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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 standard free energy of formation of H_2O_((l)). The standard enthalpy of formation of H_2O_((l))is 285.85 kJ and standard entropies of H_(2(g)), O_(2(g)) and H_2O_((l))are130.5, 205.0 and 70.3 J.K^(-1) "mole"^(-1)respectively. Temperature is 25 C |
| Answer» SOLUTION :`-237.36 KJ "MOLE"^(-1)` | |
| 2. |
Calculate standard Gibbs free energy for the reaction : Zn+Cu^(2+)toZn^(2+)+Cu,E_("cell")^(@)=1.10V |
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Answer» Solution :`E_("cell")^(@)=1.10V` For the cell reaction, `Zn+CU^(2+)TOZN^(2+)+Cu` n=2 `F=96,500C` `DeltaD^(@)=-nFE_("cell")^(@)` `=-2xx96500xx1.10` `=-2123000J` `=-212.3kJ` |
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| 3. |
Calculate standard free energy of formation of H_2O_((l)). The standard enthalpy of formation of H_2O_((l))is 285.85 kJ and standard entropies of H_(2(g)), O_(2(g)) and H_2O_((l))are130.5, 205.0 and 70.3 J.K^(-1) "mole"^(-1)respectively. |
| Answer» SOLUTION :`-237.36 KJ "MOLE"^(-1)` | |
| 4. |
Standard electrode potential for Sn^(4+)//Sn^(2+) couple is+ 0.15 V and that for the Cr^(3+) //Crcoule is -0. 74 V. These two coules in their standard state are connected to made a cell. The cell potential wil be . |
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Answer» SOLUTION :GIVEN : `E_(SN^(4+)//Sn^(2+))^(@)=0.15V` `E_(Cr^(3+)//Cr)^(@)=-0.74V` `E_("cell")=E_(Sn^(4+)//Sn^(2+))^(@)-E_(Cr^(3+)Cr)^(@)` `=0.15-(-0.74)` `=0.89V.` |
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| 5. |
Calculate standard free energy change at 300K for a reaction whose equilibrium constant is 1 times 10^4 at 300K. |
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Answer» Solution :`Delta G^@=-2.303 RT log K_p` `=2.303 times 8.314 times 300 times log times 10^4` `=-22976.57 J OR -22.976 KJ` |
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| 6. |
Calculate spin magnetic momentum of M_((aq))^(+2) ions. (Z=27) |
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Answer» 3.80BM So, for M ions `=[Ar] 3d^(7) 4s^(2)` So, for `M^(+2)` ions = `[Ar] 3d^(7)`  Here, UNPAIRED ELECTRONS are 3. `:.` Magnetic momentum `mu = SQRT(n(n+2)) = sqrt(3(3+2)) = 3.87BM` |
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| 7. |
Calculate spin magnetic momentum of M_((aq))^(+2) ions. (Z=26) |
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Answer» 4.89 BM So, for M ions `= [Ar] 3d^(6) 4S^(2)` So, for `M^(+2)` ions `= [Ar] 3d^(6) 4s^(0)`  Here, UNPAIRED ELECTRONS are 4. `:.` Magnetic momentum `mu= SQRT(n(n+2))` `=sqrt(4(4+2)) = 4.89BM` |
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| 8. |
Calculate specific activity of a radioactive nucleus having average life of (10^(15))/(216) sec. radius of the nucleus equal to 8 fermi meter. ["Given : radius of nucleus" =4/3 xx10^(-13) A^(1//3) "cm where" 'A' "represents mass number" N_(A) = 6xx10^(23) ] ["Express your answer in terms of"10^(8) "dps"//g.] |
