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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. |
The iron(s) that act/s as an oxidising agent in solution is/are |
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Answer» `TL^(+)` and `AL^(3+)` |
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| 2. |
The iron atom, how many electrons atom have n=3 and l= 2 ? |
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Answer» 2 |
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| 3. |
The IP_(1) of Li is 5.41eV and E.A of Cl is -3.61eV. The DeltaH in KJ/mole for the reaction Li(g) + Cl(g) rarr Li^(+)(g) +Cl^(-)(g) is 1.736 xx 10^(x) KJ/mole. The value of 'x' is |
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Answer» |
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| 4. |
The IP_(1) of His 13.6 eV it is expoxedinduced radiation .Find the wavelengthof these ijnduced radiation |
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Answer» Solution :`E_(1)` of H atom `= - 13.6 eV`1 Energy givento H atom `= (6.625xx 10^(-34) xx 3.0 xx 10^(8))/(1028 xx 10^(-10))` `= 1.933 xx 10^(-18) J = 12.07 eV` Energy of H atom after excitation `= - 13.6 + 1207 = - 1.53 eV` `:. E_(N) = (E_(1))/(n^(2))` `:. n^(2) = (-13.6)/(-1.53) = 9 :. n = 3` The electron in H atom is excited to third shell :. induced `lambda_(1)= (HC)/(E_(3) - E_(1))` `E_(1) = -13.6 ev, E_(3) = - 1.53 eV` `lambda_(1) = (6.626 xx 110^(-34) xx 3 xx 10^(6))/((-1.53 + 13.6) xx 1.602 xx 10^(-19)) = 1028 xx 10^(-10) m` `:. lambda = 1028 Å` `1` induced `lambda_(2) = (hc)/(E_(2) - E_(1)))` `E_(1) = -13.6 eV, E_(2) = -(13.6)/(4) eV` `:. lambda_(2)= (6.625 xx 10^(-34) xx 3 xx 10^(6))/((-(13.6)/(4) +13.6)xx1.602 xx 10^(-`19))` `:. induced lambda_(3) = (hc)/(E_(3) - E_(2))` `E_(1) = - 13.6 eV,E_(2) = - (13.6)/(4) eV` `E_(3) = -(13.6)/(9) eV` `:. lambda_(3) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/((-(13.6)/(4) + (13.6)/(4)) xx1.602 xx 10^(-19))` `= 6568 xx 10^(-10) m = 6568 Å` |
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| 5. |
TheI.P. Of Nais 5.48eV, theionizationpotentialof K willbe |
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Answer» SAMEAS THATOF Na |
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| 6. |
The ions of NaF and MgO all have the same number of electrons and internuclear distances are about the same (235 pm and 215 pm ). Why are then the melting points of NaF and MgO so different (992^@C and 2642^@C)? |
| Answer» SOLUTION :In `Mg^(2+)O^(2-)` , both the ions carry two units of charge whereas in `Na^+ F^-` , each ion carries only one unit charge. HENCE, electrostatic forces of attraction in MGO are STRONGER. | |
| 7. |
The ions of NaF and MgO all have the same number of electrons and internuclear distances are about the same (235 pm and 215 pm). Why are then the melting points of NaFand MgO so different (992^(@) C and 2642^(@)C) ? |
| Answer» SOLUTION :In ` MG^(2+)O^(2-)`, both ions CARRY TWO units of charge whereas in ` Na^(+) F^(-)` , each ions carries only ONE unitcharge. Hence, eelctrostatic forces of attraction in MgO are stronger. | |
| 8. |
The ions O^(-2),F^(-),Mg^(2+) and Al^(3+) are isoelectronic. Their ionic radii show |
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Answer» a decrease from `O^(2-)` to `F^(-)` and then increase from `Na^(+)` to `AL^(3+)` |
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| 9. |
The ionizationpotential ofhydrogen is13.6 0 eV. Calculate the energyin Kj required to produce 0.1mole mg H^(+) ions. Given1 eV =96 .49 kJ mol^(-1) |
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Answer» |
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| 10. |
The ionization potential of nitrogen is more than that of oxygen because of |
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Answer» the GREATER attraction of the electrons by the nucleus |
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| 11. |
The ionization enthalpy of hydrogen atom is 1.312xx10^6 "J mol"^(-1). The energy required to excite the electron in the atom from n=1 to n=2 is |
