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.

Calculate molefractionof solutein its 2 molalaqueoussolution .

Answer»


SOLUTION :GIVEN :Concentrationof solution=2 molal = 2 m
Hence 2 molesof a solutearepresentin 1000gramswater .
`:. W_(H_(2)O) =100 g`
`:.` Numberof molesof water `=n_(H_(2)O) =(W_(H_(2)O))/(M_(H_(2)O))`
`= (1000)/(18)`
`=55.56 `MOL
`:. ` Totalmolesin solution`=n_(H_(2)O) + n_(solute)`
`=55.56 +2`
` = 57 .56 `mol
`:. ` Mole fractionof a solute
`=("Numberof molesof solute" )/("Totalmolesin solution ")`
`:. X_("solute")=(2)/(56.56) = 0.0347`
Discussion
2.

Calculate mole fraction of solute in an aqueous 4m solution assuming the density of the solution as 1.0g//mL

Answer»


Solution :(i) Let us consider one litre of sodium thiosulphate solution
`:.` wt of the solution `=` density `XX` volume (mL)
`=1.25xx1000=1250g`
Wt. of `Na_(2)S_(2)O_(3)` present in 1L of the solution
`=` molarity `xx` mol.wt.
`=3xx158=474g`
Wt `%` of `Na_(2)S_(2)O_(3)=(474)/(1250)xx100=37.92%`
(ii) Wt. of SOLUTE `(Na_(2)S_(2)O_(3))=474g`
Moles of solute `=(474)/(158)=3`
Wt. of SOLVENT `(H_(2)O)=1250-474=776g`
Moles of solvent `=(776)/(18)=43.11`
`:.` mole fraction of `Na_(2)S_(2)O_(3)=(3)/(3+43.11)=0.063`
(iii) Molality of `Na_(2)S_(2)O_(3)= ("moles of " Na_(2)S_(2)O_(3))/("wt.of solvent in grams")xx1000`
`=(3)/(776)xx1000=3.865m`
`:.1` mole of `Na_(2)S_(2)O_(3)` contains 2 moles of `Na^(+)` ions and 1 mole of `S_(2)O_(3)^(2-)` ions
`:.` molality of `Na^(+)=2xx3.865=7.73m`
Molality of `S_(2)O_(3)^(2-)=3.865m`
Discussion
3.

Calculatemolefractionof HCIin asolutioncontaining24.8% of HCIby mass.

Answer»


Solution :Given : %by weightofHCI = 24.8 ,
Molefractionof HCI= `X_(HCI)= ?`
Consider100 gramHCIsolution .
`:.` Weightof a solute(HCI) =24.8 G
Weightof ASOLVENT(water)= 100 -24.8 = 75.2 g
`:. ` Molesof solvent= `n_(H_(2)O) = .(W_(H_(2)O))/(M_(H_(2)O))=(75.2)/(18) =4.178 mol H_(2)O`
Molesof solute`=n_(H_(2)O) =(W_(H_(2)O))/(36.5) =(24.8)/(36.5)=0.6795 mol`
`:. ` Totalmolesin solution`= n_(H_(2)O) + n_(HCI)`
`=4. 178+ 0.6795 =4. 8575`
Molefractionof HCI `X_("HCI") =("molesof HCI")/("totalmolesin solution")`
`= (0.6795)/(4.8575)= 0.1398`
Discussion
4.

Calculate mole fraction, molarity and molality of C_(2)H_(5)OH solution which is 50% by weight of C_(2)H_(5)OH in H_(2)O and has a density of .9144 g per "cc".

Answer»


ANSWER :`0.281`, `9.93M`, `21.7m`
Discussion
5.

Calculate molality of 2.5 g of ethanoic acid (CH_(3)COOH) in 75 g of benzene.