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| 9. |
Calculate solubility of MnS in a buffer solution of given pH.K_(sp) of MnS and K_(a_(1))&K_(a_(2)) for H_(2)S are given . |
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Answer» Solution :Let the NEW solubility of `MnS=x` `[Mn^(2+)]=x`=INITIAL concentration of `S^(2-)` ions but free `S^(2-)` ions will be less because some of the `S^(2-)` ions will react with `H^(+)` from buffer to form `HS^(-)` and `H_(2)S` `[Mn^(2+)]=x=underset("Free")[S^(2-)]+[Hs^(-)]+[H_(2)S]....(1)` Let us calculate the value of `[HS^(-)]`&`[H_(2)S]` in terms of free `[s^(2-)]` ion.For that consider: `HS^(-)hArr H^(+)+S^(2-)"",,,,,,""H_(2)ShArr H^(+)+HS^(-)` `K_(2)=([H^(+)][S^(2-)])/([HS^(-)])"",,,,""K_(1)=([H^(+)][HS^(-)])/([H_(2)S])` `[HS^(-)]=([H^(+)][S^(2-)])/(K_(2))....(2)` and `[H_(2)S]=([H^(+)][S^(-)])/(K_(1))=([H^(+)][S^(2-)])/(K_(1)K_(2)]....(3)` PUT (2) &(3) in (1) `x=[S^(2-)](1+([H^(+)])/(K_(2))+([H^(+)]^(2))/(K_(1)K_(2)))` `x=(K_(sp))/([Mn^(2+)])(1+([H^(+)])/(K_(2))+(H^(+)]^(2)/(K_(1)K_(2)))` `x=sqrt(K_(sp)(1+([H^(+)])/(K_(2))+([H^(+)]^(2))/(K_(1)K_(2))))` |
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| 10. |
Calculate [SO_4^(2-) ]in 0.15 M H_2SO_4solution if the first step of ionisation is complete and the second step K_2 = 1.02 xx 10^(-2) |
| Answer» SOLUTION :`8.9 XX 10^(-3)` | |
| 11. |
Calculate single electrode potential of Pt,Cl_(2)//Cl^(-) (0.01)(E^(@)= +1.36V) |
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| 12. |
Calculate similtaneous solubility of silver thiocyanate and silver bromids in water .Given that K_(sp) of silver thiocyanate=10^(-12) and K_(sp) of silver bromids =5xx10^(-13) respcectively. |
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Answer» Solution :Let the simultaneous solubility of `AGSCN` be `x` and that of `AgBr` is `y`.Then: `AgSCNhArr underset(x+y)(AG^(+))+underset(x)(SCN^(-))"",,,,,,,,,,,""AgBrhArr underset(x+y)(Ag^(+))+underset(y)(Br^(-))` `10^(-12)=XX(x+y)`....(i) `5xx10^(-13)=y(x+y)`....(ii) on SOLVING we get `x=2y` so `y=4.08xx10^(-7)M` and `x=8.16xx10^(-7)M` |
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| 13. |
Calculate rms speed of oxygen molecules in the lungs at normal body temperature, 37^(@)C. |
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Answer» Solution :We have, `C = SQRT((3RT)/(M))` `= sqrt((3 XX 8.314 xx 10^(7) xx 310)/(32)) = 4.92 xx 10^(4) cm//s` |
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| 14. |
Calculate rms speed of N_(2) at 298 K in metre/second. |
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Answer» SOLUTION :In SI units, R = 8.314 J/K/mol, T = 298 K and M = 0.028 kg/mol `THEREFORE C = SQRT((3RT)/(M))` |
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| 15. |
Calculate relative rate of effusion of O_(2) to CH_(4) through a container containing O_(2) and CH_(4) in 3:2 mass ratio: |
