Saved Bookmarks
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 pH of 0.005 M H_(2)SO_(4). |
|
Answer» SOLUTION :`pH=-log_(10)[H^(+)]-log_(10)[0.005xx2]=-log_(10)[0.01]=-log_(10)[1XX10^(-2)]` pH = `2-log1=2-0=2`. |
|
| 2. |
Calculate pH and pOH of 0.03 M NaOH solution. |
| Answer» SOLUTION :12.48 and 1.52 | |
| 3. |
Calculate pH and degree of hydrolysis of 0.01 MCH_3COONa. [K_h=5.6xx10^(-10)] |
| Answer» SOLUTION :pH=8.88 , `h=7.48xx10^(-5)` | |
| 4. |
Calculate percentage of each elements of Nitroproane. |
|
Answer» |
|
| 5. |
Calculate oxidation number of oxygen in H_(2)O_(2) (ii) Write the de-Broglle equation . |
|
Answer» Solution :`H_(2)O_(2):` `H_(2)ulO_(2)` `2(+1) + 2x =0` `2+2x =0` `2x =-2` x=1 Thereforeoxidation number of `O` in `H_(2)O_(2)` is -1 (ii) 1. Louis de Broglie extended the concept of dual nature oflight to all forms of matter. To QUANTIFY this relation, he derived an equation for the wavelength of a matter wave. 2. He combined the following two equations of energy of which ONE represents wave character (hu) and the other represents the particle nature `(mc^2)` . Planck.s quantum hypothesis:. E = hv .............. (1) Einsteins MASS-energy relationship `E=mc^2` ......... (2) From (1) and (2) `hv=mc^2` `hc/lambda = mc^2` `thereforelambda = h/mc ....... (3)` The equation (3) represents the wavelength of photons whose momentum is given by me. (Photons have zero rest mass). For a particle of matter with mass m and moving with a velocity v, the equation (3) can be written as `lambda = h/mV` ......... (4) |
|
| 6. |
Calculate oxidation number of oxygen in potassium ozonide. |
|
Answer» Solution :`KO_(3)rArr1(K)+3(O)=0` `therefore1(+1)+3(O)=0` `thereforeO=-1/3` |
|
| 7. |
Calculate oxidation number of oxygen in H_(2)O_(2) . |
|
Answer» Solution :(i)hydrogen peroxide `(H_(2)O_(2))` `2(+1) + 2x = 0 , rArr 2x = - 2x - 2 , rArr x = - 1 ` (ii)de-Broglie combined the following two equations of energyof which one reqresents wave character (Hu) and the other REPRESENTS the particle NATURE `(MC^(2))` |
|
| 8. |
Calculate oxidation number of nitrogen and chlorine in NOClO_(4). |
Answer» SOLUTION :
|
|
| 9. |
Calculate oxidation number of Fe in Fe_((0.94))O. |
|
Answer» SOLUTION :`Fe_((0.94))O:0.94(FE)+1(O)=0` `therefore0.94(Fe)+1(-2)=0` `therefore0.94(Fe)-2=0` `thereforeFe=200/94` |
|
| 10. |
Calculate oxidation number of .C. in glucose. |
|
Answer» SOLUTION :Glucose `rArrC_(6)H_(12)O_(6)` `therefore6(C)+12(H)+6(O)=0` `therefore6(C)+12(1)+6(-2)=0` `thereforeC=0` |
|
| 11. |
Calculate OH^(-) ion concentration in 0.08M solution of it. |
|