Answer»

Solution :Molar mass of `C_(2)H_(2)O_(2) : 12xx2+1xx4+16xx2`
`= 60 g MOL^(-1)`
Moles of `C_(2)H_(4)O_(2)=(2.5g)/(60 g mol^(-1))=0.0417 mol`
Mass of benzene in `kg = 75 g//1000 g kg^(-1)`
`= 75xx10^(-3)kg`
Molality of `C_(2)H_(4)O_(2)=("Moles of " C_(2)H_(4)O_(2))/("kg of benzene")`
`= (0.0417 mol xx 1000 g kg^(-1))/(75 g)`
`= 0.556 mol kg^(-1)`
Discussion
6.

Calculate molar conductivity of 0.02M solution. [Lamda_(m)=10^(3)" S "cm^(2)mol^(-1)]

Answer»


ANSWER :`2.06xx10^(-3)" S "CM^(-1)`
Discussion
7.

Calculate minimum number of hydrogen atoms that should be present in a sample so that all possible transitions between fourth energy level and ground level can take place.

Answer»


Solution :For `n_(2) = 4` to `n_(1) = 1`
MINIMUM NUMBER of H atoms = 4 to show all TRANSITION
Discussion
8.

Calculate maximum mass of CaCl_(2) produced when 2.4 xx 10^(24)atoms is taken with 96 litre of Cl_(2) gas at 380 mm pressure and at 27^(@)C. [R : 0.08 atm L/mol-K & N_(A) = 6 xx 10^(23) ]

Answer»


ANSWER :222 GM
Discussion
9.

Calculate mass of water present in 499 gm CuSO_(4).5H_(2)O ? (Atomic mass : Cu = 63.5, S= 32, O = 16 , H = 1 )

Answer»


ANSWER :180 G
Discussion
10.

Calculate mass of phosphoric acid requiredto obtain 53.4 g pygrophosphoric acid. 2H_(3)PO_(4) to H_(4)P_(2)O_(7) + H_(2)O

Answer»


ANSWER :`58.8` G
Discussion
11.

Calculate mass of O atoms in 6 gm CH_(3)COOH ?

Answer»


ANSWER :`3.2g`
Discussion
12.

Calculate mass % of He in a mixture of O_(2) and He gas at 3 atm and 27^(@)C havin density (5)/(3) gm/litre [Take : R=0.08 atm litre mol^(-1)k^(-1)]

Answer»

Solution :PM=dRT for MIXTURE
`3xxM_("avg")=((5)/(3))xx0.08xx300`
`M_("avg")=(5xx8xx3)/(3xx3)=(40)/(3)`
`M_("avg")=(100)/((x)/(4)+(100-x)/(32))=(40)/(3)RARR(100xx32xx4)/(32x+400-4x)=(40)/(3)`
`32xx10xx3=28x+400`
`=28x=960-400=560`
x=20%
`(20)/(10)=2`
Discussion
13.

Calculate mass defect in the following reaction ._(1)H^(2) + ._(1)H^(3) rarr ._(2) He^(4) + ._(0)n^(1) (Given : mass H^(2) = 2.014, H^(3) = 3.016, He = 4.004, n = 1.008amu)

Answer»

0.018 AMU
0.18 amu
0.0018 amu
1.8 amu

Solution :MASS LOSS = mass of REACTANT - mass of PRODUCT
`= (2.014 + 3.016) - (4.004 + 1.008)`
`= 5.030 - 5.012 = 0.018` amu
Discussion
14.

Calculate magnitude of work involved in the reversible cyclic process ABCD if a diatomic ideal gas is involved. [Given 2^(5/7) = 1.6, ln 2 = 0.7] Express answers in bar-lit and round off to next integer.

Answer»


ANSWER :3
Discussion
15.

Calculate the magnetic moment of Mn^(2+) ion. [Atomic number of Mn=25]

Answer»

SOLUTION :Magnetic moment `(mu)= SQRT(n(n+2))`
`mu = sqrt(5(5+2)) = sqrt(35) = 5.923 M`.
Discussion
16.