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Answer» `(3sqrt(2))/(4)` `=(3)/((2)/(2))=sqrt((16)/(32))` `=(3)/(4sqrt(2))` |
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| 16. |
Calculate relative rates of effusion of SO_(2) to CH_(4), in the mixture with initial molar ratio (n_(SO_(2)))/(N_(CH_(4)))=(8)/(1) |
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Answer» `2:1` `p PROPN` `(P_(SO_(2)))/(P_(CH_(4)))=(8)/(1)` `(r_(SO_(2)))/(r_(CH_(4)))=(8)/(1)sqrt((16)/(64))=(8)/(2)=4:1` |
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| 17. |
Calculate q, W, Delta E and Delta H when 100 g of CaCO_(3) is converted into its aragonite form given density of caicite =2 g// c c and density of aragonite =2.5 g// c c |
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Answer» Solution :`{:(CaCO_(3),hArr,CaCO_(3),,),("CALCITE",,"Aragonite",DeltaH=2kJ//"mole",):}` Generally for `{:("solid"rarr,"Solide",,,),("solid"rarr,"Liquid",,,),("solid" rarr,"Liquid",,,):}` transitions `W lt lt q` so, `DeltaE ~= q= DeltaH` while for gaseous conservation for example `{:("Solid",rarr,"gas",,),("Liquid",rarr,"gas",,):}` `q=Delta H !=Delta E` , as `W` will be siginificant `{:(S.No,Process,w,q,Delta U,Delta H,"Molar heat capacity"),(1,"Isochoric",0,nC_(vm)(T_(f)-T_(i)),q_(v),nC_(p m)(T_(f)-T_(i)),C_(vm)),(2,"isobaric",-P_(ext)(V_(2)-V_(1)),nC_(p m)(T_(f)-T_(i)),q+w,nC_(p m)(T_(f)-T_(i)),C_(p.m)),(3,"ISOTHERMAL",underset(W_(irrev)=-Pext(V_(f)-V_(i)))(W_(rav) = nRT In((V_(f))/(V_(i)))),-w,0,0,PM infty),(4,"Adidatic",(P_(f)V_(f)-P_(1)V_(1))/(alpha-1),0,w,0,0),(5,"Polytropic",(P_(f)V_(f)-P_(i)V_(i))/((n-1)),Delta U-w,nC_(vm)(T_(2)-T_(1)),nC_(p.m)(T_(2)-T_(1)),C_(vm)+(R)/(1- gamma)):}` |
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| 18. |
Calculate Q,W,Delta E and DeltaH for the isothermal reversible expansion of 1 mole of an ideal gas from initial pressure of 1.0 bar to final pressure of 0.1bar at constant temperature of 273K. |
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| 19. |
Calculate Q value of the following nuclear reaction 13^(Al^(27)) + 2^(He^(4)) rarr 14^(Si^(30)) + 1^(H^(1))+ Q. The exact mass of 13^(Al^(27)) is 26.9815 amu, 14^(Si^(30)) is 29.9738, 2^(He^(4)) is 4.0026 amu and 1^(H^(1)) is 1.0078 amu. |
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Answer» SOLUTION :`TRIANGLE m`=(29.9738+1.0078)-(26.9815+4.0026) =00.0025 AMU Q =0.0025 xx 931 MEV =2.329 MeV |
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| 20. |
Calculate Q value of the following nuclear reaction 13^(Al^(27)) + 2^(He^(4)) rarr 14^(Si^(30)) + 1^(H^(1))+ Q. The exact mass of 13^(Al^(27)) is 26.9815 amu, 14A^(Si^(30)) is 29.9738, 2^(He^(4)) is 4.0026 amu and 1^(H^(1)) is 1.0078 amu. |
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Answer» Solution :`triangle m`=(29.9738+1.0078)-(26.9815+4.0026) =00.0025 amu Q =0.0025 xx 931 MEV =2.329 MeV |
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| 21. |