Answer» Solution :For a weak BASE, `[OH^(-)]=sqrt(K_(B)xxC)=sqrt(1.75xx10^(-5)xx0.08)` `=1.183xx10^(-3)"mol "dm^(-3)` |
|
| 12. |
Calculate [OH^(-)] if pOH = 8.3. |
|
Answer» SOLUTION :`pOH=-log_(10)[OH^(-)],8.3=-log_(10)[OH^(-)]` TAKING antilog on both the sides `[OH^(-)]` = anti LOG(-8.3), `[OH^(-1)]` = antilog(-9-8.3+9) = antilog`(0.7)xx10^(-9)` = `5.012xx10^(-9)mol//dm^(3)`. |
|
| 13. |
Calculate [OH^(-)] if pH = 5.284. |
|
Answer» Solution :`pH=-log_(10)[H^(+)],[H^(+)]="ANTI"LOG[-5.284]="anti"log[6-5.284-6]` `[H^(+)]="anti"log(0.716-6)="anti"log[0.716]xx10^(-6)=5.2xx10^(-6)mol//dm^(3)` `[H^(+)][OH^(-)]=10^(-14),SO[OH^(-)]=10^(-14)/([H^(+)])=10^(-14)/(5.2xx10^(-6))=0.192xx10^(-8)mol//dm^(3)` |
|
| 14. |
Calculate [OH^-] and pH of 0.001 M = [H^+] containing solution. |
| Answer» SOLUTION :`[OH^-]=1XX10^(-11)` , pH=3.0 | |
| 15. |
Calculate number of total atoms and molecules of 4 Lit SO_(2) gas at 350 K temperature and 10^(3) pa pressure. [R=8.3144 J K^(-1)mol^(-1)]. |
|
Answer» No. of molecules `= 8.275x10^(20)`, No. of ATOMS `= 2.483xx10^(21)` |
|
| 16. |
Calculate number of sigma (sigma) and pi (pi) bonds in the above structures (i-iv). |
|
Answer» Solution :`SIGMA` bonds : 33,`pi` bonds : 2 `sigma` bonds : 17,`pi` bonds : 4 `sigma` bonds : 23,`pi` bond : 1 `sigma` bonds : 41,`pi` bond : 1 |
|
| 17. |
Calculate moles of 0.224 L at STP of ideal gas ? Calculate moles of Ideal H_(2) gas. (Molar volume = 22.4 L acceptable). |
| Answer» SOLUTION :0.01 MOLE. | |
| 18. |
Calculate molecular mass of glucose (C_(6)H_(12)O_(6)) molecule. |
|
Answer» `=6(12.11u) + 12(1.008u) + 6(16.00 U)` `=(72.066u) + (12.096u) + (96.00u)` `=180.162u` |
|
| 19. |
Calculate molecular mass of glucose (C_6H_12O_6) molecule. |
| Answer» Solution :Molecular mass of GLUCOSE `(C_6H_12O_6)=6(12.011u)+12(1.008u)+6(16.00u)=(72.066u)+(12.096u)+(96.00u)=180.162u` | |
| 20. |
Calculate molarity of 28% w/w H_(2)SO_(4) solution. Density of H_(2)SO_(4) solution is 1.202 g mL^(-1). |
|
Answer» |
|
| 21. |
Calculate molality of 1 litre solution of 93% H_(2) SO_(4) by volume. The density of solution is 1.84 g mL^(-1). |
|
Answer» Solution :Given `H_(2) SO_(4)` is 93% by volume `:.` WEIGHT of `H_(2) SO_(4) = 93 g` Volume of solution `= 100 ML` Weight of solution `= 100 xx 1.84 = 184 g` Weight of water `= 184 - 93 = 91 g` Molality `(m) = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (93 xx 1000)/(98 xx 91) = 10.42` |
|
| 22. |
Calculate mass percent of different elements present in Fe_(2)O_(3) ( Foric Oxide) |
|
Answer» |
|
| 23. |
What is the mass of photonhavingwavelength3.6 A (h= 6.6 xx 10^(34)kgm^(2) s^(-1) ,c=3 xx 10^(8) ms^(-1)) |
|