Calculate magnetic moment of Fe^(3+) in [Fe(CN)_(6)]^(3+) and in [Fe(H_(2)O)_(6)]^(3+) Arrange following complexes in decreasing order of magnetic moment [Ni(H_(2)O)_(4)]^(2+), [Ni(CN)_(4)]^(2-), [Fe(CN)_(6)]^(3-), [Fe(CN)_(6)]^(4-)

Answer»

Solution : In `[Fe(CN)_(6)]^(3-)`, `Fe^(3+)`ion has only one unpaired electron, Thus, magnetic momentof `Fe^(3+)` will be `sqrt(3)`. i.e., 1.732 B. M. `(mu_(s) = sqrt(N(n+2)))` B.M., where, nis NUMBER of unpaired ELECTRONS). In `[Fe(H_(2)O)_(6)]^(3+) Fe^(3+) ion` has 5 unpaired electrons, hence, its magnetic moment will be `sqrt(35) B.M., i.e., 5.92 B.M`
(b) `[Fe(CN)_(6)]^(4-) = `[Ni(CN)_(4)]^(2+) lt [Fe(CN)_(6)]^(3-) lt [Ni(H_(2)O)_(4)]^(2+)`
`In`[Fe(CN)_(6)]^(4-)` and `[NI(CN)_(4)]^(2)`, central atom has no unpaired electron, hence, their magnetic moment is zero. In `[Fe(CN)_(6)]^(3+)` there is only one electron with central Fetion, hence, its magnetic moment will be `1.732 B.M` In `[Ni(H_(2)O)_(4)]^(2+)` there
are two unpaired electrons with `Ni^(2+)` and hence, it.s magnetic moment will be sqrt8 B.M`
Discussion
17.

Calculate magnitude of net work in following cyclic process for 1 mole of an ideal monoatomic gas (in cal). (Given: B rarr C isothermal process: D rarr A adiabatic process : T_(A) = 300 K, T_(D) = 75, T_(D) = 75 K, R = 2cal//"mole"-K: "ln" 2 = 0.7)

Answer»


ANSWER :1605
Discussion
18.

Calculate magnetic moment of Fe^(+3) ions. (Fe= 26)

Answer»

5.9BM
0.59BM
59 BM
590BM

Solution :ELECTRONIC configuration of `FE^(+3) = [Ar] 3d^(5) 4s^(0)` So, d-orbital
here, 5 unpaired ELECTRONS are PRESENT in d-orbitals so, magnetic momentum `mu = sqrt(n(n+2))`
`=sqrt(5(5+2))= 5.9BM`
Discussion
19.

Calculate Lamda_(m)^(@) for CaCl_(2) and MgSO_(4): Value of lamda_(m)^(@):

Answer»

Solution :We know from Kohlrausch law that: Limiting molar CONDUCTIVITY of electrolyte=sum of limiting molar conductivity of IONS present in it.
(i) Calculation for limiting molar conductivity of `CaCl_(2)`:
`CaCl_(2)toCa_((aq))^(2+)+2Cl_((aq))^(-)`
`therefore Lamda_(m)^(@)(CalCl_(2))=lamda_(m)^(@)(Ca^(2+))+2lamda_(m)^(@)(Cl^(-))`
`=[119.0+2(76.3)]" S "cm^(2)mol^(-1)`
`=(119.0+152.6)"S "cm^(2)mol^(-1)`
`=271.6" S "cm^(2)mol^(-1)`
(ii) Calculation for limiting molar conductivity of `MgSO_(4):`
`therefore Lamda_(m)^(@)(MgSO_(4))=lamda_(m)^(@)(MG^(2+))+lamda_(m)^(@)(SO_(4)^(2-))`
`=[106.0+160]" S "cm^(2)mol^(-1)`
`=266.0" S "cm^(2)mol^(-1)`
Discussion
20.

