(+)+e^(-)TOAG)`
`AghArrAg_(a)^(+)`
From nearest EQ,
`E_(cell)=E_(cell)^(@)-(0.059)/(1)log(([Ag^(+)]_(a))/([Ag^(+)]_(c)))implies0=0-(0.059)/(1)log(([Ag^(+)]_(a))/([Ag^(+)]_(c)))`
`therefore[Ag^(+)]_(a)=[Ag^(+)]_(c)implies(K_(sp)" of " AGBR)/([Br^(-)])=(K_(sp)" of " AgCl)/([Cl^(-)])`
or, `(5xx10^(-13))/(10^(-10))=([Br^(-)])/([Cl^(-)])=([Br^(-)])/([Cl^(-)])=(1)/(200)`
Discussion
22.

Consider the cell Ag|AgBr_((s)), Br^(-)||AgCl_((s)), Cl^(-)|Ag at 25^(@)C. The solubility product and AgCl and AgBr are 1 xx 10^(-10) and 5 xx 10^(-13) respectively. For what ratio of the concentrations of Br^(-) and Cl^(-) ions would the e.m.f. of the cell be zero ?

Answer»


ANSWER :`(1)/(200) ;`
Discussion
23.

Consider the below given statements about alkaline earth metals. (i) Solubility of sulphates decreases down the group. (ii) Solubility of hydroxides decreases down the group (ii) Thermal stability of carbonates increases down the group (iv) Basic nature of oxides increases down the group. The correct one are.

Answer»

 (i). (II) & (ILL)
(i), (ii) & (IV)".
(i) & (iv) only
(ii) only

Answer :B
Discussion
24.

Consider the cell Ag | AgBr | KBr || KCl | AgCl | Ag with EMF 0.059 V . Assume that [Br^(-)] = [Cl^(-)] . Here , conclusion inferred may be

Answer»

The ratio of the simultaneous solubilities of AgCl and AgBr in pure water is 1000
It is concentration cell
CHANGE in concentration of KCl will not EFFECT EMF
`K_(sp) (AgCl) gt K_(sp) (AgBr)`

SOLUTION :`E_(RP) = E_(SRP)^(0) -(0.0591)/(1) LOG""((Ag^(Ɵ))_a)/((Ag^(oplus))_e) ,0.059 = 0-(0.0591)/(1) log ""(K_(sp) AgBr[Br^(-)])/(K_(sp)AgCl[Cl^(-)]) = 10^(-1)`
` (K_(sp)AgBr)/([Br^(Ɵ)]) xx 10 = (K_(sp) AgCl)/([Cl^(Ɵ)]), K_(SP) [Br^(Ɵ)] = [Cl^(Ɵ)] ` Then ` K_(sp) AgCl to K_(sp) AgBr`
Discussion
25.

Consider the Born - Haber cycle for the formation of a ionic compound given below and identify the compound (Z) formed.

Answer»

MX
`M^(+)X_((G))^(-)`
`M^(+)X^(-)`
`M^(-)X_((s))^(-)`

Solution :The energy released when the requisite number of positive and negative IONS are condensed into crystal toform onemole of the compound is CALLED LATTICE energy.
`M_((g))^(+)+X_((g))^(-)overset(DeltaH_(5))tounderset((Z))(MX_((s)) DeltaH_(5)to`), Lattice energy.
In OPTION (a) state of MX is not given.
Discussion
26.

Consider the basicity of the following aromatic amines: (I) aniline (II) p-nitroaniline (III) p-methoxyaniline (IV) p-methylaniline the correct order of decreasing basicity is:

Answer»

IIIgtIVgtIgtII
IIIgtIVgtIIgtI
IgtIIgtIIIgtIV
IVgtIIIgtIIgtI

Answer :A
Discussion
27.