Answer» `6.135 xx 10^(29) kg` `V= c= 3XX 10^(8) MS^(-1) ` Velocityof photon ACCORDINGTO de - Broglie`lambda = (h )/(m V)` `m = (h)/( lambda v) = (6.626 xx 10^(34))/( 3.6 xx 10^(10)xx 3 xx 10^(8))` `=6.135 xx 10^(29) kg` |
|
| 24. |
Calculate magnitude of change in Gibbs free energy (inkJ) of a reaction occurruong at 500 K: 2O_(3)(g)hArr3O_(2) When a sample of ozonide oxygen (having average molecular mass of the mixture =(128)/(3)is taken at a pressure of 3 bar. Round off your answer to next highest integer. [Given DeltaG_(f)^(@) "of "O_(3)(g)=145 kJ/"mole" and500R ln 2=2880.8J//mol.] |
|
Answer» |
|
| 26. |
Calculate K_h and pH of 0.1 M NH_4Cl solution .[K_w=1xx10^(-14) , K_(NH_4OH)=1.75xx10^(-5)] |
| Answer» SOLUTION :`K_h=5.7xx10^(-10)` , PH = 5.12 | |
| 27. |
What is the density of carbondioxide at STP? |
|
Answer» Solution :At STP , the mass of 22.4 lit of carbondioxide is 44g `"Densiy"=("mass")/("volume")` DENSITY of `CO_(2)` at STP `=(44)/(22.4)` Density of `CO_(2)` at S.T.P.= 1.96`gL^(-1)` |
|
| 28. |
Calculate (i) the volume of one molecule of water. (ii) the radius of a water molecule assuming the molecule to be spherical. (Given that the density of water is 1 g/cm^3. |
| Answer» SOLUTION :VOLUME `=2.991 xx 10^(-23) mL`, radius `=1.926 xx 10^(-8)` CM | |
| 29. |
Calculate (i) First excitation energy of the electron in the hydrogen atom. (ii) Ionization energy of the hydrogen atom |
|
Answer» Solution :The energy of the ELECTRON in the nth shell of HYDROGEN atom is given by `E_(n) = - (2pi^(2) me^(4))/(n^(2) H^(2)) = - (1.312 xx 10^(6))/(n^(2)) J mol^(-1)` (i) First excitation energy is the amount of energy required to excite the electron from n = 1 (ground state) to n = 2 (first excited state) `Delta E = E_(2) - E_(1) = - (1.313 xx 10^(6))/(2^(2)) - (-(1.312 xx 10^(6))/(1^(2))) = - 3.28 xx 10^(5) + 13.12 xx 10^(5) J mol^(-1)` `= +9.84 xx 10^(5) J mol^(-1)` (ii) IONIZATION energy is the amount of energy required to REMOVE the electron from `n = 1 " to " n = oo`, i.e., `Delta E = E_(oo) - E_(1) (-1.312 xx 10^(6)) = +1.312 xx 10^(6) J mol^(-1)` |
|
| 30. |
Calculate : (i) 8.56xx10^(6) and 10.64xx10^(5) addition (ii) 3.33xx10^(-3) and 5.80xx10^(-4) substraction |
|
Answer» `=9.624xx10^(6)` (II) `3.33xx10^(-3) - 0.58xx10^(-3)=(3.33-0.58)xx10^(3)` `=2.75xx10^(-3)` |
|
| 31. |
Calculate heat of solution of NaCl from the following data Hydration energy of Na^(+) = -389kJ mol^(-1) Hydration energy of Cl^(-) = -382kJ mol^(-1) Lattice energy of NaCl = +776 kJ mol^(-) |
|
Answer» =-389 -382 =-771 kj/mol =[-771-(-776)]=5kj/mole |
|
| 32. |
