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. |
If r_(Na^+) and r_(Cl^-) represent the radii of Na^+ and Cl^- ions respectively and n is the number of NaCl units per unit cell, derive an expression for molar volume of the crystal in terms of r_(Na^+), r_(Cl^-) and n. |
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Answer» Solution :Edge of the unit cell (a)=`2(r_(Na^+)+r_(Cl^-))` VOLUME of the unit cell= `a^3=8(r_(Na^+)+ r_(Cl^-))^3` Volume occupied by 1 mole of NaCl (MOLAR volume ) =`(8(r_(Na^+)+r_(Cl^-))^3)/nxxN_0` where `N_0` = Avogadro's number =`6.023xx10^23` |
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| 2. |
Ifr_(Na^(+)) and r_(Cl^(-))represent the radii ofNa^(+) and Cl^(-) ions respectively and n is the number of NaCl units per unit cell, derive an expression for molar volume of the crystal in terms ofr_(Na^(+)) , t_(Cl^(-)) and n. |
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Answer» Solution :Edge of the unit cell (a) = ` 2 (r_(Na^(+)) + r_(Cl^(-))` Volume of the unit cell = ` a^(3) = 8 (r_(Na^(+)) + r_(Cl^(-)))^(3)` Volume occupied by 1 mole of NACL (molar volume) = `( 8 (r_(Na^(+)) + r_(Cl^(-))^(3)))/N XX N_(0)` where ` N_(0)` = Avogadro's number = ` 6.023 xx 10^(23)` |
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| 3. |
If RMS velocity of CH_4 at 27^@C is 0.2ms^(-1) the RMS velocity at 927^@C |
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Answer» `4MS^(-1)` |
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| 4. |
If RMS velocity of carbon dioxide is 4.4 xx 10^(4)cm s^(-1) at given temperature, find the RMS velocity of ethane at the same temperature |
| Answer» SOLUTION :`5.33 XX 10^4cms^(-1)` | |
| 5. |
If r.m.s. speed of gaseous molecules is x cm/sec at a pressure of p atm r.m.s. at a pressure of 2p atm and constant temperature will be |
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Answer» x As TEMPERATURE is constant, Pv is constant. Hence, r.m.s. is constant even if pressure is doulbed. Alternatively, `u=sqrt((3P)/(d))`. At constant TEMP. , if P is doulbed, d is doubled so that u remians same. |
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| 6. |
If r_(max) represents the longest maximum probability region and ltr_(max)gt represents avergae r_(max), then which of the following holds true for multi-electron species? |
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Answer» `r_(max) "order":3sgt3pgt3d` |
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| 7. |
If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane. |
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Answer» Solution :According to the reaction : `underset("n-Butane")(CH_(3)CH_(2)CH_(2)CH_(3))underset("light")overset(Cl_(2), 298K)to underset("2-Chlorobutane")(CH_(3)-underset(Cl)underset(|)CH-CH_(2)CH_(3)) + underset("1-Chlorobutane")(CH_(3)CH_(2)CH_(2)CH_(2)Cl)` The RELATIVE rates of two isomeric chlorobutanes will be equal to their number of types of H.s (`1^(@), 2^(@)` or `3^(@)`) and their relative rates of substitution. `("1-chlorobutane")/("2-chlorobutane")= ("No. of" 1^(@)H)/("No. of" 2^(@)H) xx ("Reactivity of" 1^(@) H)/("Reactivity of" 2^(@)H)` `=(6)/(4) xx (1)/(3.8)=(6)/(15.2)` Now, if x is the percentage of 1-chlorobutane, then Percentage of 2-chlorobutane = 100 - X `:. (x)/(100-x)=(6)/(15.2)` 15.2 x = 600 - 6X 21.2 x = 600 x = 28 % `:.` 1-Chlorobutanee = 28% and 2-chlorobutane = 72% |
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| 8. |
If R_(H) represents Rydberg constant, then the energy of the electron in the ground state of hydrogen atom is |
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Answer» `- (hc)/(R_(H))` `bar(v) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`....(i) According to Bohr model, for an electronic TRANSITION, `Delta E = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` But `Delta E = hv = (hc)/(LAMDA) = hc bar(v)` `:. hc bar(v) = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` or `bar(v) = - (2pi^(2) me^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`...(ii) COMPARING eqns. (i) and (ii) `R_(H) = - (2pi^(2) me^(4))/(ch^(3)) :. E_(1) = - (2pi^(2) me^(4))/(h^(2)) = - R_(H) ch` |
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| 9. |
If .R_(H). is the Rydberg constant, then the energy of an electron in the ground state of hydrogen atom is |
