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.

For complete neutralization of HCl with NaOH, the heat of neutralization is

Answer»

`+13.70 kJ mol^(-1)`
`-13.70 kJ mol^(-1)`
`-57.32 kJ mol^(-1)`
`-57.32 kJ mol^(-1)`

Solution :HEAT of NEUTRALIZATION of strong ACID and strong base is equal to the `-57.32 kJ mol^(-1)`.
Discussion
2.

For complete combustion of ethanol, C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l),the amount of heta producedas measured in bomb calorimeter, is 1364.47 kJ mol^(-1) at 25^(@). Assuming ideality the Enthalpy of combustion. Delta_(C)H, for the reaction will be(R=8.314 kJ mol^(-1))

Answer»

`-1366. 95 kJ mol^(-1)`
`-1361.95 kJ mol^(-1)`
`-1460.50 kJ mol^(-1)`
`-1350.50 kJ mol^(-1)`

SOLUTION :`C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`
`DeltaU=-1364.47 KJ//mol`
`DeltaH=DeltaU+Deltan_(g)RT`
`Deltan_(g)=-1`
`DeltaH=-1364.47+[(-1xx8.314xx298)/(1000)]`
`=-1364.47-2.4776`
`=-1366.94 KJ//mol`.
Discussion
3.

For complete combustion of ethanol, C_(2)H_(5)OH(l) + 3O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(l) the amount of heat produced as measured in bomb calorimeter is 1364.47 kJ mol^(-1) at 25^(@)C. Assuming ideality the enthalpy of comustion, Delta_(c)H for the reaction will be (R = 8.314 kJ mol^(-1))

Answer»

`- 1366.95 kJ mol^(-1)`
`- 1361.95 kJ mol^(-1)`
`- 1460.50 kJ mol^(-1)`
`- 1350.50 kJ mol^(-1)`

Solution :`C_(2)H_(5)OH(l) + 3O_(2)(G) rarr 2CO_(2)(g) + 3H_(2)O(l)`
`Delta n_(g) = 2 - 3 = - 1`
:. `Delta H = - 1364.47 + (1) xx 8.314 xx 10^(-3) xx 298`
`= - 1366.93 kJ mol^(-1)`
Discussion
4.

For colourless solution obtained in column-3 , the incorrect option is :

Answer»

3,b,R
4,b,P
3,a,Q
4,b,R

Solution :`Cu^(2+)+"excess"CN^(-)to underset("colourless SOL")([Cu(CN)_4]^(3-))`
`Cu^(+) , 3d^(10) implies sp^3` , diamagnetic & tetrahedral
`ZN^(2+)+"excess"NH_3(aq)tounderset("Colourless solution")([Zn(NH_3)_4]^(2+))`
`Zn^(2+), 3d^10 implies sp^3` tetrahedral diamagnetic
Discussion
5.

For coagulation of AS_(2)S_(3) colloid, the coagulation value is minimum for

Answer»

`AlCl_(3)`
`KI`
`BeCl_(2)`
`NaNO_(3)`

Solution :`AlCl_(3)`
Discussion
6.

For Co(II), (Choose incorrect statement):

Answer»

tetrahedral complexes are GENERALLY formed with monodentate anionic ligands LIKE `N_(3)^(-), OH^(-)` etc.
planar complexes are formed with BIDENTATE monoanionic like dmg, o-aminophenoxide etc.
planar complexes are also formed with a neutral bidentate ligands likeethylenediamine.
none of these

Answer :D
Discussion
7.

For coagulation of 20ml ofa negative sol Xml of 1M NaCl , Y ml of 1M BaCl_2 and Z ml of AlCl_3 are required separately . What is the correct order of X.Y and Z, values ?

Answer»


ANSWER :X > Y > Z
Discussion
8.

For coagulation of 10ml of a positive sol, the volumes of IM each NaCl, Na_(2),SO_(4),Na_(3)PO_(4) and Na_(4)[Fe(CN)_(6)] required separately are P, Q, R and S ml respectively. Arrange P, Q, R and S in the descending order.

Answer»

Solution :For coagulation of a POSITIVE sol, negative IONS are required.
As PER Hardy-Schulze rule greater the valence of the flocculating ions, greater is the flocculating power and lesser volume is required.
Thus the correct order of the volume of solutions is ` P GT Q gt R gt S` .
Discussion
9.