Calculate Lambda_(m)^(oo) for AgCl given that : Lambda_(m)^(oo)AgNO_(3)=133.4" ohm"^(-1)cm^(2)"equiv"^(-1) Lambda_(m)^(oo)KCl=149.9" ohm"^(-1)cm^(2)"equiv"^(-1) Lambda_(m)^(oo)KNO_(3)=145.1" ohm"^(-1)cm^(2)"equiv"^(-1)

Answer»


SOLUTION :According to Kohirausch's law :
`Lambda_(m)^(oo)AGCL=[Lambda_(m)^(oo)AgNO_(3)]+[Lambda_(m)^(oo)KCl]-[Lambda_(m)^(oo)KNO_(3)]`
`[133.4+149.9-145.1]=138.2" ohm "^(-1)cm^(2)EQUIV^(-1)`.
Discussion
21.

wedge^(@)._(m) for CaCl_(2) and MgSO_(4) from the given data. lambda_(Ca^(2+))^(@)=119.0S cm^(2)mol^(-1) ltbr. lambda_(Cl^(c-))^(@)=76.3S cm^(2)mol^(-1) lambda_(Mg^(2+))^(@)=106.0S cm^(2)mol^(-1) lambda_(SO_(4)^(2-))^(@)=160.0 cm^(2)mol^(-1)

Answer»

Solution :`Lambda_(m(CaCl_(2)))^(@)=Lambda_(m(CA^(2+)))^(@)+2Lambda_(m(Cl^(-)))^(@)`
`=119+(2xx76.3)="271.6 S CM"^(2)" MOL"^(-1)`
`Lambda_(m(MgSO_(4)))^(@)=Lambda_(m(Mg^(2+)))^(@)+2Lambda_(m(SO_(4)^(2-)))^(@)`
`=106+160="266 S cm"^(2)" mol"^(-1)`
Discussion
22.

Calculate Lambda_(HOAc)^(@) using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25^@C.

Answer»

`517.2`
`552.7`
`390.7`
`217.5`

SOLUTION :`(Lambda_(oo))_(HOAC) = [(Lambda^@)_(HCl) + (Lambda^@)_(NAOAC)] - (Lambda^@)_(NACL)`
`(426.2 + 91) - (126.5) = 390.7`.
Discussion
23.

Calculate Lambda_(HO Ac)^(oo) using appropriate molar conductances of the electrolytes listed above at infinite dilution in H_(2) O at 25^(@) C

Answer»

`217.5`
`390.7`
`552.7`
`517.2`

SOLUTION :`Lambda_(HCL)^(OO) = 426.2 "" … (i)`
`Lambda_(AcONA)^(oo) = 91.0 "" … (ii)`
`Lambda_(NaCl) = 126.5 "" …(iii)`
`Lambda_(AcOH)^(oo) = (i) + (ii) - (iii) = [426.2 + 91.0 - 126.5] = 390.7`
Discussion
24.

Calculate K_(p) for thereaction C_(2)H_(4(g))+ H_(2(g)) hArrC_(2) H_(6(g)) Delta G^(@)=- 100 kJmol^(-1)" at" 25^(@) C

Answer»

<P>

SOLUTION :`K_(p)`for thereaction`=3.39 XX 10^(17)`
Discussion
25.

CalculateK_(p)for thereactionat 240^(@)C NOCl_((g)) to 2NO_((g))+ Cl_(2(g)), Delta G^(@) = 7. 38 kJ m

Answer»

<P>

SOLUTION :`K_(p) = 0.017 ATM`
Discussion
26.

Calculate K_(p) for the equilibrium, NH_(4)HS_((s))iffNH_(3(g))+H_(2)S_((g)) if the total pressure inside the reaction vessel is 1.12 atm at 105^(@)C.

Answer»

`0.56`
`1.25`
`0.31`
`0.63`

Solution :`{:(,NH_(4)HS_((s)),iff,NH_(3(g)),+,H_(2)S_((g))),("LET initial moles",1,,0,,0),("At equilibrium",(1-x),,x,,x):}`
TOTAL GASEOUS moles at equilibrium = `x+x=2x`
We know `K_(p)=p_(NH_(3))xxp_(H_(2)S)`
but partial pressure (p) = mole fraction `xx` total pressure (P)
`K_(p)=((x)/(2x)xxP)((x)/(2x)xxP)=((P)/(2))^(2)=((1.12)/(2))^(2)=0.3136`
`II^(nd)` method : Both `NH_(3)andH_(2)S` have same number of moles at equilibrium so have same mole fraction and thus equal partial pressures.
i.e., `p_(NH_(3))=p_(H_(2)S)=(1.12)/(2)`
`K_(p)=p_(NH_(3))xxp_(H_(2)S)=(1.12)/(2)xx(1.12)/(2)=0.3136`
Discussion
27.