Calculate heat capacity of a diatomic ideal gas (Molar mass=11 gm//mol) if it is subJected to a process such that pressure exerted is directly proportional to cube of the volume.[R=2cal//mol K] |
|
Answer» `5cal//mol K` |
|
| 33. |
Calculate [H^+]of 0.2 M HCN in 1 M KCN solution. [K_(HCN)=4xx10^(-10)] |
| Answer» SOLUTION :`8XX10^(-11)` | |
| 35. |
Calculate [H^(+)] if pOH = 9.23. |
|
Answer» Solution :`pH+pOH=14""pH=14-pOH=14-9.23=4.77` `pH=-log_(10)[H^(+)]=-4.77=LOG[H^(+)]` Taking ANTILOG on both the sides `[H^(+)]` = ANTI log (-4.77) = anti log[+5 - 4.77 - 5] = anti log[0.23]`xx10^(-5)` = `1.698xx10^(-5)mol//dm^(3)`. |
|
| 36. |
Calculate [H^(+) ]in a 0.1 M solution of dichloroacetic acid (K_a =5 xx 10 ^(-2))that also contains 0.1 M sodium dichloroacetae. Neglect hydrolysis of sodium salt. If [H^(+) ]= yxx 10 ^(-2)then what is the value of y ? |
|
Answer» ` pH =2 -log 5 rArr [H^(+)] =5 xx 10 ^(-2) ` |
|
| 37. |
Calculate gas constant (R) value in litre. Bar. /K/ mole. |
|
Answer» Solution :For EQUATION `R=(PV)/(nT)` substitute n=1 mole, `T=273.15 K, P "bar" v=22.7 L` `:. R=((1"bar")(22*7L))/((1"mol")(273*15K))` |
|
| 38. |
Calculate frequency of yellow radiation having wavelength 5800 overset@A |
| Answer» SOLUTION :CALCULATION of the FREQUENCY `v=c/lambda=(3xx10^8m^(-1))/(5800xx10^(-10M))=5.172xx10^14 s^(-1)` | |
| 39. |
Calculatefrequencyand wavenumberof 3.6Awavelengthof photon . |
| Answer» SOLUTION :`v=8.333 xx 10^(16) s^(1)VEC( v) =2.778 xx 10 ^(9) m^(-1)` | |
| 40. |
Calculate formal charge on each O-atom ofO_(3) molecule. |
Answer» Solution : Lewis structure of ` O_(3)` is : The ATOMS have been numbered as 1, 2 and 3. Formal charge on end O-atom numbered `1 = 6 - 4 - (1)/(2) (4) = 0 ` Formal charge on end O-atom numbered 2 = ` 6 - 2 - (1)/(2) (6) = + 1` Formal charge on end O-atom numbered ` 3 = 6 - 6 - (1)/(2) (2) = - 1` Hence, we represent ` O_(3)`along with formal charges as 
|
|
| 41. |
Calculate formal charge of the central atom in ozone molecule. |
|
Answer» Solution :CENTRAL O ATOM in Lewis structure has ONE IONE pair and three bond pairs. Formal charge `=6-(2)-1//2(6)=+1` |
|
| 42. |
Calculate equivalent weight of KMnO_(4) in Acidic, Basic and Neutral medium. |
|
Answer» SOLUTION :Acidic medium : `MnO_(4)^(-)+5e^(-)toMn^(+2)" "thereforeM/5` Basic medium : `MnO_(4)^(-)+3E^(-)toMnO_(2)" "thereforeM/3` NEUTRAL medium : `MnO_(4)^(-)+E^(-)toMnO_(4)^(-2)" "thereforeM/1` |
|
| 43. |
Calculate equivalent conductivity of 1 MH_(2)SO^(4) whose conductivity is 26 xx 10^(-2) "ohm"^(-1) cm^(-1) |
|
Answer» 260 Concentration `= 1 MH_(2)SO_(4)=98 g//L` Eq. wt. of `H_(2)SO_(4)=49` `therefore` No. of gram equivalents per litre `=(98)/(49)=2` Equivalent conductance `=( K xx 1000)/(C)` `=(26xx10^(-2) xx 1000)/(2)` `=130 Omega^(-1) cm^(2) "mol"^(-1)` |