For [Co_(2)(CO)_(8)], what is the toal number of metal-carbon bonds and number of metal-metal bonds.

Answer»

10,1
8,2
8,1
10,0

Solution :STRUCTURE of `[Co_(2)(CO)_(8)]`

TOTAL M-C bonds=10, Total M-M bonds=1
Discussion
10.

For coagulating 200 mL of arsenious sulphide sol, 10 mL of 1M NaCl solution is required. Find out the flocculation volum of NaCl.

Answer»

Solution :10 mL of 1 M NaCl solution contain 10 milli-moles
Now, 200 mL of arsenious sulphide `(As_(2)S_(3))` REQUIRE for COAGULATION NaCl = 10 milli-moles
100 mL of solution require for coagulation `NaCl=(10)/(100)xx1000=50` milli-moles
Discussion
11.

For [Co (en)_3] Cl_3 : What type of stereoisomerism does it exhibit ?

Answer»

SOLUTION :OPTICAL ISOMERISM.
Discussion
12.

For clear water ,its BOD should be less than

Answer»

50 ppm
17 ppm
10ppm
5ppm

Answer :D
Discussion
13.

For [Co (en)_3] Cl_3 : Give the IUPAC name.

Answer»

SOLUTION :TRIS (ethane-1, 2-diamine) COBALT (III) CHLORIDE
Discussion
14.

For cis platin which option is incorrect ?

Answer»

It is used for cancer treatment
EAN is 86
Hybridisation is `dsp^(2)`
its DIPOLE MOMENT `NE0`

Solution :`[overset(+2)PT(NH_(3))_(2)Cl_(2)]
EAN=78-2+8=84`
Discussion
15.

For chloroform and acetone or for a solution of chloroform and acetone if p_s [observed(actual)] is compared with p_s [Theoretical (Raoult)] then which of the following is/are true ?

Answer»

`p_(s("ACTUAL"))LT p_(s("Reoult"))`

Solution :Chloroform forms HYDROGEN bond with acetone

Due to hydrogen bond formation vapour pressure of the solution BECOME less than expected RESULTS.
Discussion
16.

For chrome plating the electrolytic bath contains

Answer»

`HClO_4` and CONC. `H_2SO_4`
`CHROMIC` `ACID` and conc. `H_2SO_4`
`K_2Cr_2O_7`
Chromic sulphate

Answer :B
Discussion
17.

For chemical reactions, the calculation of change in entropy is normally done

Answer»

At constant pressure
At constant temperature
At constant temperature and pressure both
At constant volume

Solution :CALCULATION of change in ENTROPY is DONE at constant temperature and pressure both.
Discussion
18.

For CH_4 + Cl_2overset(hv)(to) product, rate of reaction at t=0 is given by rate = (-d)/(dt) [CH_4] = 2 xx 10^(-4) M-"min"^(-1) ,then rate of the reaction at t = 20 min is

Answer»

`"rate"=(-d)/(dt)[CH_(4)]=10^(-4)`
`"rate"=(-d)/(dt)[Cl_(2)]=2XX10^(-4)`
`"rate"=(-d)/(dt)[CH_(4)]=4xx10^(-4)`
`"rate"=(-d)/(dt)[Cl_(2)]="ZERO"`

Answer :B
Discussion
19.

For CH_3-CH_2-overset(CH_3)overset(|)C=CH_2 the only correct combination is

Answer»

<P>(III)(iv)(P)
(III)(i)(P)
(III)(ii)(S)
(III)(iv)(Q)

SOLUTION :`CH_3-CH_2-oversetoverset(CH_3)(|)C=CH_2overset(B_2H_6//H_2O_2.OH^(-))to`Anti-Markonikov's product
Discussion
20.

For CH_3-C=CH, the only incorrect combination is :

Answer»

<P>(I)(iii)(R)
(I)(IV)(S)
(I)(ii)(Q)
(I)(i)(P)

Solution :TERMINAL Alkynes do not undergo Birch Reduction
Discussion
21.