Calculate K_p and K_cof the reaction: I_2 iff 2Iwhen 1.513 xx 10^(-3)mole of iodine isheated to 1073 K if its vapour occupies a volume of 249.3 xx 10^(-6) m^3at 5.81 xx 10^4 Pa.

Answer»

SOLUTION :0.01113, 0.1264
Discussion
28.

Calculate K_(H) when 2 milimoles of O_(2) dissolve in 540 mL of water at 27^(@)C temperature. (P_(O_(2))=2xx10^(-8)" bar")

Answer»


ANSWER :`3XX10^(-4)` BAR
Discussion
29.

Calculate K_(c ) for the reversible process given below, if K_(p)=167andT=800^(@)C. CaCO_(3(s))iffCaO_((s))+CO_(2(g))

Answer»

<P>`1.95`
`1.85`
`1.89`
`1.60`

SOLUTION :`K_(p)=K_(c)(RT)^(DELTAN)`
`Deltan=n_(P)-n_(R)=1`
`K_(p)=K_(c)(0.0821xx1073)`
`K_(c)=(167)/(0.0821xx1073)=1.89`
Discussion
30.

Calculate K_(b) of water when 1 mole of the solute is dissolved in 1000g of water. The latent heat of vaporisation of water is 539.9 calories per gram.

Answer»


ANSWER :`0.514`
Discussion
31.

Calculate K_c for the reaction: A(g) +B(g) iff 2C(g) , if 1 mole of A, 1.4 moles of B and 0.50 mole of C are placed in a one-litre vessel and allowed to reach equilibrium. The equilibrium concentration of C is 0.75 mole per litre.

Answer»


ANSWER :0.5
Discussion
32.

Calculate K_(a) of acetic acif if its 0.05 N solution has equivalent conductance of 7.36 mho cm^(2) at 25^(@)C. "" (Lambda_(CH_(3)COOH)^(@) = 290.7)

Answer»

Solution :DEGREE of dissociation (x) `= (Lambda_(c))/(Lambda_(0)) = (7.36)/(390.7) = 0.0188`.
For the equilibrium
`{:(0.05,0,0,,"Initial concn. (moles/litre)"),(CH_(3)COOH =, CH_(3)COO^(-)+,H^(+),,),(0.05(-1x),0.05 x,0.05x,,"Equilibrium concentration"):}`
(for `CH_(3)COOH, 0.05 N = 0.05 M`)
`K_(a) = (0.05 x xx 0.05 x)/(0.05(1-x))`
Since x is very small,
`K_(a) = 0.05 x^(2) = 0.05 xx (0.0188)^(2)`
`= 1.76 xx 10^(-5)` mole/litre.
Discussion
33.

Calculate isoelectric point (PI) for following basic amino acid if:

Answer»

SOLUTION :
`PI=(pK_(a_2)+pK_(a_3))/(2)=(8.95+10.53)/(2)`
`=(19.48)/(2)`
`=9.74`.
Discussion
34.

Calculate instantaneous rate (r_(inst)) on the base of example problem-1 by graphically .Calculate 600 s,250 s,350s and 450s.

Answer»

Solution :`r_(INST)` is DETERMINED for any smallest time dt is as under .
When `Deltat to0` or `r_(inst)=-(d[R])/(dt)=(d[P])/(dt)`
So, dt =(When `Deltat` APPROACHES zero)
For given Butyl chloride determine
`r_(inst)=(d[C_(4)H_(9)Cl])/(dt)` ,draw graph of `[C_(4)H_(9)Cl]to t` (time) and determine the instantaneous rate by calculating the slope value of any time
To calculate value of slope value of any time .That touches the curves and TAKE `DeltaR` and `Deltat` in graph.