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Answer» `6.56xx10^5 "J MOL"^(-1)` `=1.312xx10^6xx3/4=0.984xx10^6 =9.84xx10^5` J/mol |
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| 13. |
The ionization energy of hydrogen in excited state is +0.85 eV. What will be the energy of the photon emitted when it returns to the ground state? |
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Answer» SOLUTION :ENERGY of H-atom in the ground state `= - 13.6 eV` Ionization energy of HYDROGEN in excited state equal to `+0.85 eV` MEANS Energy of H-atom in the excited state `= - 0.85 eV` `:.` Energy EMITTED `= -0.85 eV - (-1.6 eV) = 12.75 eV` |
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| 14. |
The ionization energy of hydrogen is |
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Answer» GREATER than INERT gases |
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| 15. |
The ionization energy of hydrogen atom is -13.6 eV. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro's constant =6.022xx10^23) |
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Answer» `1.69xx10^(-20)` J We know that energy required for excitation `DeltaE=E_2-E_1=-3.4-(-13.6)=10.2 eV` Therefore energy required for excitation of ELECTRON per atom `=10.2/(6.02xx10^23)=1.69xx10^(-23)` J |
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| 16. |
The ionizationenergyof hydrogenatomis 1.312 xx 10^(6) J mol^(-1)calculatethe energyrequirefor jumpof electron from n=1 ton=2 |
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Answer» `9.56 xx 10^(5) j//mol` `=1.312 xx 10^(6)((1)/(1)-(1)/(4))` `=9.84 xx 10^(5)J mol^(-1)` |
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| 17. |
Ionization potential of hydrogen atom is 13.6 eV hydrogen atoms in the ground state are excited by monochromatic radiation of photo energy 12.1 eV. According to Bohr's theorgy, the spectral lines emitted by hydrogen will be |
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Answer» N=3 to n=1 |
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| 18. |
Theionizationenergyhydrogenatom is13.6ev .Theionizationenergyof He |
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Answer» 13.6 eV `E_(H) = (13.6 Z^(2)H)/(n^(2)H)` |
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| 19. |
The ionization constant of weak acid H_3PO_4 in three step are respectively K_a(1), K_a(2) and K_a (3). Give increasing order of these constant. Give reason. |
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Answer» SOLUTION :`K_a(3) lt K_a(2) lt K_a(1)` It is EASIER to form conjugate base by REMOVAL of proton from neutral atom therefore `K_a (1)` is maximum. As, negative charge of conjugate base is more the removal of positively charge proton from it is difficult. As a RESULT negative charge increase then `K_a`, decreases VALUE of `K_a(1) to K_a(2) to K_a(3)` decreases . |
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| 20. |
The ionization constant of phenol is 1.0xx10^(-10) .What is the concentration of phenolate ion is 0.05 M solution of phenol ? What will beits degree of ionization if the solution is also 0.01 M in sodium phenolate ? |
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Answer» Solution :The CONCENTRATION of Phenolate ion `(C_6H_5OH)` in 0.05 M Phenol `(C_6H_5O^-)` is =`ALPHA` `{:(,C_6H_5OH_((aq)) + H_2O hArr,C_6H_5O_((aq))^(-) + , H_3O_((aq))^(+)),("Initial :", 0.05 M, -,-),("Change in equilibrium :", -alpha,+alpha,+alpha),("At equili. M:" ,(0.05-alpha),alpha,alpha) :}` `K_a=([C_6H_5O^-][H_3O^+])/([C_6H_5OH])=1.0xx10^(-10)` `therefore alpha^2/(0.05-alpha)=1.0xx10^(-10)` `therefore alpha^2/0.05 = 1.0xx10^(-10)` (`because alpha lt lt` 0.05 , So take 0.05 - `alpha`=0.05) `therefore alpha^2= 0.05xx10^(-10) = 5.0xx10^(-12)` `therefore alpha=sqrt(5.0xx10^(-12))=2.236xx10^(-6) M = [C_6H_5O^-]` If 0.01 M sodium phenolate degree of ionization COMPLETE ionization of sodium phenolate is under. `{:(C_6H_5ONa to , C_6H_5O^(-)+ , Na^+),(0.01 M,0.01 M,0.01M):}` If dissociation degree of phenol is equal to x . So, `therefore` Dissociated phenol = `C xx x`= (0.05 )x `therefore` At