Consider the Arrhenius equation given below and mark the correct option. k="A e"^(-E_(a)//"RT")

Answer»

Rate CONSTANT increases exponentially with increasing activation energy and DECREASING temperature
Rate constant decreases exponentially with increasing activation energy and decreasing temperrature
Rate constant increases exponentially with decreasing activation energy and decreasing temperature
Rate constant increases exponentially with decreasing activation energy and increasing temperature

Solution :`K="A E"^(-E_(a)//"RT")`. As `E_(a)` increases, `-E_(a)//"RT"` decreases and so does k or conversely as `E_(a)` decreases, k will increase exponentially. As T increases, `E_(a)//"RT"` decreases or `-E_(a)//"RT"` increases and so does k.
Discussion
28.

Consider the Arrhenius equation given below and mark the correct option. k=A e^(-E_(a)//RT)

Answer»

Rate constant INCREASES exponentially with increasing ACTIVATION energy and decreasing TEMPERATURE.
Rate constant decreases exponentially with increasing activation energy and decreasing temperature.
Rate constant increase exponentially with decreasing activation energy and decreasing temperature.
Rate constant increases exponentially with decreasing activation energy and increasing temperature.

Solution :Arrhenius EQUATION:
`k=A e^(-E_(a)//RT)`
`therefore k PROP e^(-1//T)` and `kprop e^(e^(-E_(a)))`
`therefore k prop T`
`therefore` Rate constant increases exponentialy with decreasing activation energy and increasing temperature.
Discussion
29.

Consider the arrangement of circles of equal radii with their centres arranged as per the 2-dimensional lattice defined by a=b,theta=60^@ such that each circle is touching all its nearest neighbours.If all the void areas present are additionally occupied by smmaller circules of relevent size so that the void circles are just contacting their neighbours find the packing efficiency of the configuration in percent.

Answer»


Solution :PACKING FRACTION`=(1/2 PI R^2+pi(0.155R)^(2))/(1/2(2R)^2sqrt3/2)xx100=95%`
Discussion
30.

Consider the arrangement of bulbs shown in the drawing. Each of three bulbs contains a gas at pressure shown. What is pressure of system when all stopcocks are opened. Assuming that temperature remain constant. (Neglect the volume of capollary tubing connecting bulbs)

Answer»

440 torr
200 torr
360 torr
320 torr

Solution :After all stopcocks are OPENED, gases MOVE unit partial pressure becomes same in all bulbs
For `O_(2)` gas
`P_(1) V_(1) = P_(2) V_(2)`
`P_(1) = 300` torr, `P_(2)` ?
`V_(1) = 1` litre, `V_(2) = 2.5` litre
`300 XX 1 = P_(2) xx 2.5`
`P_(2) = 120` torr
For He gas
`P_(1) V_(1) = P_(2)V_(2)`
`P_(1) = 600` torr, `P_(2) =` ?
`V_(1) = 1` litre, `V_(2) = 2.5` litre
`600 xx 1 = P_(2) xx 2.5`
`P_(2) = 240` torr
For `N_(2)` gas
`P_(1)V_(1) = P_(2)V_(2)`
`P_(1) = 400` torr, `P_(2) =` ?
`V_(1) = 0.5` litre, `V_(2) = 2.5` litre
`400 xx (1)/(2) = P_(2) xx (5)/(2)`
`P_(2) = 80` torr
Pressure of SYSTEM
`= P_(O_(2)) + P_(N_(2)) + P_(He)`
`= 120 + 80 + 240`
`= 440` torr
Discussion
31.

Consider the amines: 1. NH_(3) 2. RNH_(2) 3. C_(6)H_(5)NH_(2) The correct sequence of basicity is :

Answer»

`3lt2lt1`
`3lt1lt2`
`1lt2lt3`
`2lt3lt1`.

ANSWER :B
Discussion
32.

Consider the acid strength of the carboxylicacids : (i) PhCOOH (ii) o-NO_(2)C_(6)H_(4)COOH(iii)p-NO_(2)C_(6)H_(4)COOH (iv)m-NO_(2)C_(6)H_(4)COOH

Answer»

`i gtiigtiiigtiv`
`ivgtiiigtiigti`
`iigtiiigtivgti`
`iigtivgtiiigti`.

ANSWER :C
Discussion
33.