If time t=600 sec. is TAKEN
`r_(inst)(600s)=(0.0165-0.037 mol L^(-1))/((800-400)s)`
If t=250 s then `r_(inst)(250)=1.22xx10^(-4)mol L^(-1)s^(-1)`
If t=350 s then `r_(inst)(350)=1.0xx10^(-4)mol L^(-1)s^(-1)`
If t=450 s then `r_(inst)(450)=6.4xx10^(-5)mol L^(-1) S^(-1)`
Discussion
35.

Calculate (i)(5.7xx10^(6))xx(4.2xx10^(5))""(ii)(5.7xx10^(6))xx(4.2xx10^(-3))""(iii)(5.7xx10^(-4))xx(4.2xx10^(-2))

Answer»

SOLUTION :`(i)(5.7xx10^(6))XX(4.2xx10^(5))=(5.7xx4.2)(10^(6+5))=23.94xx10^(11)`
`(ii)(5.7xx10^(6))xx(4.2xx10^(-3))=(5.7xx4.2)xx(10^(6-3))=23.94xx10^(3)`
`(iii)(5.7xx10^(-4))xx(4.2xx10^(-2))=(5.7xx4.2)(10^(-4-2))=23.94xx10^(-6)`
Discussion
36.

Calculate : (i) Ratio of [Ag(NH_(3))_(2)]^(+) and [Ag^(+)] in 0.1M NH_(3) solution (ii) Ratio of [Ag(S_(2)O_(3))_(2)]^(3-) and [Ag^(+)] in 0.1M S_(2)O_(3)^(2-) solution. Given that the stability/formation constants (K_(f)) for [Ag(NH_(3))_(2)]^(+) and [Ag(S_(2)O_(3))_(2)]^(3-) are 1.7xx10^(7) and 1.0xx10^(13) respectively.

Answer»

Solution :(i) The equilibrium is :
`Ag^(+)+2NH_(3)overset(K_(f))iff[Ag(NH_(3))_(2)]^(+)`
`therefore` STABILITY CONSTANT, `K_(f)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=1.7xx10^(7)` (Given)
or `([Ag(NH_(3))_(2)]^(+))/([Ag^(+)])=1.7xx10^(7)xx[NH_(3)]^(2)=1.7xx10^(7)xx(0.1)^(2)=1.7xx10^(5)`
(ii) The equilibrium is : `Ag^(+)+2S_(2)O_(3)^(2-)overset(K_(f))iff[Ag(S_(2)O_(3))_(2)]^(3-)`
`therefore` Stability constant, `K_(f)=([Ag(S_(2)O_(3))_(2)]^(3-))/([Ag^(+)][S_(2)O_(3)^(2-)]^(2))=1.0xx10^(13)`(Given)
`therefore ([Ag(S_(2)O_(3))_(2)]^(3-))/([Ag^(+)])=1.0xx10^(13)xx[S_(2)O_(3)^(2-)]^(2)=1.0xx10^(13)(0.1)^(2)=1xx10^(11)`
Discussion
37.

Calculate (i) 4.56xx10^(3)+2.62xx10^(2)""(ii)4.5xx10^(-3)-2.6xx10^(-4).

Answer»

SOLUTION :`(i)4.56xx10^(3)+2.62xx10^(2)=45.6xx10^(2)+2.62xx10^(2)=(45.6+2.62)xx10^(2)=58.22xx10^(2)`
(II) `4.5xx10^(-3)-2.6xx10^(-4)=4.5xx10^(-3)-0.26xx10^(-3)=(4.5-0.26)xx10^(-3)=4.24xx10^(-3)`
Discussion
38.

Calculate how much current is necessary to produce hydrogen gas at the rate of 1 cc per second under standard conditions.

Answer»

Solution :`because` 1 equivalent of substance is produced by 1 mole of electric charge
`therefore` 11200 cc (i.e., 1 equivalent) of hydrogen is produced by 96500 coulombs
`therefore` 11200 cc of hydrogen is produced PER second by 96500 AMPERES (Q = I.t)
`therefore` 1 cc of hydrogen is produced per second by `(96500)/(11200)` AMP
= 8.61 amp.
Discussion
39.

Calculate how many methane molecules and how many hydrogen and carbon atoms are there in 25.0 g of methane?