equilibrium [Phenol]=(0.05-0.05x) but x is very less, So , [Phenol] `approx` 0.05=C At equilibrium `[H_3O^+]` in [Phenol] =`Cxxx`= 0.05 x (Phenolet ion at equilibrium)=(Phenolet formed from phenol)+(Phenolet of sodium phenolet ) `[C_6H_5O^-]=(C alpha+0.01)` `=(0.05 alpha+0.01)` `approx` 0.01 (`because alpha` is very less) This values put in table : `{:(,C_6H_5OH_((aq))+H_2O_((l)) hArr, C_6H_5O_((aq))^(-)+,H_3O_((aq))^(+)),("At equili.",0.05 M, 0.01 M, x(0.05)):}` `K_a=([C_6H_5O^-][H_3O^+])/([C_6H_5OH])=1.0xx10^(-10)` `therefore K_a=((0.01)(0.05x))/((0.05))=1.0xx10^(-10)` `therefore x=(1.0xx10^(-10))/0.01 = 1.0xx10^(-8)` |
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| 21. |
The ionization constant of propanoic acid is 1.32xx10^(-5). Calculate the degree of ionization if its solution is 0.05 M. What will be its degree of ionization if the solution is 0.01 M in HCl also? |
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Answer» Solution :Assuming `alpha` to be very small,applying the formula directly, we have `alpha=sqrt(K_(a)//c)=sqrt((1.32xx10^(-5))//0.05)=1.62xx10^(-2)` `CH_(3)CH_(2)CO OH hArrCH_(3)CH_(2)CO O^(-)+ H^(+)` In presence of HCL, equilibrium will shift in the backward direction, i.e., concentration of `CH_(3)CH_(2)CO OH` will increase, i.e., amount dissociated will be LESS. If c is the INITIAL concentration and x is the amount now dissociated,then at equilibrium `[CH_(3)CH_(2)CO OH]=c-x, [CH_(3)CH_(2)CO O^(-)]=x, [H^(+)]=0.01 + x ` `:.K_(a) = (x(0.01xx x))/(c-x)~=(x(0.01))/(c) or (x)/(c) =(K_(a))/(0.01)=(1.32xx10^(-5))/(10^(-2))=1.32xx10^(-3)`. But `(x)/(c) ` means `alpha`. Hence, `alpha = 1.32xx10^(-3)`. |
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| 22. |
The ionization constant of propanoic acid is 1.32 xx 10^(-15). Calculate the degree of ionization if its solution is 0.05 M. What will be its degree of ionization if the solution is 0.01 M in HCl solution. |
| Answer» SOLUTION :`1.62 XX 10^(-2) , 1.32 xx 10^(-3)` | |
| 23. |
The ionizationconstant of phenol is 1.0xx10^(-10). What is the concentration of phenate ion in 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01 M in sodiumphenate ? |
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Answer» Solution :`{:(,C_(6)H_(5)OH,hArr,C_(6)H_(5)O^(-),+,H^(+),,),("INITIAL",0.05M,,,,,,),("After disso.",0.05-x,,x,,x,,):}` `:. K_(a) = (x xx x)/(0.05-x) = 1.0xx10^(-10)` (Given) or `(x^(2))/(0.05) = 1.0xx10^(-10)` or `x^(2)=5xx10^(-12) or x = 2.2 xx 10^(-6) M` In presence of 0.01 `C_(6)H_(5)ON a`, suppose y is the AMOUNT of phenol dissociated, then at equilibrium `[C_(6)H_(5)OH]=0.05-y~=0.05, [C_(6)H_(5)O^(-)]=0.01+y~=0.01M, [H^(+)]=y M` `:. K_(a)=((0.01)(y))/(0.05)=1.0xx10^(-10) ` (Given) or `y=5xx10^(-10)` `:. alpha = (y)/(c) = (5xx10^(-10))/(5xx10^(-2))=10^(-8)`. |
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| 24. |
The ionization constant of nitrous acid is 4.5xx10^(-4). Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. |
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Answer» Solution :Sodium nitrite is a SALT of weak acid , strongbase. HENCE, `K_(h)=2.22xx10^(-11)=K_(w)//K_(a)=10^(-14)//(4.5xx10^(-4))` `h=SQRT(K_(h)//c)=sqrt(2.22xx10^(-11)//0.04)=sqrt(5.5xx10^(-10))=2.36xx10^(-5)` `{:(,NO_(2)^(-),+,H_(2)O,hArr,HNO_(2) ,+,OH^(-)),("Initialconc.",c,,,,,,),("After hydrolysis",c-ch,,,,ch,,ch):}` `[OH^(-)]=ch=0.04xx2.36xx10^(-5)=9.44xx10^(-7)` `pOH=-log (9.44xx10^(-7))=7-0.9750=6.03` `pH = 14 - pOH = 14 - 6.03 = 7.97`. or directly, as `NaNO_(2)` is a salt of weak acid and strong base, `pH=(1)/(2) [pK_(w)+pK_(a)+logC]` `pK_(w)=-log K_(w)=14` `pK_(a)=-logK_(a)=-log (4.5xx10^(-4))=-(0.65-4)=3.35` `log C = log 0.04 = log 4 XX 10^(-2) = - 2 + 0.6021 = - 1.3979 ~=-1.40` `:. pH = (1)/(2) [14+3.35-1.40]=7.97` |
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| 25. |
The ionization constant of nitrous acid is 4.5xx10^(-4) . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis . |
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Answer» Solution :CALCULATE pH of salt sodium nitrite solution :Nitrous Acid and sodium Nitrite combine and from sodium nitrite. This mixture is acidic buffer solution. DEGREE of hydrolysis (h) of `NaNO_2` (Sodium nitrite) : Ionic equilibrium in nitrous acid and `K_a` `HNO_(2(AQ)) hArr H_((aq))^(+) + NO_(2(aq))^(-)` `K_a=([H^+][NO_2^-])/([HNO_2])=4.5xx10^(-4)` Hydrolysis and CONSTANT of hydrolysis of `NO_2^(-)` : `{:(NO_2^(-) + H_2O hArr , HNO_2+ ,OH^-),((C-x)=C, x=C , x=C),((0.04),,):}` `K_h=([HNO_2][OH^-])/([NO_2^-])` `=((x)(x))/0.04 =x^2/0.04` ...(a) Put `(H^+)` in Numerator and Denominator in this expression , So, `K_h=([HNO_2][OH^-][H^+])/([NO_2^-][H^+])` But `[OH^-][H^+]=K_w` and `([HNO_2])/([NO_2^-][H^+])=1/K_a` `therefore K_h=K_w/K_a` `=(1.0xx10^(-14))/(4.5xx10^(-4))=2.222xx10^(-11)` Put value of `K_h` is equation (a), `2.222xx10^(-11)=x^2/0.04` `therefore x=sqrt(0.04xx2.222xx10^(-11))` `=sqrt(0.8888xx10^(-12))` `=0.9428 xx10^(-6)` `=9.428xx10^(-7)` =C `therefore h=x/C` `=(9.428xx10^(-7))/0.04` `=235.7xx10^(-7)` `=2.357xx10^(-5)` |
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| 26. |
Theionization constantofNH_(4)^(+) in water is 5.6 xx 10^(-10)at 25^(@)C. The rate constant for reaction of NH_(4)^(+) and OH^(-)to form NH_(3) and H_(2)O "at"25^(@)C "is " 3.4xx10^(10)litre"mol"^(-1) "sec^(-1)". Calculate the rate constant for proton transfer from proton transfer from water to NH_(3). |
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Answer» Solution :`NH_(4)^(+) + H_(2)O hArr NH_(4)OH + H^(+) , K_(a) = 5.6xx10^(-10)` `NH_(3) + H_(2)O UNDERSET(k_(b))overset(k_(f))hArr NH_(4)^(+) + OH^(-) , k_(b) = 3.4xx10^(10)` Ai,=m, To FIND `k_(f)` we know that for a conjugate acid-base pair `K_("acid")xxK_("base") = K_(w), i.e., K_(a)xxK_(b)=K_(w)` `:. K_(b) = (k_(w))/(K_(a))=(10^(-14))/(5.6xx10^(-10))` But `K_(b) = (k_(f))/(k_(b))` `k_(f) = K_(b) xx k_(b)=(10^(-14))/(5.6xx10^(-10))xx 3.4xx10^(10)=0.607xx10^(6)=6.07xx10^(5)` |
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| 27. |
The ionization constant of HF is 3.2xx10^(-4). Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all the species present (H_(3)O^(+), F^(-) andHF) in the solution and its pH . |
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Answer» Solution : `{:((i),HF,+,H_(2)O ,hArr,H_(3)O^(+),+,F^(-)),("Initial CONC.",0.02 M ,,,,,,),("Eqm. conc.",0.02-0.02 alpha=0.02(1-alpha),,,,0.02 alpha,,0.02 alpha):}` `K_(a)=([H_(3)O^(+)][F^(-)])/([HF]):. 3.2 xx 10^(-4)=((0.02 alpha)^(2))/(0.02(1-alpha))=(0.02 alpha^(2))/(1-alpha)~=0.02 alpha^(2)`(Neglecting `alpha` in comparison to 1) `:. alpha^(2)=(3.2xx10^(-4))/(0.02)=1.6xx10^(-2) or alpha = 0.12` Note : If `alpha` is not NEGLECTED in comparison to 1, solve as follows (solution of quadratic equation) : `0.02 alpha^(2) = 3.2xx10^(-4) - 3.2 xx 10^(-4) alpha or 2xx10^(-12) alpha^(2) + 3.2 xx 10^(-4) alpha - 3.2xx10^(-4) = 0` or `alpha^(2) + 1.6xx10^(-2) alpha -1.6xx10^(-2) = 0` `:. =+ 0.12 and -0.12`. Neglecting -ve value, `alpha = 0.12` (ii) EQUILIBRIUM concentrations : `[HF]=0.02(1-0.12)=1.76xx10^(-3)M` `[H_(3)O^(+)]=[F^(-)]=0.02xx0.12=2.4xx10^(-3)M` (iii) `pH = -log [H_(3)O^(+)]=-log(2.4xx10^(-3))=2.62` |
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| 28. |
The ionization constant of HF is 3.2xx10^(-4). Calculate the degree of dissociation of HF in its 0.02 M solution.Calculate the concentration of all species present (H_3O^(+), F^- and HF ) in the solution and its pH. |
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Answer» Solution :In aqueous solution of HF, two proton transfer reactions occurs like, (i)Ionization of HF (ii) Self ionization of water. (i)`HF_((aq))+H_2O_((ll)) hArr H_3O_((aq))^(+) + F_((aq))^(-)` (ii)`H_2O_((l)) + H_2O_((l)) hArr H_3O_((aq))^(+)+ OH_((aq))^(-)` `K_a (i) =3.2xx10^(-4)` and `K_w`(ii) = `1.0xx10^(-14)` At `K_a (i) gt gt K_w (ii)`, So, main EQUILIBRIUM in solution is ionization of HF. Suppose , dissociation amount of HF= `alpha` Initial [HF]=0.02 M `therefore` Dissociated [HF]=`(0.02alpha)M` ` therefore` At equilibrium [HF]=(Initial - Dissociated ) =(0.02 - 0.02 `alpha`) `=0.02 (1-alpha)M` `H_3O^+` and `F^-` produce equal to the amount of the HF dissociate . `therefore` At equilibrium `[H_3O^+]=[F^-] 0.02 alpha M` These are SHOWN in the following table. `{:("Equilibrium :",HF_((aq))+,H_2O_((l)) hArr, H_3O_((aq))^(+) +, F_((aq))^(-)),("Initial M:", 0.02, -,Zero,Zero),("Change:", -0.02alpha,-,+0.02alpha,+0.02 alpha),("Equili (M)", 0.02 - 0.02 alpha, , 0.02 alpha,0.02 alpha):}` Equilibrium constant for ionization of HF. `K_a=([H_3O^+][F^-])/"[HF]"` `therefore 3.2 xx10^(-4) = ((0.02 alpha)(0.02alpha))/(0.02 (1-alpha))=(0.02 alpha^2)/(1-alpha)` `therefore 3.2xx10^(-4) (1-alpha)=0.02 alpha^2` `therefore 0.02 alpha^2 + 3.2 xx10^(-4) alpha -3.2 xx10^(-4) = 0.0` `therefore alpha^2 + 0.016 alpha - 0.016 =0.0` For this quadratic equation a=1,b=0.016 and c=-0.16 `therefore alpha =(-b+sqrt(b^2-4ac))/(2a)` `=(-0.016 pm sqrt((0.016)^2 - 4(1)(-0.016)))/(2(1))` `=(-0.016 pm sqrt(0.000256+0.064))/2` `=(-0.016 pm 0.2535)/2` `=0.2375/2` `-0.2695/2` =0.1187 OR -0.1347 `approx` Dissociation degree or negative value is not possible . At equilibrium `[H_3O^+] = [F^-]=0.02alpha` `=0.02xx0.1187` = 0.002374 `=2.374xx10^(-3)` M `approx 2.4 XX 10^(-3)` M At equilibrium [HF]=`(0.02-0.02 alpha)` =0.02-0.02(0.12)=(0.02-0.0024) =0.0176 `=1.76xx10^(-2)` M pH=log `[H_3^+O]`=log `(2.4xx10^(-3))` =-(0.3802-3)=-(-2.6198) =+2.6198 `approx` 2.62 OR `K_a=(H_3O^+)^2/C_o=3.2xx10^(-4)` `[H_3O^+]=sqrt(K_aC_o)=sqrt(3.2xx10^(-4)xx0.02)` `=sqrt(6.4xx10^(-6))= 2.53xx10^(-3)` M So, `[H_3O^+]=[F^-]=2.53xx10^(-3)` M = 0.00253 M [HF]=(0.02-0.00253)=0.01747 `approx` 0.017 M pH=-log `(2.53xx10^(-3))` = 2.5969 `approx` 2.6 |
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| 29. |
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8xx10^(-4), 1.8xx10^(-4) and 4.8xx10^(-9) respectively . Calculate the ionization constants of the corresponding conjugate base. |
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Answer» SOLUTION :`F^-` is a conjugate base of HF, its ionization constant `K_b` is, `K_b=K_w/K_a=(1.0xx10^(-14))/(6.8xx10^(-4))= 1.47xx10^(-11)` `HCOO^-` is a conjugate base of HCOOH , its ionization constant `K_b`, `K_b=K_w/K_a=(1.0xx10^(-14))/(1.8xx10^(-4))=5.556xx10^(-11)` `CN^-` is a conjugate base of HCN, its ionization constant `K_b` is , `K_b=K_w/K_a=(1.0xx10^(-14))/(4.8xx10^(-9))= 2.08xx10^(-6)` |
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| 30. |
Theionization constant of HF, HCOOH and HCN at 298 K are 6.8xx10^(-4), 1.8xx10^(-4)and 4.8xx106(-9) respectively. Calculate the ionization constant of the corresponding conjugate base. |
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Answer» SOLUTION :For `F^(-) , " " K_(B) = K_(w)//K_(a) = 10^(-14)//(6.8xx10^(-4))=1.47 XX 10^(-11) ~= 1.5 xx 10^(-11)`. For `HCO O^(-) , " " K_(b) = 10^(-14)//(1.8xx10^(-4))=5.6xx10^(-11)` For `CN^(-), " " K_(b) = 10^(-14) // (4.8xx10^(-9))=2.08 xx 10^(-6)` |
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| 31. |
The ionization constant of fromic acid is 1.8xx10^(-4). Calculate the ratio of sodium formate and formic acid in a buffer of pH 4.25 . |
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Answer» `log . (["Salt"])/(["ACID"])=pH - pK_(a) = 4.25 - 3.74 = 0.51 or (["Salt"])/(["Acid"])="ANTILOG " 0.51 = 3.24`. |
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| 32. |
The ionization constant of formic acid is 1.8xx10^(-4). Around what pH will its mixture with sodium formate give buffer solution of highest capacity ? |
| Answer» Solution :Buffer solution of HIGHEST capacity is formed at which `PH = pK_(a)=- LOG (1.8xx10^(-4) ) = 3.74`. | |
| 33. |
The ionization constant of dimethylamine is 5.4xx10^(-4). Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M NaOH ? |
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Answer» Solution :`ALPHA = sqrt(K_(b)//C)=sqrt((5.4xx10^(-4))//(2xx10^(-2)))=0.164` In presence of 0.1 M NaOH, if x is the amount of dimethyl amine dissociated, `{:(,(CH_(3))_(2)NH,+H_(2)O,hArr,(CH_(3))_(2)NH^(+)OH,+,OH^(-),),("Initial conc.",0.02 M,,,,,,),("After disso.",0.02-x,,,x,,0.1+x,),(,~=0.02,,,,,~=0.1,):}` `K_(b)=(x(0.1))/(0.02) or (x)/(0.02)=(K_(b))/(0.1)=(5.4xx10^(-4))/(10^(-1))=5.4xx10^(-3)` i.e., ` alpha = 5.4 xx 10^(-3) :. % ` ionized = 0.54. |