Calculate 'Q' for last electron of Ga. where Q=n+l+ maximum possible value of 'm'. |
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Answer» SOLUTION :`Ga(31) 1s^(2)2s^(2)2p^(6)3S^(2)3p^(6)4s^(2)3D^(10)4P^(1)` n=4 l=1 n+1+m=6 m=1 |
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| 22. |
Calculate pOH of 0.001 M NH_(4)OH. When it is 1% dissociated in the solution |
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Answer» 5 `RARR 0.001 xx (1)/(100) = 1 xx 10^(-5), pOH = -log[OH^(-)]` `rArr log[1 xx 10^(-5)], pOH = 5`. |
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| 23. |
Calculate pH using following cell : underset(1 atm)(Pt(H_2)|underset(1 atm)(H^+(xM)|Pt(H_2)) if E_("cell")=0.2364V |
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| 24. |
Calculate pH of HCl solution at 298K temperature for the following cell: Pt_((S))|H_(2)("1 bar")|HCl (xM)||Ag^(+)(0.01M)|Ag_((S)) |
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| 25. |
Calculate pH of HCl solution at 298 K temperature for the following cell : Pt_((S))|H_(2)("1 bar")|HCl (xM)||Cu^(2+)(0.02M)|Cu_((S)) |
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| 26. |
Calculate pH of basic solution having OH^(-)as ( i) 10^(-2) M and(ii) 10^(-8) M |
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Answer» Solution :(i) `10^(-2)` M `OH^(-)` `OH^(-) = 10^(-2) , POH = - log [ OH ] =2 ` PH = 14-2 = 12 (ii) `10^(-8) M OH^(-)` In PRESENCE of `10^(-8) M OH^(-)` , water dissociates as `H_(2) O rarr H^(+) + OH^(-)` `( c-x ) `x`( 10^(-8)+ x )` `K_( w ) = [ H^(+)] [OH^(-) ] = ( 10^(-8) + x ) = 10^(-14)` Which reduce to QUADRATIC equation `x = ( - 10^(-8) +- sqrt( 10^(-8) + 4 xx 10^(-14)))/( 2)` `:. x = 0.95 xx 10^(-7) mol //L` `= [OH ] = ( 10^(-8) +x ) = 10^(-8) + 0.95 xx 10^(-7)` `= 1.05 xx 10^(-7)` POH `= - log [OH] = - log ( 1.05 xx 10^(-7)) = 6.9788 ` `PH = 14 - POH - 6.9788 = 7.0212` |
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| 27. |
Calculate pH of a basic soluion OH^(-) as (i) 10^(-2) M ,and(ii) 10^(-8) M |
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Answer» Solution :(i) `10^(-2) " M " OH^(-)` ` OH^(-) = 10^(-2) , POH = -log [OH^(-)] = 2 ` ` pH = 14-2 =2` (ii)` 10^(-6) " M " OH^(-)` In presenceof `10^(-8) " M " OH^(-) ` , WATER dissociates as `{:(H_(2)O,hArr,H^(+),+,OH^(-)),((C-x),,x,,(10^(-8)+x)):}` `K_(W) = [H^(+)] [OH]= (10^(-8)+x) x = 10^(-14)` Which reduces to quadratic equation ` x = (-10^(-8)pmsqrt(10^(-8)+4xx10^(-14)))/2` ` :.x = 0.95 xx10^(-7)` mol/L `[OH]= (10^(-8) +x) = 10^(-8) +0.95 xx10^(-7)` ` =1.05 xx10^(-7)` `pOH = - log [OH] = -log (1.05 xx10^(-7)) = 6.9788` ` pH = 14 - pOH = 14 - 6.9788 = 7.0212` |
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| 28. |
Calculate pH of a 0.1 M Na_2 HPO_4solution. K_1 , K_2and K_3for H_3PO_4 are 7.1 xx 10^(-3), 6.3 xx 10^(-6)and 4.5 xx 10^(-13) respectively. Which approximation is necessary for the calculation? |