Answer»


SOLUTION :No. of MOLES of methane`=(25.0)/(16.0)`
ONE molecule of methane contains one carbon atom and four hydrogen ATOMS.
Discussion
40.

Calculate how long a hydrogen atom will remain on the surface of a solid at 298 K if its desorption activatino energy is (a) 15 kJmol ^(-1) (b) 150 kJ mol^(-1). Assume that tau_(0)=10^(-13) s. Also calculate the results at 1000 K. What do you conclude from your results ?

Answer»

Solution :`tau=tau_(0)e^(E_(a)//RT)`
In `tau=ln tau_(0)+(E_(a))/(2.303RT)`
At 298 K
(a) `log tau =log10^(-3)+(15000)/(2.303xx8.314xx298)=-13+2.6289=bar(11).6289`
`therefore tau=` Autilog `bar (11).6289=4.255xx10^(-11)`
(b) `logtau log 10^(-13)+(150000)/(2.303xx8.314xx298)=-13+26.289=13.289`
`tau=` Antilog `=13.289=1.909xx10^(3)s~~600.000` years
At 1000 K
(a) `log tau =log 10^(-13)""+(15000)/(2.303xx8.314xx1000)=-13+0.7834=bar(13).7834`
`tau=` Antilog `bar (13).7834 =6.073xx10^(-13)~~6.1xx10^(-13)s`
(b) `logtau=-13+7.834=bar 6. 834`
`tau=` Antilog `bar6.834=6.823xx10^(-6)s~~6.8xx10^(-6)s`
Conclusion. (i) The time for a hydrogen atom to remain on the surface of the adsorbent depends upon desorption activation energy and temperature.
(ii) Greater the desorption activation energy `(E_(a)).` greater teh time. Greater the temperature, lesser the time.
Discussion
41.

Calculate heat of the following reaction at constant pressure, F_(2)O(g)+H_(2)O(g)rarrO_(2)(g)+2HF(g) The heats of formation of F_(2)O(g),H_(2)O(g) and HF(g) are 5.5 kcal-57kcal and -64 kcal repectively.

Answer»

Solution :GIVEN that,
(i) `F_(2)(g)+(1)/(2)O_(2)(g) to F_(2)O(g),DeltaH=-5.5kcal`
(ii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-57.8kcal`
(III) `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g) to HF(g),DeltaH=-64.2kcal`
`F_(2)O` and `H_(2)O` in EQNS. (i) and (ii) and in the equation given in the problem are on the opposite sides, while HF in Eqn. (iii) and in the equation given in the problem is on the same sides.
THUS applying, `[-Eqn. (i) - Eqn.(ii)+2xxEqn. (iii)]`, we get
`-F_(2)(g)-(1)/(2)O_(2)(g)-H_(2)(g)-(1)/(2)O_(2)(g)+H_(2)+F_(2)(g) to `
`-F_(2)O(g)-H_(2)O(g)+2HF(g),`
`DeltaH=-5.5-(-57.8)+2xx(64.2)`
or `F_(2)O(g)+H_(2)O(g) to O_(2)(g)+2HF(g),DeltaH=-76.1kcal.`
Discussion
42.

Calculate [H^(+)] in a 0.20M solution of dichloriacetic acid (K_(a)=5xx10^(-2)) that also contains 0.1M sodium dichloroacetate.Neglect hydrolysis of sodium salt.

Answer»

Solution :`{:(,CHCl_(2)COOHhArr,CHCl_(2)COO^(-),+,H^(+)),("Before diassociation",0.2,-,,-),("After DISSOCIATION",(0.2-x),x,,x),(,CHCl_(2)COONararr,CHCl_(2)COO^(-),+,Na^(+)),(,,0.1,,0.1):}`
for the dissociation of acid
`K_(a)=5xx10^(-2)=([CHCl_(2)COOH^(-)][H^(+)])/([CHCH_(2)COOH])or 0.05=([0.1+x][x])/([0.2-x])`
`x=0.05` or `[H^(+)]=0.05M`
Discussion
43.