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| 34. |
The ionization constant of chloroacetic acid is 1.35xx10^(-3). What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution ? |
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Answer» Solution :`ClCH_(2)CO OH hArr ClCH_(2)CO O^(-)+ H^(+),K_(a)=([ClCH_(2)CO O^(-)][H^(+)])/([ClCH_(2)CO OH])=([H^(+)]^(2))/(C)` `[H^(+)]=sqrt(K_(a)xxc)=sqrt(1.35xx10^(-3)xx0.1)=1.16xx10^(-2)M` `pH=-log (1.16xx10^(-2))=2-0.06=1.94` Sodium salt of chloroacetic acid is a salt of strong base and weak acid . Hence, `pH = - (1)/(2) [ log K_(W)+logK_(a)-logc]` `:. pH = -(1)/(2) [log 10^(-14)+log 1.35xx10^(-3)-log 0.1]` `=-(1)/(2) [-14+(-3+0.1303)-(-1)]=7+1.44-0.5=7.94` |
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| 35. |
The ionization constant of chloroacetic acid is 1.35xx10^(-3) . What will be pH of 0.1 M acid and its 0.1 M sodium salt solution. |
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Answer» Solution :(i)PH of ACID solution `[H^+]=sqrt(K_axxc)` `K_a=1.35xx10^(-3),c=0.1` M `[H^+]=sqrt(1.35xx10^(-3)xx0.1)` `=sqrt(1.35xx10^(-4))=1.16xx10^(-2)` `pH=-log [H^+]` `=-log (1.16xx10^(-2))` =2-0.064 =1.936 (ii)pH of 0.1 M sodium salt solution. Sodium salt of chloro acetic acid is salt of WEAK acid and strong base. HENCE, `pH=1/2[pK_w+pK_a+logc]` `K_a=1.35xx10^(-3)` `pK_a=-log K_a` `=-log (1.35xx10^(-3))` =-(0.1303-3)=2.8697 `pK_w=-log 10^(-14)` =14 c=0.1 M, log c = log (0.1)=-1 `pH=1/2[14+2.8697-1]` =7.935. |
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| 36. |
The ionization constant of benzoic acid is 6.5 xx 10^(-5) at 298 K temperature. Calculate pH of its 0.15 M solution. |
| Answer» SOLUTION :`[H^+] = 3.12xx10^(-3)` M, PH = 2.5 | |
| 37. |
The ionization constant of aniline and acetic acid and water at 25^(0)C are respectively 3.83 xx 10^(-10), 1.75 xx 10^(-5) and 1 xx 10^(-14). Calculate the percentage hydrolysis of aniline acetate in a decinormal solution |
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Answer» `54.95` `alpha = sqrt((K_(H))) = sqrt(((K_(W))/(K_(a) xx K_(b))))"......"(i)` Where `K_(q) = 10^(-14)` `K_(a) = 1.75 xx 10^(-5), K_(b) = 3.83 xx 10^(-10)` `alpha = sqrt(((10^(-14))/(1.75 xx 10^(-5) xx 3.83 xx 10^(-10)))) = 1.22` Which is not possible because, degree of of hydrolysis may be notbe greater than `1`. Thus, we have no avoid the APPROXIMATION. `(alpha)/(1-alpha) = sqrt((K_(h))) = 1.22 , alpha = (1.22)/(2.22) = 0.5495` `%` hydrolysis `= alpha ' 100` `= 0.5495 xx 100 = 54.95%` |
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| 38. |
The ionization constant of benzoic acid is 6.46xx10^(-5) and K_(sp) for silver benzoate is 2.5xx10^(-13). How many times is silver benzoate more soluble in a buffer of pH3.19 compared to its solubility in pure water ? |
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Answer» Solution :`C_(6)H_(5)CO Oag rarr C_(6)H_(5)CO O^(-) + Ag^(+)` Solubility in water. Suppose solubility in water = x mol `L^(-1)`. Then `[C_(6)H_(5)CO O^(-)]=[Ag^(+)] = x "mol" L^(-1) .:. X^(2)=K_(sp) or x = sqrt(K_(sp))=sqrt(2.5xx10^(-13))=5xx10^(-7)` mol ` L^(-1)` Solubility in buffer of pH = 3.19 `pH = 3.19 ` means ` - log (H^(+)]=3.19 or log [H^(+)]=-3.19 = bar(4).81 or[H^(+)] = 6.457 xx 10^(-4)M` `C_(6)H_(5)CO O^(-)` ions now conbine with the `H^(+)` ions to from benzoic acid but `[H^(+)]` remains almost constant because we have buffer solution . Now `C_(6)H_(5) CO OH hArr C_(6)H_(5)CO O^(-) + H^(+)` `:. K_(a)=([C_(6)H_(5)CO O^(-)][H^(+)])/([C_(6)H_(5)CO OH]) or([C_(6)H_(5)CO OH ])/([C_(6)H_(5)CO O^(-)])=([H^(+)])/(K_(a))=(6.457xx10^(-4))/(6.46xx10^(-5))=10` ...(i) Suppose solubility in the buffer solution isy mol `L^(-1)` . Then as most of the benzote ions are converted into bezoic acid molecules (which remain almost IONIZED), we have `y = [Ag^(+)] = [ C_(6)H_(5)CO O^(-)]+[C_(6)H_(5)CO O^(-)] + [ C_(6)H_(5)CO OH]=[C_(6)H_(5)CO O^(-)] + 10 [C_(6)H_(5)CO O^(- )]=11[C_(6)H_(5)CO O^(-)] " " `(using eqn. (i) ) `:.