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Answer» Solution : As `K_3` for `H_3PO_4` is very LOW compared to `K_1`assume no acidic ionisation. CONSIDER only the equilibrium, `HPO_4^(2-) + H_2O IFF H_2PO_4^(-) + OH^-` (10,10) |
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| 29. |
Calculate pH of 10^-8 M H_2SO_4 (ii) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4. (iii) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 HCl with 50 ml 0.1 M NaOH |
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Answer» Solution :`H_2SO_4 leftrightarrow^(H_2O)2H_3O^++SO_4^(2-)` `10^-8M 2 times10^-8 10^-8M` In this case the concentration of `H_2SO_4` is very lowand HENCE `[H_3O^+]` from water cannot be neglected. `therefore[H_3O^+]=2times10^-8` (from `H_2SO_4)+10^-7` (from water) `=10^-8(2+10)` `=12 times10^-8=1.2 times10^-7` `pH=-log_10[H_3O^+]` `=-log_10(1.2 times 10^-7)` `=7-log_10 1.2` `=7-0.0791=6.9209` (ii) pH of the solution=5.4 `[H_3O^+]`=antilog of (-pH) =antilog of (-5.4) =antilog of (-5+0.6)=6.6 `=3.981 times 10^-6` i.e, `3.98 times 10^-6 MOL dm^-3` (iii) No. of moles of HCl=`0.2 times 50 times 10^-3=10 times 10^-3` No. of moles of NaOH=`0.1 times 50 times 10^-3=5 times 10^-3` No. of moles of HCl after mixing=`10 times 10^-3 -5 times 10^-3` `=5times 10^-3` After mixing total volume =100 ML `therefore` COncentration of HCl in moles per litre `=(5 times 10^-3 MOLE)/(100 times 10^-3L)` `[H_3O^+]=5times10^-2M` `pH=-log(5times10^-2)` `=2-log5` `=2-0.6990` =1.30 |
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| 30. |
Calculate pH of 10^-7 M HCl |
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Answer» Solution :If we do not consider `[H_3O^+]` from the ionisation of `H_2O`. Then `[H_3O^+]=[HCL]=10^-7M` i.e., pH=7, which is a pH of a neutral solution, We know that HCl solution is acidic whatever may be the CONCENTRATION of HCl i.e., the pH value should be less than 7, In this CASE the concentration of the acid is very LOW `(10^-7M)`. Hence, the `H_3O^+ (10^-7M)` formed due to the AUTO ionisation of water cannot be neglected. So, in the case we should consider `[H_3O^+]` from ionisation of `H_2O` `[H_3O^+]=10^-7` (from HCl) +`10^-7` (from water) `=10^-7 (1+1)` `=2 times 10^-7` `ph=-log_10[H_3O^+]` `=-log_10(2 times 10^-7)=-[log 2+log 10^-3]` `=-log 2-(-7),log_10 10` `=7-log 2` `=7-0.3010=6.6990` =`6.70` |
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| 31. |
Calculatepackingefficinecy in thecrystal of body -centredcubic (bcc) structure. |
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Answer» Solution :Body -centred cubic (bcc) structure: In thisunit cell, THEREARE 8 atoms at 8 cornersand one atom atthe body centre . `therefore` Total NUMBER of atoms `= 8 xx (1)/(8) + 1 = 1 + 1 = 2` If a is the edgelengthof cubicunit cell and r is theradius of an atom them, in bcc unitcell `r =(sqrt(3))/(4) xxa` VOLUMEOF unit cell `=a^(3) = ((4)/(sqrt(3))xx r)^(3) = (64)/(3sqrt(3))r^(3)` Volumeof 1 atom `= (4)/(3)pir^(3)` `therefore` Packing fraction `=("Volumeoccupiedby 2atoms")/("Volume of unit cell")` `=(2xx(4)/(5)pir^(3))/((64)/(3sqrt(3))r^(3))` `= (sqrt(3)pi)/(8)` `=(sqrt(3) xx 3.142 )/(8)` ` = 0.68` `therefore` Packingefficinecy `= 0.68 xx 10= 68%` `therefore` %of voidspace= 100-68 = 32% . |