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Answer» `(R_(H)C)/(h)` |
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| 10. |
If relative masses of He is taken as one unit, what is that of magnesium? |
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Answer» Solution :RELATIVE MASSES of He and Mg are RESPECTIVELY 4 and 24. Compared to He as 1, that of Mg is `(24)/(4)=6` When we use atomic masses of elements in calculations, we use average atomic masses. |
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| 11. |
if relative mass of He is taken as one unit, What is that of calcium? |
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Answer» SOLUTION :Relative MASSES of He and Ca are respectively 4 and 40. COMPARED to He as 1 that of Ca is `(40)/(4) = 10` |
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| 12. |
If reaction A+BhArrC+D, taken place in 5 liter close vessel, the rate constant of forward reaction is nine times of rate of backward reaction. If initially one mole of each reactantpresent in the container, then find the correct option//is. |
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Answer» `([C])/([B])=(3)/(1)` `K_(F)=9K_(b) K_(C)=(K_(F))/(K_(b))=(9)/(1)` `K_(C)=(5//5)^(2)/(((1-x)/(5))^(2))=(9)/(1) (x)/(1-x)=3` `x=3-3x` `4x=3` `x=(3)/(4)=0.75` `[A]=(0.25)/(5xx100)=5xx10^(-2) "mole" L^(-1) [B]=[A] C=(0.75)/(500)=15xx10-2 "mol" L^(-1)` |
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| 13. |
The radius or second stationary orbit in Bohr's atom is R. The radius of the third orbit in the same atom will be |
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Answer» R/3 |
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| 14. |
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ? |
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Answer» 2.55 eV `=(13.6)xx(3)^2 [1/(2^2) - 1/(4^2)] = 13.6 xx (27)/(16) = 22.95 eV` Energy needed to REJECT electron from N = 2 level in H - atom`=13.6 xx 1^2 [1/(2^2) - 1/(oo^2)]` `=(13.6)/(4) eV = 3.4 eV` `Delta E = 22.95 - 34 = 19 .55 eV` |
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| 15. |
If radiation correcsponding to second line of “Balmer series” of Li^(2+) ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be: |
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Answer» 2.55 eV |
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| 16. |
If pressure is P, temperature is T and gas constant is R. For an ideal gas, then the moles per litre of gas will be |
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Answer» <P>`(RT)/(P)` |
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| 17. |
If pressure is applied on the "ice"hArr "water" equilibrium , more of …….. will be formed. |
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Answer» |
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| 18. |
If pK_(b) for fluoride ion at 25^(@)C is 10.83, the ionisation constant of hydrofluoric acid at this temperature is |
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Answer» `1.74xx10^(-5)` ` pK_(a)+pK_(b)=pK_(w)` `:. pK_(a)+10.83=14 "or " pK_(a)=3.17` `- log K_(a)=3.17` or log `K_(a)=-3.17=bar(4).83 "or" K_(a)=6.76xx10^(-4)`. |
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| 19. |
If pK_(a) of acetic acid and pK_(b) of ammonium hydroxide are 4.76 each, the pH of ammonium acetate is |
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Answer» 7 |
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| 20. |
If pK_(a) of a weak acid is 5, then pK_(b) of the conjugate base will be |
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Answer» Hence `pK_(b)=14-pK_(a) = 14 - 5 = 9`. |
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| 21. |
If P^(ka) of acetic acid and P^(kb)of ammonium hydroxide are 4.76 each. Find the pH of ammonium acetate. |
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Answer» ` pH=(1)/(2)[PK_w+PK_a-PK_b]` ` = (1)/(2)[14 +4.76 - 4.76 ]=7` |
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| 22. |
If photon of the wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of 1.5 xx 10^(7) ms^(-1), calculate the energy with which it is bound to the nucleus. |