Calculate [H^+], [H_2PO_4^-], [HPO_4^(2-)] and [PO_4^(3-)] in a 0.01 M solution of H_3PO_4. Take K_1=10^(-3), K_(2)=10^(-8), K_(3)=10^(-13)

Answer»


ANSWER :`[H^(+)]=[H_(2)PO_(4)^(-)], [PO_(4)^(3-)]=3.7 XX 10^(-19)`
Discussion
44.

Calculate [H^(+)] " and "% dissociation of 0.1 M solution of ammonium hydroxide solution . The ionisation constant for NH_(4)OH " is " K_(b) = 2.0 xxxx 10^(-5) .

Answer»

`7.09xx10^(-12)M, 3%`
`7.09xx10^(-12)M, 1.4%`
`9.02xx10^(-12)M, 2.4%`
`9.02xx10^(-12)M,3%`

ANSWER :B
Discussion
45.

Calculate frequency of the spectral line when an electron from 5th Bohr orbit jumps to the second Bohr orbit in a hydrogen atom.

Answer»


ANSWER :`6.906xx10^(14)HZ`
Discussion
46.

Calculate freezing point of solution prepared by dissolving 1.8 gm glucose in 500 gm water. K_(f) value for solvent is 1.8 K kg mol^(-1).

Answer»


ANSWER :272.964 K
Discussion
47.

Calculate equivalent conductivity at infinite dilution of the salt KOOC-COONa. Given the ionic molar conductivities at infinite dilution are: lambda^(oo)(C_(2)O_(4)^(2-))=148.5S cm^(2)mol^(-1), lambda^(oo)(K^(+))=50.1 S cm^(2) mol^(-1) lambda^(oo)(Na^(+))=73.5 S cm^(2) mol^(-1) and

Answer»

`136.05 S cm^(2)EQ^(-1)`
`544.2 Scm^(2) eq^(-1)`
`272.1 S cm^(2) eq^(-1)`
`27.21 S cm^(2) eq^(-1)`

Solution :Number of faraday `(q_(F))=(9.65xx60xx1.5)/(96500)=9XX10^(-3)F`
`2e^(-)+CU^(+2)rarrCu`
`""(9xx10^(-3))/(2)xx64=0.288`
`4OH^(-)rarrO_(2)+2H_(2)O+4e^(-)`
`""(9xx10^(-3))/(4)xx32=0.072g`
Total wt. loss `=0.288+0.072=0.360g`
Discussion
48.

Calculateequilibriumconstantfor thereaction Ni_((s))+ 2A_((aq))^(+)to Ni_((aq))^(+)+ 2Ag at 25^(@)C E_(Ni^(2+) | Ni)^(@)=- 0.25V and E_(Ag^(+) | Ag)^(@) = 0.799 V

Answer»

SOLUTION :`K= 2.754xx 10^(35)`
Discussion
49.

Calculate equilibrium constant for the following reaction : Zn+CuSO_(4) rarr ZnSO_(4) +Cu E_(Zn//Zn^(2+))^(@)=0.765" volt,"E_(Cu^(2+)//Cu)^(@)=0.347 volt

Answer»


ANSWER :`1.862xx10^(37)`
Discussion
50.

Calculate equilibrium constant for I_(2)+I^(-) hArr I_(3)^(-) at 298 K from the following information : {:(I_(2) (aq.)+2e^(-) rarr 2I^(-),,E^(@)=0.6197" volt"),(I_(3)^(-)+2e^(-) rarr 3I^(-),,E^(@)=0.5355" volt"):}

Answer»


Solution :`{:(I_(2)+2E^(-) RARR 2I^(-),""E^(@)=0.6197" volt"),(3I^(-) rarr I_(3)^(-)+2e^(-)," "E^(@)=-0.5355" volt"),(bar(I_(2)+I^(-) HARR I_(3)^(-),)""E^(@)=0.6197-0.5355),(""=0.0842" volt"):}`
`K="antilog" [(NE^(@))/0.0591]="antilog"[(2xx0.0842)/0.0591]=706.9`
Discussion