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : NH_4Cl hArr NH_3+HClI_2hArr 2I N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2 With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., A_n(g)hArr nA(g) t=0" " a" " 0 t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha. a-a alpha naalpha Total no. of moles =a-a alpha+n a alpha =[1+(n-1)alpha]a Observed molecular weight or molar mass of the mixture M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))=Molar mass of gas A_n The equation alpha=(D-d)/((n-1)d) is correctly matched for :

Answer»

`A hArr nB//2+nC//3`
`A hArr nB//3+(2n//3)C`
`A to(n//2)B+(n//4)C`
`A hArr (n//2)B+C`

Solution :`ALPHA=(D-d)/((n-1)d) IMPLIES (n-1)alpha=D/d-1`
`implies D/d=1+(n-1)alpha`
Hence ONE mole of reactant should PRODUCE total n moles of product
Discussion
22.

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : NH_4Cl hArr NH_3+HClI_2hArr 2I N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2 With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., A_n(g)hArr nA(g) t=0 " "a" " 0 t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha. a-a alpha naalpha Total no. of moles =a-a alpha+n a alpha =[1+(n-1)alpha]a Observed molecular weight or molar mass of the mixture M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))=Molar mass of gas A_n x(degree of dissociation) varies with D/d in the above reaction according to :

Answer»




SOLUTION :`D/d=1+(n-1)X`
`implies D/d=1+x"" [n=2]`
which is a line with positive slope and X INTERCEPT =1
Discussion
23.

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : NH_4Cl hArr NH_3+HClI_2hArr 2I N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2 With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., A_n(g)hArr nA(g) t=0"" a"" 0 t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha. a-a alpha naalpha Total no. of moles =a-a alpha+n a alpha =[1+(n-1)alpha]a Observed molecular weight or molar mass of the mixture M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))=Molar mass of gas A_n A sample of mixture of A(g),B(g) and C(g) under equilibrium has a mean molecular weight (observed ) is 80. The equilibrium is underset((mol.wt.=100))(A(g))hArrunderset((mol.wt.=60))(B(g))+underset((mol.wt.=40))(C(g)) Find the degree of dissociation alpha for A(g)

Answer»

`0.25`
`0.5`
`0.75`
`0.8`

SOLUTION :`A(g) HARR B(g)+C(g)`
`M_(Ob)=M_(th)/(M+(n-1)ALPHA)IMPLIES 80=100/(1+alpha)implies =1/4=0.25`
Discussion
24.

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : NH_4Cl hArr NH_3+HClI_2hArr 2I N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2 With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., A_n(g)hArr nA(g) t=0 " " a"" 0 t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha. a-a alpha naalpha Total no. of moles =a-a alpha+n a alpha =[1+(n-1)alpha]a Observed molecular weight or molar mass of the mixture M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))=Molar mass of gas A_n If the total mass of a mixture in the above case is 300 gm, the moles of C(g) present are.

Answer»

`1/4` MOLE
`4/3` mole
`3/4` mole
None

Solution :mole of C`=a ALPHA`
where a = MOLES of 'A'`=300/100 IMPLIES ` moles of `C=3xx0.25=3/4`
Discussion
25.

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : NH_4Cl hArr NH_3+HClI_2hArr 2I N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2 With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., A_n(g)hArr nA(g) t=0 " " a" "0 t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha. a-a alpha naalpha Total no. of moles =a-a alpha+n a alpha =[1+(n-1)alpha]a Observed molecular weight or molar mass of the mixture M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))=Molar mass of gas A_n The K_P for the reaction N_2O_4 hArr 2NO_2 is 640 mm at 775 K.The percentage dissociation of N_2O_4 at equilibrium pressure of 160 mm is :

Answer»

<P>`80%`
`30%`
`50%`
`70%`

SOLUTION :`{:(,N_2O_4""hArr,2NO_2),("Mole before equilibrium",1,),("Mole at equilibrium",(1-x),2x):}`
`K_p=(4x^2)/((1-x))xx[P/(sumn)]^(Deltan)implies 640=(4x^2)/((1-x))160/((1+x))`
`4=(4x^2)/((1-x))` or `1-x^2=x^2` or `2x^2=1`
`:. "" x^2=1//2 or x=0.707=70.7%`
Discussion
26.

For cell reaction, Zn+Cu^(2+)toZn^(2+)+Cu, cell representation is

Answer»

`Zn|Zn^(2+)||CU^(2+)|Cu`
`Cu|Cu^(2+)||Zn^(2+)|Zn`
`Cu|Zn^(2+)||Zn|Cu^(+)`
`Cu^(2+)|Zn||Zn^(2+)|Cu`

Solution :Electrode on which OXIDATION OCCURS is written on L.H.S. and the other on the R.H.S. as REPRESENTED by
`Zn|Zn^(2+)||Cu^(2+)|Cu`.
Discussion
27.