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| 32. |
Calculate packingefficieny in the crystal of facecentred cubic (fcc) sturcture. |
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Answer» Solution :Face - centred cubic (FCC) structre: In the unit cell, there are 8 atoms at 8 cornersand 6 atoms at 6 facecentres. NUMBEROF atoms due to 8 corners `= 8 xx (1)/(8) = 1` Number of atomsdue to 6 facecentre `= (1)/(2) xx 6 = 3` `therefore` Totalnumberof atoms in the unit cell = 1+3=4 If a is the edge lenghtof the unit cell and r is the radius OFTHE atoms , then in fccstructure, `r = (a)/(2sqrt(2))`. Hence `a = sqrt(2)` Volumeof unit cell `= a^(3) = (2sqrt(2r))^(3) = 16 sqrt(2r)^(3)` Volumeof unit cell`=a^(3) = (2qrt(2r))^(3) = 16 sqrt(2r)^(3)` Volumeof 1 atom `= (4)/(3)pir^(3)` Packing FRACTION `= ("Volume occupiedby 4 atoms")/("Volumeof unit cell")` `=(4((4)/(3)pir^(3)))/(16sqrt(2)pir^(3))` `=(16//3pir^(3))/(16sqrt(2)r^(3))` `=(pi)/(3sqrt(2))` `=(3.142)/(3xx1.414)` ` = 0 .74` `therefore` Packingefficiency `= 0.74 xx 100 = 74%` `therefore` of void SPACE= 100 - 74 = 26% |
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| 33. |
Calculate packing efficiency in face-centred cubic (fee) structure. |
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Answer» Solution :Face-centred cubic (Fee) structure: In this UNIT CELL, there are 8 atoms at 8 corners and 6 atoms at 6 face centres. Number of atoms due to 8 corner `= 8 XX (1)/(8) = 1` Number of atoms due to 6 face centres `= (1)/(2) xx 6 = 3` `THEREFORE` Total number of atoms in the unit cell= 1 + 3 = 4 If a is the edge length ofthe unit cell and ris theradius of the atom, the in fcc structure, `r = (a)/(2sqrt(2))`. Hence `a= 2sqrt(2)r` Volumeof unit cell `=a^(3) = (2sqrt(2r))^(3) = 16 sqrt(2r)^(3)` Volumeof 1 atom ` = (4)/(3) pir^(3)` Packingfraction `=("VOLUME occupiedby 4 atoms")/("Volumeof unit cell")` `=(4((4)/(3)pir^(3)))/(16 sqrt(2)r^(3)) = (16//3pir^(3))/(16sqrt(2)r^(3))=(pi)/(3sqrt(2))` `=(3.142)/(3xx 1.414) = 0.74` `therefore` Packing efficiency= 0.74 `xx` 100 = 74 % `therefore` % of void space = 100 - 7 4 = 26 % . |
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| 34. |
Calculate packing efficiency in HCP and CCP structures. |
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Answer» Solution : LET edge length be "a" and face DIAGONAL AC = b In `triangleABC`, `AC^2 = b^2 = BC^2 + AB^2` `= a^2 + a^2 = 2a^2` `b = sqrt(2)a` If r is the radius of the sphere We get `b = 4r = sqrt(2)a` OR `a = (4r)/(sqrt2) = 2sqrt(2)r` The number of spheres per unit CELL in CCP structure is 4. Hence total volume of four spheres is equal to `4 xx (4/3 pi r^3)` and the volume of cube is equal to `a^3` ` implies a^3 = (2sqrt(2)r)^3` % Packing EFFICIENCY =`("Volume occupied by four spheres in unit cell" xx 100)/("Volume of the unit cell")` `=(4xx(4/3pir^3)xx100)/(2sqrt2r)^3` =74% |