Consider some facts about decomposition of H_(2)O_(2). It is catalysed by iodide ion in acidic medium. It is second order reaction with respect to both H_(2)O_(2).andI^(-1). Rate equation of this particular reaction will be Rate=(-d[H_(2)O_(2)]^(2))/(dt)=k[H_(2)O_(2)]^(2)[I^-] It completes in two steps and both steps are unimolecular elementary reactions. Which of the above written facts are correct, regarding decomposition of H_(2)O_(2)? Choose the correct option.

Answer»

I and II
II and III
III and IV
I and Iv

Solution :Decomposition of `H_(2)O_(2)`
`2H_(2)O_(2)overset(1^(-))underset("Alkaline medium")rarr2H_(2)O+O_(2)`
Rate EQUATION is rate `(-d[H_(2)O_(2)])/(dt)=K[H_(2)O_(2)][1^(-)]`
This reaction is first order w.r.t. both `H_(2)O_(2) and 1^(-).` It MAKES place in two steps and both are bimolecular elementary reactions.
Discussion
34.

Consider thfollowing statement I Q=lt II charge is required for reductin or reduction depends on the stoichiometry of electrode reaction III charge on 1 electron =1.6021xx10^(-19) C IV charge on one mole of elctron =1.6021xx10^(-19) C (V) Quantity of electricity is coulomb VI 1F =96500 C mol^(-1)which of the statements (s) given above is /areincorrect ? choose the correct option

Answer»

I and II
II and III
IVand V
VI and I

Solution :Q=It
Qis in Columbs when l is in ampere and t isin second
CHARGEON one electrone =`1.6021 xx1d0^(-19)` C
Charge on one MOLE of electron =`96487 C mol^(-1)`
`1 F = 96500 C mol^(-1)`
Quantity of electricityis calledfaraday
Discussion
35.

Consider separate solutions of 0.500 M C_(2)H_(5)OH(aq), "0.100 M "Mg_(3)(SO_(4))_(2)(aq),"0.250 M KCr(aq) and 0.125 M "Na_(3)PO_(4)(aq)" at "25^(@)C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

Answer»

`"0.500 M "C_(2)H_(5)OH (aq)` has the highest OSMOTIC pressure
They all have the same osmotic pressure
`"0.100 M "Mg_(3)(PO_(4))_(2)` has the highest osmotic pressure
`"0.125 M "Na_(3)PO_(4)(aq)` has the highest osmotic pressure

Solution :Osmotic pressure for ELECTROLYTES, `pi=iCRT`
`pi(C_(2)H_(5)OH)=1xx0.5RT=0.5RT,`
`pi(Mg_(3)(PO_(4))_(2))=5xx0.100RT=0.5RT`
`pi(KBr)=2xx0.25RT=0.5RT,`
`pi(Na_(3)PO_(4))=4xx0.125RT=0.5RT`
Thus, all solution have the same osmotic pressure.
Discussion
36.

Consider separate solutions of 0.500 M C_(2)H_(5)OH(aq), 0.100 M Mg_(3)(PO_(4))_(2)(aq),0.250M KBr(aq) and 0.125 M Na_(3)PO_(4)(aq) at 25^(@)C. Which statement is True about theses solutions, assuming all salts to be strong electrolytes ?

Answer»

`0.500 M C_(2) H_(5)OH(AQ)` has the highest osmotic pressure
They all have the same osmotic pressure
`0.100 MM g_(3)(PO_(4))_(2)(aq)` has the highest osmotic pressure
`0.125 M Na_(3)PO_(4)` (aq) has the highest osmotic pressure

Solution :`0.5 M C_(2)H_(5)OHimpliespi =0.5 RT`
`0.1 M Mg_(3)(PO_(4))_(2)impliespi=iCRT=(5xx0.1)RT=0.5RT`
`0.250M Kbrimpliespi=iCRT=(2xx0.250)RT=0.5RT`
`0.125M Na_(3)PO_(4)impliespi=iCRT=(4xx0.125)RT=0.5RT`
Discussion
37.