For CaCO_(3)(s)iffCaO(s)+CO_(2)(g)" at "977^(@)C, DeltaH=176 " kJ mol, then "DeltaE is

Answer»

180 KJ
186.4 kJ
165.6 kJ
160 kJ

Solution :`Deltan=1-0=1`
`DeltaE=DeltaH+DeltanRT`
`DeltaE=+176-1xx(8.314)/(1000)xx1240=165.6 kJ`.
Discussion
28.

For carbylamine reaction, we need alcoholic KOH and

Answer»

any PRIMARY AMINE and CHLOROFORM 
aromatic primary amine and chloroform 
ALIPHATIC primary amine and chloroform 
any amine and chloroform 

Answer :A
Discussion
29.

For C_(6)H_(5)CHO which of the following is incorrect

Answer»

On OXIDATION it YIELDS benzoic acid
It is used in perfumery
It is an AROMATIC aldehyde
On REDUCTION yields phenol

Solution :
On reduction it gives benzylalcohol and not phenol.
Discussion
30.

For C_(6)H_(5)CHO which of the following is incorrect ?

Answer»

On oxidationit YIELDS BENZOIC acid
It is USED in perfumery
It is an AROMATIC aldehyde
On REDUCTION yields phenol

Answer :D
Discussion
31.

For C_(12)H_(22)O_(11(aq))overset(H^(+))(to) Products, the concentration of sucrose changes from 0.06M to 0.03M in 30 minutes. Then, concentration of sucrose at the end of 60 minutes will be

Answer»

ZERO
`0.015M`
`0.09M`
`0.12M`

ANSWER :B
Discussion
32.

For, both lanthanides and actinides, which of the following statement is false ?

Answer»

Both involves the FILLING of(N - 2) F subshell
In atoms of both the series, three outer shells are partly filled while the remaining are completely filled
Some of them are electronegative in nature
Their cation with unpaired electrons are paramagnetic in nature

ANSWER :C
Discussion
33.

ForProducts, the concentration of sucrose changes from 0.06 M to 0.03 M in 30 minutes. Then, concentration of sucrose at the end of 60 minutes will be

Answer»

ZERO
0.015 M
0.09 M
0.12 M

Answer :B
Discussion
34.

For BaCl_(2) xH_(2)O, if 2.1 gm of compound gives 2 gm of anhydrous BaSO_(4) upon treatement with H_(2)SO_(4). Then calculate value of 'x'

Answer»

1
2
3
4

Solution :POAc on Ba
`n_(BaCl_(2)).xxH_(2)O=n_(BaAO_(4))`
`(2.1)/(208+18x)=(2)/(233)`
`48.3=416+36xrArr75.3=36x`
x=2.03 LTBR. `xapprox2`
Discussion
35.

For azimuthal quantum number 1 = 2, the maximum number of electrons will be :

Answer»

2
6
10
14

Answer :C
Discussion
36.

For azimuthal quantum number 1=3, the maximum number of electrons will be:

Answer»

2
6
Zero
14

Answer :D
Discussion
37.

For Ato Products, pressure of A t = 0 and t = 10 min respectively are 760 mm & 740 mm, rate of reaction during this time interval is (inmm/min)

Answer»

2
200
20
`(20)/(10xx760)`

ANSWER :A
Discussion
38.

For AtoB, a first order reaction, k=10^(-2)"min"^(-1). If initial concentration of A is 1M, rate of the reaction after 2.31 hrs will be ("in M-min"^(-1))

Answer»

`2.5XX10^(-4)`
`5xx10^(-3)`
`2.5xx10^(-1)`
`2.5xx10^(-3)`

ANSWER :D
Discussion
39.

For Ato Products (order = 1), rate of reaction is 4xx10^(-2)"M-s"^(-1) when [A] = 0.5 M, then, rate constant of the reaction is ("in sec"^(-1))

Answer»

`0.08`
`0.02`
`4XX10^(-2)`
`0.05`

ANSWER :A
Discussion
40.

For artificial transmutation of nuclei, the most effective one is

Answer»

Proton
Deuteron
Helium nuclei
Neutron

Answer :D
Discussion
41.