[C_(6)H_(5)CO O^(-)]=(y)/(11)` `K_(sp)=[C_(6)H_(5)CO O^(-)][Ag^(+)]` i.e., `2.5xx10^(-3)=(y)/(11) xx y or y^(2)=2.75 xx 10^(-12) or y=1.66xx10^(-6) :. (y)/(x)=(1.66xx10^(-6))/(5xx10^(-7))=3.32`. Note that in case of salts of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as follows : TAKING EXAMPLE of `C_(6)H_(5)CO O Ag`, we have `C_(6)H_(5)CO O Ag hArr C_(6)H_(5)CO O^(-) + Ag^(+)` In acidic solution, the anions (`C_(6)H_(5)CO O^(-)` in thepresent case) undergo protonation in presence of acid . Thus, `C_(6)H_(5)CO O^(-)` ions are removed. Hence, EQUILIBRIUM shifts forward producing more `Ag^(+)` ions. ALTERNATIVELY, as `C_(6)H_(5)CO O^(-)` ions are removed, `Q_(sp)` decrease. In order tomaintain solubility product equilibrium `(Q_(sp) = K_(sp)), Ag^(+)` ion concentration must increase. Hence, solubilityis more. |
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| 39. |
The ionization constant of benzoic acid is 6.46xx10^(-5) and K_(sp) for silver benzoate is 2.5xx10^(-13) . How many times is silver benzoate more soluble in a buffer of pH 3.19 comparedto its solubility in pure water ? |
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Answer» Solution :Solubility of silver benzoate in pure water (S) : `{:(C_6H_5COOAg_((s)) hArr , C_6H_5COO_((aq))^(-) + , Ag_((aq))^(+)),(S,"S M","S M"):}` `K_(sp)=[C_6H_5COO^-][Ag^+]=(S)(S)=S^2` `therefore S=SQRT(K_(sp))=sqrt(2.5xx10^(-13))=sqrt(25xx10^(-14))` `=5.0xx10^(-7)` `C_6H_5COOAg`, the solubility in pure water `[H^+]` in BUFFER (pH=3.19): pH=-log `[H^+]`=3.19 so log `[H^+]` = -3.19 `therefore [H^+]`= Antilog (-3.19)= `6.4566xx10^(-4)` M `approx 6.46xx10^(-4)` M Ionicequilibrium and CONCENTRATION in Benzoic acid : `C_6H_5COOH_((aq)) + H_2O hArr C_6H_5COO_((aq))^(-) + underset(6.46xx10^(-4))(H_3O_((aq))^(+))` `K_a=([C_6H_5COO^-][H_3O^+])/(C_6H_5COOH)` `therefore ([C_6H_5COOH])/([C_6H_5COO^-])=([H_3O^+])/K_a=(6.46xx10^(-4))/(6.46xx10^(-5))=10.0` `therefore [C_6H_5COOH]=10(C_6H_5COO^-)` Solubility of silver benzoic in Buffer (x) : Suppose solubility of silver benzoate in buffer (Benzoic acd + Silver benzoate) = x mol `L^(-1)` `(i)underset"(x)"(C_6H_5COOHAg_((s)))hArr, underset"x M"(C_6H_5COO_((aq))^(-))+underset"x M"(Ag_((aq))^(+))` (ii)`C_6H_5COOH+H_2O hArr C_6H_5COO_((aq))^(-) + H_3O_((aq))^(+)` Equilibrium (ii), produce `C_6H_5COOH` by the effect of common ion `C_6H_5COO^-` and REACTION is reverse . `therefore [Ag^+]=x=[C_6H_5COO^-]=[C_6H_5COOH]` but According to (i)`[C_6H_5COOH]=10[C_6H_5COO^-]` So, `x=[C_6H_5COO^-]+10[C_6H_5COO^-]` `=[C_6H_5COO^-](1+10)` `[Ag^+]=x=11[C_6H_5COO^-]` `therefore` in Buffer `[C_6H_5COO^-]=x/11` Calculation of x : `C_6H_5COOAg_((s)) hArr C_6H_5COO^(-) + Ag^+` here `[Ag^+] =x` and `[C_6H_5COO^-]=x/11` `therefore K_(sp)=[C_6H_5COO^-][Ag^+]` `therefore 2.5xx10^(-13) =(x/11) (x) =x^2/11` `therefore x = sqrt((2.5xx10^(-13))xx11)` `=sqrt(2.5xx1.1xx10^(-12))` `=sqrt(2.75xx10^(-12))` `=1.6583xx10^(-6)` `approx 1.66xx10^(-6)` M solubility in buffer Solubility in water =S= `5.0xx10^(-7)` = less VALUE Solubility in buffer = x= `1.66xx10^(-6)` M = more value `therefore "Solubility of silver benzoate in buffer"/"Solubility of silver benzoate in water"=(1.66xx10^(-6))/(5.0xx10^(-7))` =3.32 Thus, 3.32 times more solubility of silver benzoate in buffer . |
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| 40. |
The ionization constant of ammonium hydroxide is 1.77xx10^(-5) at 298 K. Hydrolysis constant of ammonium chloride is |
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Answer» `6.50xx10^(-12)` |
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| 41. |
The ionization constant of ammonium hydroxide is 1.77 xx 10^(-5) at 298 K. Hydrolysis constant of ammonium chloride is : |
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Answer» `6.5xx10^(-12)` `K_h=K_w/K_b=(1.0xx10^(-14))/(1.77xx10^(-5))` `=5.65xx10^(-10)` |
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| 42. |
The ionization constant of acetic acid is 1.74xx10^(-5). Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH . |
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Answer» Solution :`{:(,CH_(3) CO OH,hArr,CH_(3)CO O^(-),+,H^(+),,),("INITIAL",C,,0,,0,,),("At eqm.",C-C alpha,,C alpha,,C alpha ,,):}` `K_(a)=(C alpha. C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha) ~= C alpha^(2)` `alpha = sqrt((K_(a))/(C))` (or directly by Ostwald DILUTION law) `=sqrt((1.74xx10^(-5))/(0.05))=sqrt(3.48xx10^(-4))=1.86xx10^(-2)` `[CH_(3)CO O^(-)] = C alpha = 0.05 xx (1.86xx10^(-2))=9.3xx10^(-4)M` `[H^(+)]=C alpha = 9.3 xx 10^(-4)` `pH = - log [H^(+)]=- log (9.3xx10^(-4))=3.03` |
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| 43. |
The ionization constant of acetic acid is 1.74xx10^(-5) . Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentrationof acetate ion in the solution and its pH. |
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Answer» Solution :This is WEAK acid , So, following equilibrium established in `CH_3COOH` solution. Dissociation degree = `ALPHA` So, `[H^+] = [CH_3COO^-] = 0.05 alpha` `{:(,CH_3COOH_((aq)) hArr, H_((aq))^(+) + , CH_3COO_((aq))^(-)),("MOLARITY in initial", 0.05,0,0),("Change in equili.",-0.05 alpha,+0.05 alpha,+0.05alpha),("M at equilibrium",(0.05-0.05alpha),0.05alpha,0.05alpha),(,=0.05(1-alpha) approx 0.05M,,):}` `K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.74xx10^(-5)` `therefore ((0.05alpha)(0.05alpha))/(0.05) =1.74xx10^(-5)` `therefore 0.05 alpha^2 = 1.74xx10^(-5)` `therefore alpha=sqrt((1.74xx10^(-5))/0.05)=sqrt(3.48xx10^(-4))` `therefore alpha`= Degree of dissociation `=1.865xx10^(-2)` =0.01865 `[H^+]=[CH_3COO^-]` `=0.05 XX alpha= 0.05 xx 0.01865 = 9.327xx10^(-4)` M pH=-log `[H^-]` =-log `(9.327xx10^(-4))` =-(0.9697-4)=-(-3.03)=3.03 |
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| 44. |
The ionization constant for acetic acid is |
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Answer» Three TIMES LARGER than deuteric acid |
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| 45. |
The ionisation potential of Hydrogen is 2.17 xx 10^(-11) erg/atom. The energy of the electron in the second orbit of the hydrogen atom is |
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Answer» `-(2.17 xx 10^(-11))/(2) ` erg/atom |
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| 46. |
The ionisation of hydrochloric in water is given below : HCl (aq) + H_(2)O (l) hArr H_(3)O^(+) (aq) + Cl^(-)(aq) Label two conjugate acid-base pairs in this ionisation. |
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Answer» SOLUTION :(i) HCl (ACID), `Cl^(-)` (conjugate BASE) (II) `H_(2)O` (base), `H_(3)O^(+)` (conjugate acid). |
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| 47. |
The ionisation of hydrochloric in water is given below: HCl_((aq))+H_2O_((l)) hArr H_3O_((aq))^(+) +Cl_((aq))^(-)Label two conjugate acid-base pairs in this ionisation. |
Answer» Solution : NOTE: If Bronsted acid is a strong acid then its CONJUGATE base is a weak base and vice-versa. Generally, the conjugate acid has one EXTRA PROTON and conjugate base has one LESS proton |
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| 48. |
The ionisation enthalpy of sulphur is "1014kJmol"^(-1). If its electronegativity is 2.4, what is its electron gain enthalpy? |
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Answer» SOLUTION :In the SCALE, electronegativity (E.N) is given in terms of ionisation, enthalpy `(I_1)` and ELECTRON GAIN enthalpy `(E_1)` as `E.N. = (I_1 + E_1)/(544)` SUBSTITUTING the values, `2.4 - (1014 + E_1)/(544)` Electron affinity `= E_1 = (544 xx 2.4) - 1014 = 290` Electron gain enthalpy of sulphur = `- 290 kj mol^(-1)` |
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| 49. |
The ionisation enthalpy of sodium is 5.14 eV. How many k cal of energy is required to ionise all atoms present in one gram of gaseous Na atoms? |
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Answer» Solution :1e V ATOM`""^(-1)` = 23 k CAL `mol."^(-1)` Energy required to ionise all atoms of 23 GRAMS (one mole ) of gaseous Na atoms `= 23 XX 5.14 k cal ` Energy required for ionisation of all atoms present in one gram of gaseous Na atom `= (23 xx 5.14)/(23)` `=5.14` k cal . |
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