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Answer» Solution :Energy of the INCIDENT photon `= (hc)/(lamda) = ((6.626 xx 10^(-34) JS) (3.0 xx 10^(8) MS^(-1)))/((150 xx 10^(-12) m)) = 13.25 xx 10^(-16) J` Energy of the electron ejected `= (1)/(2) mv^(2) = (1)/(2) (9.11 xx 10^(-31) kg) (1.5 xx 10^(7) ms^(-1))^(2) = 1.025 xx 10^(-16)J` Energy with which the electron was bound to the nucleus `= 13.25 xx 10^(-16)J - 1.025 xx 10^(-16)J` `= 12.225 xx 10^(-16)J = (12.225 xx 10^(-16))/(1.602 xx 10^(-19)) eV = 7.63 xx 10^(3) eV` |
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| 23. |
If pH values of pure water at 25^@ C and 35^@ C are respectively 7 and 6, calculate Delta H for the formation of water from H^+ and OH^-. |
| Answer» SOLUTION :`-354kj MOL^(-1)` | |
| 24. |
If phenyl magnesium bromide and acetaldehyde are the reactants, the product formed after hydrolysis would be |
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Answer» BENZYL alcohol |
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| 25. |
If pH of a saturated solution of Ba(OH)_(2) is 12, the value of its K_((sp)) is |
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Answer» `5.00 xx 10^(-7) M^(3)` Hence, `[OH^(-)]=10^(-2)M` `Ba(OH)_(2) hArr Ba^(2+)+2OH^(-)` `[OH^(-)]=10^(-2)M` `:. Ba^(2+)=(10^(-2))/(2) M=5xx10^(-3)M` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` `=(5xx10^(-3))(10^(-2))^(2)=5xx10^(-7) M^(3)` |
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| 26. |
If pH of a saturated solution of Ba(OH)_2 is 12, the value of its K_((Sp)) is : |
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Answer» `4.0xx10^(-6) M^3` pH=12 or pOH=2 `[OH^-]=10^(-2)` M `{:(Ba(OH)_2 to , Ba^(+2)+, 2OH^-),(,0.5xx10^(-2), 10^(-2)):}` [`because` CONCENTRATION of `Ba^(2+)` is half of `OH^-` ] `K_(sp)= [Ba^(+2)][OH^-]^2` `=[0.5xx10^(-2)][1.0xx10^(-2)]^2` `=0.5xx10^(-6)=5.0xx10^(-7) M^3` |
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| 27. |
If penta -2,4 - dione is treated with DCl//D2O,isotopic exchange occurs via ketoenol tautomerism. By how many grams , will the molar mass increase from the increase from the strarting compound. |
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Answer» |
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| 28. |
If parent element of Na^(+)(X) and parent element of Mg^(+)(Y), then what is true for X and Y ? |
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Answer» X is positive ELEMENT, Y is negative element. |
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| 29. |
If P is the pressure and d is the density of gas, then P and d are related as : |
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Answer» <P>`P ALPHA 1"/"d` |
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| 30. |
If oxygen diffuses at a rate of 6cm^(3),"sec"^(-1) through a fine hole, find the rate of diffusion of carbon dioxide gas under the same conditions. |
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Answer» |
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| 31. |
If other factors being same, the ionisation energy are in the order of |
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Answer» `s lt p lt d lt F` |
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| 32. |
if only oneelectronof anatomget exitedstillmorelineobservein spectrumWhy ? |
| Answer» SOLUTION :In anatomhigherenergylevelsare moresoevenone ELECTRONIS thereit cantransitsfrommorethanone energylevel. Hencelinesareobserved. | |
| 33. |
If one organic compound has C & H respectively 92.3% and 7.7% then its empirical formula is..... |
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Answer» Solution :`C= 92.3% ,H= 7.7%` `C:H= (92.3)/(12)= (7.7)/(1) = 7.7:7.7=1:1` Empirical formula `=CH` |
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| 34. |
If one of the atoms in I are completely converted to get compound in II |
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Answer» |
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| 35. |
If one mole of H_(3)PO_(x) is completely neutralized by 40 g of NaOH, select the correct statements (s): |
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Answer» `x=2` and ACID is MONOBASIC |
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| 36. |
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then |
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Answer» `DeltaHgtDeltaU` |
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| 37. |