For artificial ripening of fruit which of the following is used ?

Answer»

TESTOSTERONE 
Insulin 
ETHYLENE 
Estrogen

Answer :C
Discussion
42.

For are reaction Ato product, initially 10 mole of A is taken. In 80 minutes 9.375 mole of A is found to be reacted. The rate constant of the reaction is represented as 3.5xx10^(-x)"min"^(-1). What is the value of X

Answer»


SOLUTION :`ATO` PRODUCTS `K=2.303/80xxlog(10/0.625)=2.303/80xxlot((100xx100)/625)=.03465\3.465xx10^(-2)`
Discussion
43.

For any given series of spectral lines of atomic hydrogen let Deltabar(v)=bar(v)_("max") fl bar(v)_("min")be the difference in maximum and minimum frequencies in

Answer»

`4:1`
`9:4`
`5:4`
`27:5`

Solution :`bar(v)PROP DeltaE`
For H - atom
`bar(v)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For Lyman SERIES,
`bar(v)" (MAX)"=13.6(1-(1)/(oo))`
`bar(v)" (min)"=13.6(1-(1)/(4))`
`therefore""bar(v)_("max")-bar(v)_("min")=13.6((1)/(4))`
For Balmer series,
`bar(v)(max)=13.6((1)/(4)-(1)/(oo))`
`bar(v)(min)=13.6((1)/(4)-(1)/(6))`
`therefore bar(v)_("max")-bar(v)_("min")=13.6((1)/(9))`
`(Deltabar(v)_("Lyman"))/(Deltabar(v)_("Balmer"))=(9)/(4)`
Discussion
44.

For any first order reaction following observation is made. If at temperature (T) half life of the reaction is 6930 sec. and at temperature (T') half life of the reaction is 0.693 mu sec. then, calculate (T')/(T).

Answer»

2
4
6
8

Answer :A
Discussion
45.

For any electron, the magnetic moment is …………….. To the orbital angular moment but ………………. To the orbital angular momentum.

Answer»

EQUAL, proportional
anti PARALLEL, proportional
parallel, INVERSELY proportiional
different, proportiional

Answer :B
Discussion
46.

For any chemical reaction ,value of slope of In Kto (1)/(T) graph will be…….

Answer»

`-(E_(a))/(2.303)`
`-(E_(a))/(R)`
`-(E_(a))/(2.303R)`
`-E_(a)`

ANSWER :B
Discussion
47.

Foranychemicalreactionchemiststry to findoutwhichthings?

Answer»

SOLUTION :For any chemical reactions chemists try to find out
(i)The feasibility of a chemical reaction which can be predicted by thermodynamics at constant pressure and temprature at constant pressure and temprature is `DeltaG` negatie than the reaction is spontaneous and if `DeltaG` is positive (`gt`0) than reaction is not feasible.
(ii)Extant to whicha reaction will PROCEED can be determind from chemical EQUILIBRIUM.
(iii)Speed of a reaction will proceed can be determined from chemical equilibrium .
(iii) Speed of a reaction i.e. TIME TAKEN by a reaction to reach equilibrium.
Discussion
48.

For anionic hydrolysis, pH is given by

Answer»

`pH=1/2 pK_w -1/2pK_b -1/2 LOG C`
`pH=1/2 pK_w +1/2 pK_a -1/2log C`
`pH=1/2 pK_w +1/2 pK_a +1/2 log C `
`pH=1/2pK_w +1/2 pK_a -1/2 pK_b`

SOLUTION :`pH=1/2 pK_w +1/2 pK_a +1/2 log C `
Discussion
49.

For anion of oxyacid, formed by removal of one water molecule from Boric acid , the only one correct combination is :

Answer»

(II)(i)(S)
(II)(IV)(R)
(III),(iii),(R)
(I),(ii),(R)

Solution :`undersetunderset("Oxidation number of (+3)")("BORIC acid")(H_3BO_3)overset(-H_2O)toundersetunderset("Oxidation number of (+3)")("META Boric acid")(HBO_2)toundersetunderset("Oxidation number of (+3)")("Meta Borate ion")(BO_2^(-))`
Discussion
50.

For anion of CaS_2O_8 , the only correct combination is :

Answer»

(II)(i)(S)
(III)(iii)(R)
(IV)(ii)(R)
(II)(iv)(S)

SOLUTION :
Discussion