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| 35. |
Calculate packing efficiency in face centred cubic (fcc) structure. |
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Answer» Solution : Face centred cubic (fcc) structure: In this UNIT cell, there are 8 atoms at 8 corners and 6 atoms at 6 face centres. Number of atoms due to 8 corners `=8xx(1)/(8)=1` Number of atoms due to 6 face centres `=(1)/(2)xx6=3` `therefore` TOTAL number of atoms in the unit cell =1+3=4 If a is the edge length of the unit cell and r is the radius of the atom, then in fcc structure, `r=(a)/(2sqrt2)`. Hence `a=2sqrt2r` Volume of unit cell `a^(3)=(2sqrt2r)^(3)=16sqrt2r^(3)` Volume of 1 atom `=(4)/(3)pir^(3)` `"Packing FRACTION"=("Volume occupied by 4 atoms")/("Volume of unit cell")` `=(4((4)/(3)pir^(3)))/(16sqrt2r^(3))` `=(16//3pir^(3))/(16sqrt2r^(3))` `=(PI)/(3sqrt2)` `=(3.142)/(3xx1.414)` `=0.74` Packing efficiency=`0.74xx100=74%` `therefore ` % of void space=100-74=26% |
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| 36. |
Caleulate the packing efficiency in a unit cell of Cubic Close Packing(CCP) structure. |
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Answer» Solution :For ccp structure `a = (4r)/(sqrt(2)) = 2 sqrt(2) r or r = (a)/(2 sqrt(2))` Packing efficiency = `("VOLUME occupied by four spheres in the unit CELL" XX 100 %)/("Total volume of the unit cell")` `= (4 xx (4)/(3) pi r^(3) xx 100)/((2 sqrt(2) r)^(3))%` `= ((16)/(3) pi r^(3) xx 100)/(16 sqrt(2) r^(3))%` = 74% |
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| 37. |
Calculate pπ-dπ bonds present in SO_(4)^(2-), NO^(3-)and trimer of SO_(3) . If number of pπ-dπ bonds inSO_(4)^(2-), NO^(3-)and trimer ofSO_(3) are x, y and z respectively then calculate x+y+z. |
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| 38. |
Calculate packing efficiency in body-centred cubic structure. |
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Answer» Solution : In a body-centred unit cell, the spheres LOCATED at the corners donot touch each other but they are in contact with body centred atom. In `triangleEFD` `b^2 = a^2 + a^2` `2a^2` `b = sqrt(2)a`. In `triangleAFD` `c^2 = a^2 + b^2` ` = a^2 + 2a^2` ` = 3a^2` `c = sqrt(3)a` The length of the body diagonal c is equal to 4r, where r is the radius of sphere, as all three spheres along the diagonal touch each other. Therefore,`sqrt(3)a = 4r` `a = (4r)/(sqrt3)` Or `r = (sqrt3)/(4)a` In BCC, the total number of ATOMS per unit cell are two. Hence the total volume of two atoms is`(2 xx 4/3 pi r^3)` Volume of the cube = `a^3 = ((4)/(sqrt3) r)^(3)` % PACKING efficiency =`("Volume occupied by FOUR spheres in unit cell" xx 100)/("Volume of the unit cell")` `= (2 xx (4/3) pi r^3 xx 100)/((4/(sqrt4) r)^3) = 68%`. 