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then ____ |
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Answer» `DELTAH GT DELTAU` |
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| 38. |
If one mole of a gas A (mol.wt-40) occupies a volume of 201itres, under the same conditions of temperature and pressure the volume occupied by 2 moles of gas B (mol.wt=80) is |
| Answer» Answer :D | |
| 39. |
If one atom of hydrogen weighs 1.66xx10^-24 g then mass of one atom of nitrogen is |
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Answer» `1.162xx10^(-23)g` `=14xx1.66xx10^(-24) g= 2.324 xx10^(-23) g` |
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| 40. |
If on electrophilic substitution reaction of C_(6)H_(5)Y gives m-NO_(2)C_(6)H_(4)Y then what is not y form the following ? |
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Answer» `-NH_(2)` |
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| 41. |
If on adding FeCl_(3) solution to acidified Lassaigne solution, a blood red colouration is produced, it indicates the presence of |
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Answer» S |
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| 42. |
If -O-H bond is weadker than -O-D bond, |
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Answer» The BIOLOGICAL growth of a PLANT which is fed by `D_(2)O` is flast |
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| 43. |
If NH_4Cl+NH_4OH+(NH_4)_2CO_3 Successively added than the solution is acidic or basic ? |
| Answer» Solution :The solution is BASIC, as DUE to the commonion EFFECT of `NH_4^+`the CONCENTRATION of `CO_3^(2-)`is not increase more. | |
| 44. |
If NaClis doped with 10^(-4) mol% "of" SrCl_(2)the concentration of cation vacancies will be (N_(A)=6.02times10^(23)mol^(-1)) |
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Answer» `6.02times10^(23) mol^(-1)` `therefore SrCl_(2)" droped per mole of " NaCl = 10^(-4)//100` =`10^(-6) "mole" = 10^(-6)times(6.02times10^(23))Sr^(2+) Sr^(2+)ions` `therefore"concentration of cation VACANCIES"=6.02times10^(17) mol^(-1)` |
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| 45. |
If NaCl is doped with 10^(-4) mol % of SrCl_2, the concentration of cation vacancies will be (N_A=6.02xx10^23 mol^(-1)) |
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Answer» `6.02xx10^14 "mol"^(-1)` `therefore SrCl_2` doped per mole of NaCl=`10^(-4)//100` =`10^(-6)` mole =`10^(-6)xx (6.02xx10^23) Sr^(2+)` ions `=6.02xx10^17 Sr^(2+)` ions Hence, concentration of cation vacancies =`6.02xx10^17 mol^(-1)` |
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| 46. |
If NaCl is doped with 10^(-3) mol % SrCl_2 , what is the concentration of cation vacancies ? |
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Answer» SOLUTION :Doping of NaCl with `10^(-3)` mol % `SrCl_2` means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_2` `therefore` 1 MOLE of NaCl is doped with `SrCl_2 =10^(-3)/100` mole =`10^(-5)` mole As each `Sr^(2+)` ion introduces one CATION vacancy, therefore , concentration of cation vacancies =`10^(-5)` mol/mol of NaCl =`10^(-5)xx6.02xx10^23 "mol"^(-1) =6.02xx10^18 "mol"^(-1)` |
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| 47. |
If NaCl is doped with10^(-3)mol %SrCl_(2) , what is the concentration of cation , vacancies ? |
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Answer» Solution :Doping of NACL with` 10^(-3)`mol % ` SrCl_(2)`means that 100 moles of NaCl are doped with`10^(-3) " mol of " SrCl_(2)` 1 mole of NaCl is doped with `SrCl_(2) = ( 10^(-3))/100 " mole"= 10^(-5)`mole As each `SR^(2+)`ion introduces one CATION vancancy, therefore, concentration , of cation vanancies. ` = 10^(-5) `mol/mol of NaCl ` = 10^(-5)xx 6.02 xx 10^(23)" mol"^(-1) = 6.02 xx 10^(18)"mol"^(-1)` |
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| 48. |
If Na^(+) ion is larger than Mg^(2+) ion and S^(2-) ion is larger than Cl^(-)ion, which of the following will be least soluble in water? |
| Answer» Answer :D | |
| 49. |
If n=6,the correct sequence for filling of electrons will be,………… |
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Answer» `nsrarr(n-2)frarr(n-1)drarrnp` Accoding AUFBAU PRINCIPLE, `6srarr4frarr5drarr6p` `nsrarr(n-1)frarr(n-2)drarrnp` |
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