Consider separate solutions of 0.500 M C_(2)H_(5)OH_((aq)), 0.100 M Mg_(3)(PO_(4))_(2(aq)), 0.250 M KBr_((aq)) and 0.125 M Na_(3)PO_(4(aq)) at 25^(@)C.Which statement is true about these solutions, assuming all salts to be strong electrolytes ?

Answer»

`0.125 M Na_(3)PO_(4(aq))` has the highest osmotic pressure.
`0.500 M C_(2)H_(5)OH_((aq))` has the highest osmotic pressure.
They all have the same osmotic pressure.
`0.100 M Mg_(3)(PO_(4))_(2(aq))` has the highest osmotic pressure.

ANSWER :C
Discussion
38.

Consider separate solutions of 0.500 M C_(2)H_(5)OH_((aq)), 0.100 M Mg_(3)(PO_(4))_(2)(aq), 0.125 M Na_(3)PO_(4)(aq) at 25^(@)C. Which statement is true about tese solutions, assuming all saltes to be strong electrolytes ? Solutions, assuming all salts to be strong electrolytes ?

Answer»

they all have the same osmotic PRESSURE
o.100 M `Mg_(3)(PO_(4))_(2)(AQ)` has the HIGHEST osmotic pressure osmotic pressure.
`0.125 M Na_(3)PO_(4)(aq)` has the highest osmotic pressure.
`0.50 M C_(2)H_(5)OH(aq)` has the highest osmotic pressure.

Solution :Osmotic pressure `pi=I CRT`
For all the solution, R and T are both CONSTANT :
`pipropixxc`
Osmotic pressure depends upon the product of ven't Hoff factor and molar cncentration of the solution. `{:("Solution",i,c,ixc),(C_(2)H_(5)OH(aq),1,0.5,0.5),(Mg_(3)(PO_(4))_(2)(aq),5,0.1,0.5),(KBr(aq),2,0.25,0.5),(Na_(3)PO_(4)(aq),4,0.125,0.5):}`
Since the product is constant, this means that all the soklutions have some osmotic pressure.
Discussion
39.

Consider separatesolutions of 0.500 M C_(2)H_(5)OH(aq), 0.100 M Mg_(3)(PO_(4))_(2)(aq), 0.125 M Na_(3)PO_(4)(aq)at 25 0 C. Which statement is true about these solutions, assuming all salts to be strong electrolytes ?

Answer»

They all have same osmotic pressure.
0.100 M `Mg_(3)(PO_(4))_(2)` (aq) has the highest osmotic pressure.
0.125 M `Na_(3)PO_(4)`(aq) has the highest osmotic pressure
0.500 M `C_(2)H_(5)OH`(aq) has the highest osmotic pressure.

Solution :Osmotic pressure,
`pi=icRT`
`pi[C_(2)H_(5)OH]=1xx0.500xxR XX T=0.5"RT"`
`pi[Mg_(3)(PO_(4))_(2)]=5xx0.100xxRxxT=0.5RT`
`pi[KBR]=2xx0.250xxRxxT=0.5RT`
`pi[Na_(3)PO_(4)]=4xx0.125xxRxxT=0.5RT`
Therefore, all have same osmotic pressure.
Discussion
40.

Consider separate solution of 0.5 M C_(2)H_(5)OH(aq), 0.1 M Mg_(3)(PO_(4))_(2)(aq), 0.25 M KBr(aq) and 0.125 M Na_(3)PO_(4)(aq) at 25^(@)C which statement is true about these solutions, assuming all salts to be strong electrolytes ?

Answer»

They all have the same osmotic PRESSURE
`0.1 M Mg_(3)(PO_(4))_(2)(aq)` has the highest osmotic pressure
`0.125 M Na_(3)PO_(4)(aq)` has the highest osmotic pressure
`0.5 M C_(2)H_(5)OH(aq)` has the highest osmotic pressure

Answer :A
Discussion
41.

Consider para sulphony benzene diazonium reacting with para amino phenol which of the following statements are correct :

Answer»

A coupling is an electrophilic substitution.
under basic conditions DIAZONIUM attacks at alpha POSITION.
under weakly acidic CONDITION diazonium attacks at b position.
If sulphonyl group is REPLACED with methyl group couplingtakes place

Answer :A::B::C
Discussion
42.