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| 39. |
Calculate pπ-dπ bonds present in SO_(4)^(2-) , NO^(3-) and trimer of SO_(3) . If number of pπ-dπ bonds in SO_(4)^(2-), NO^(3-) , and trimer of SO_(3) are x, y and z respectively then calculate x + y + z . |
| Answer» SOLUTION :`SO_(4)^(2-)`has 2 , `NO_(3)^(-)` has 0 and `SO_(3)` TRIMER has 6 | |
| 40. |
Calculate ^^overset(oo)HOAc using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25^(@)C |
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Answer» 390.7 `=91.0+426.2-126.5=517.2-126.5` `=390.7S CM^(2)"mol"^(-1)` |
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| 41. |
Calculate osmotic pressure of solution which is prepared by dissolving 1.75 gm glucose in 150 mL of solution at 17^(@)C. |
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| 42. |
Calculate [OH^(-)] in a 1M solution of NaOCN . K_a(HOCN) = 3.3 xx 10^(-4) . |
| Answer» SOLUTION :`5.5 XX 10^(-6) ` | |
| 43. |
Calculate of heat of formation of ethylene from the following data at 20^(@)C : H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),DeltaH=-65 kcal C(s)+O_(2)(g) to CO_(2)(g),DeltaH=-97 kcal C_(2)H_(4)(g)+3O_(2)(g) to 2CO_(2)(g)+2H_(2)O(l), DeltaH=-340 kcal |
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| 44. |
Calculate % of 'free volume' available in 1 mol gaseous water at 1.0 atm and 100^(@)C .Density of liquid water at 100^(@)C is 0.958 g//mol. |
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Answer» Solution :We know that PV = nRT Volume OCCUPIED by 1 mol gaseous WATER at `100^(@)C` and 1 atm pressure`= ( 1 xx 0.0821 xx 373 )/( 1) = 30.62 ` litre volume of 1 mol liquid water `= ( " mass")/( " density ") = ( 18) /( 0.958) = 1.88 xx 10^(-2) ` litre % of volume occupied by liquid water `= ( 1. 88 xx 10^(-2))/( 30.62 ) xx 100 = 0.0613 %` % of free volume = 100 - 0.0613 = 99.93% |
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| 45. |
Calculate % of frea volume' available in 1 mol gaseous water at 1.0 atm and 100 ^(@)C. Density of liquid water at 100^(@)C is 0.958g/mol. |
| Answer» SOLUTION :`99.93%` | |
| 46. |
Calculatenumber of m.e of theacids present in (i) 100 mL of 0.05 M oxalic acid solution , (ii) 50 mL of 0.1 M sulphuric acid solution . |
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Answer» Solution :Normality = molarity `xx` FACTORY relating mol.wt and eq.wt Normality of OXALIC acid= `0.5 xx2=1 N ` m.e of oxalicacid= normality `xx` vol. in ML = `1XX 100 = 100` (ii) Normality of sulphuric acid` = 0.1 xx2= 0.2 N ` m.e of sulphuric acid= `0.2 xx 50 = 10 ` |
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| 47. |
Calculate [NH_4^+](derived from NH_4Cl ) to prevent Mg(OH)_2from precipitating ina one-litre solution, containing 0.01 M NH, and 0.001 M Mg^(2+)ions.K_(sp)[Mg(OH)_2] = 1.2 xx 10^(-11), K_b(NH_4OH) = 1.8 xx 10^(-5). |
| Answer» SOLUTION :`1.6 XX 10^(-3)` | |
| 48. |
Calculate moles of excess reagent left and moles of product produced in the following - (i) {:(2A + 3B to 4C),(1""2):} (ii) {:(A + 2B + 3 C to 4C + 5B),(0.5" "0.8" "0.9):}s (iii) {:(3A + B to 2C),(5""2):} (iv) {:(2A+ 3B+4C to 5D + 6E),(12. " "1.8""3):} (iv) {:(5A+6B+8C+2D),(10 ""15):} |
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Answer» `{:((i),0,0.5,2,,),((II),0.2,0.2,0,1.2,1.5),((III),0,1/3,10/3,,),((IV),0,0,0.6,3,3.6),((iv),0,3,16,4,):}` |
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| 49. |
Calculate moles of C formed if 40 moles each of A and B are taken. A + 2B overset(40%)(rarr) D 2D overset(100%)(rarr) C |
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Answer» `2D overset(100%)(rarr) C` B is LB `n_(c ) = (1)/(2) xx 40 xx (40)/(100) xx (1)/(2) xx (100)/(100)` `= (1)/(4) xx 40 xx (40)/(100) = 4` moles |
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| 50. |
Calculate mole fractions of urea and water if 2.0g of urea is dissolved in 31.4g of aqueous solution. |
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Answer» |
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