Consider H_(3)PO_(2),H_(3)PO_(3),H_(3)PO_(4) and H_(4)P_(2)O_(7) . Which of the above oxoacids resultsinto two series of salts ?

Answer»

`H_(3)PO_(2)`
`He_(3)PO_(3)`
`H_(3)PO_(4)`
`H_(4)P_(2)O_(7)`

SOLUTION :Phosphorous acid ` (H_3PO_3)`is a diprotic acid . It formstwo seriesof salts such as `NaH_2PO_3`and `Na_2HPO_3` but NONE of the type `NaPO_3` with NaOH . Its structureis as
Discussion
43.

Consider given compounds & write the number of compounds which have higher HOC than

Answer»


ANSWER :4
Discussion
44.

Consider gas to be characterised by Vander Waal's equation. {:(,"Column - I",,"Column - II",),((A),Z = 1,(P),H_(2) " gas at room temperature and moderate pressure",),((B),Z gt 1,(Q),"Any gas at very low pressure and high temperature.",),((C),Z lt 1,(R),"Any gas showing more compressibility than expectedd from ideal behaviour",),((D),Z = (3)/(8),(S),"Gas at critical point",),(,,(T),"Gas at Boyle's temperature and at different pressures.",):}

Answer»


Answer :(A) = QT, (B) = PT, (C) = RS, (D) = RS
Discussion
45.

Consider following two reactions Ato"Product", (d[A])/(dt)=k_(1)[A]_(0)^(0) Bto"Product", (d[B])/(dt)=k_(1)[A]_(0)^(0) k_(a) and k_(2) are expressed are expressed in termof molarity (mol L^(-1))and time (s^(-1)) as :

Answer»

`s^(-1),M s^(-1)L^(-1)`
`M s^(-1),MS^(-1)`
` s^(-1),M^(-1)s^(-1)`
` M s^(-1), s^(-1)`

ANSWER :D
Discussion
46.

Consider following vapour pressure composition graph. Hence-

Answer»

V.P. of A = SZ
V.P. of B = ZY
V.P. of B = SY
V.P. of SOLUTION at X = SZ + SY

Answer :A::C::D
Discussion
47.

Consider following transformations! I. XeF_(5)+NaFto Na^(+)[XeF_(7)]II. 2PCI_(5) (s)to[PCI_(4)] [PCI_(5)]III A(H_(2)O)_(5)Н_(2)Oto[AI(H_(2)O)_(5)+H_(2)H_(2)O possible transformations are:

Answer»

 I, II, III
I,III
I, II
II, III

Answer :A
Discussion
48.

Consider following statements – (i) Froth floatation technique is used to concentrate sulphide ore. (ii) Froth floatation technique is based upon differential wettability of ore and impurity in water.(iii) Red Bauxite is concentrated by Bayer's process (iv) Carbon is used as reducing agent during extraction of zinc from zinc blende. Correct statements are –

Answer»

(i) and (II)
(ii) and (III)
(i),(iii) and (IV)
All are correct

Answer :D
Discussion
49.

Consider following solutions: I. 1M aqueous glucose solution. II. 1M aqueous sodium chloride solution III. 1M aqueous ammonium phosphate solution. IV. 1M benzoic acid in benzene. Select correct statements for the above solutions.

Answer»

All are ISOTONIC solutions
III is hypertonic of I, II and III
IV is HYPOTONIC of I, II and III
II is hypotonic of III but hypertonic of I and IV

Answer :B::C::D
Discussion
50.

Consider following solutions. 1. 1 M aq. Glucose 2. 1 M aq. Sodium chloride 3. 1 M benzoic acid in benzene 4. 1 M ammonium phosphate Select correct statements (s)

Answer»

all are isotonic SOLUTIONS
2 is hypotonic of 1,2,4
1,2,4 are hypertonic of 3
4 is hypertonic of 1,2,3

Answer :B::C::D
Discussion