|
| 44. |
Calculate equilibrium constant (in multiples of 10^(-80)) when sodium reduces Aluminium oxide to aluminium at 298 K. Delta G^(@)_(f) of Na_(2)O_(3(s)) at 298 K = -377 KJ "mole"^(-1) and Delta G^(@)_(f) of Al_(2)O_(3) at 298K = -1582 KJ "mole"^(-1)) |
|
Answer» `6Na + Al_(2)O_(3) hArr 3Na_(2) O + 2AL` So `Delta G` of this reaction `3(-377) - (-1582) = 451kJ` We know that `Delta G = -2.303 RT log K` log `= 79.04 rArr k = 9 xx 10^(-80)` |
|
| 45. |
Calculate enthalpy of formation of methane( CH_(4)) from the following data : (I) C (s) + O_(2)(g) rarr CO_(2)(g), Delta _(r)H^(@)= -393.5 kJ mol^(_1) (ii)H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), Delta _(r) H^(@) = - 285.8 kJ mol^(_1) (iii)CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l) , Delta _(r) H^(@)= - 890.3 kJ mol^(-1) |
|
Answer» Solution :We aimat at `: C(s) + 2H_(2)(g) RARR CH_(4)(g) , Delta_(f)H^(@) = ?` Multiplyingeqn. (ii) with 2, adding to eqn. (i) and then subtracting eqn. (iii) from the sum, i.e., operating eqn. (i) `+2 xx ` eqn. (ii) - eqn. (iii) , we get `C(s) + 2H_(2)(g) - CH_(4)(g) rarr 0,DELTA _(R) H^(@) - 393.5 + 2( -285.8) - ( - 890.3) = - 74.8 kJ mol^(_1)` or `C_(s) + 2H_(2)(g) rarr CH_(4)(g), Delta _(f) H^(@) = -74.8 kJ mol^(-1)` Hence, enthalpy of FORMATION of methane is `: Delta_(f) H^(@) = - 74.8 kJ mol^(-1)` |
|
| 46. |
Calculate enthalpy of ionisation of OH^(-) ion. Given: H_(2)O_((l)) to H_((aq))^(+) + OH_((aq))^(-) , DeltaH^(@) = -285.83 KJ |
|
Answer» SOLUTION :On adding given equations, `H _(2 (g)) + l//2 O _(2 (g)) to H _((aq )) ^(+) + OH _((aq)) ^(-) Delta H ^(0) =228.51KJ.` But for `H _((aq)) ^(+) , Delta H ^(0)0.` Therefore heat of IONISATION of `OH-is-228.51 KJ.` |
|
| 47. |
Calculate energy of one mole of photons of radiation whose frequency is 5xx10^14 Hz. |
| Answer» SOLUTION :ENERGY of one photon is given by the expression E = HV h = `6.626xx10^(-34)` Js V = `15xx10^14s^(-1)` E = `(6.626xx10^(-34)js)xx(5xx10^10s^(-1))=3.313xx10^(-19)J` Energy of one MOLE of photons `=(3.313cc10^(=19)j)xx(6.022xx10^23mol^(-1))=199.51kJmol^(-1)` | |
| 48. |
Calculateenergyof onemoleof photonsofradiationwhosefrequencyis 5xx 10^(14) Hz (h = 6.626 xx 10^(34)j) |
|
Answer» Solution :E= hv = (6.626 `xx10^(34)`J s ) `xx (5 xx 10^(14) s^(-1))` 1 molephtonn= 6.626 `xx10^(34) js` E=(6.022 `xx 10^(23)` )(3.313 `xx 10^(19)` J) E = 199. 51 KJ `mol^(-1)` |
|
| 49. |
Calculateelectronprotonand neutronin CI_(2) CI andCI^(-)(for : Z=17 and A=35 ) |
|
Answer» <P> Solution :`C1 : e^(-)(17) p (17) ,N(18)``CI^(-): e^(-)(18) ,p(17) ,n(18)` `CI_(2): e (34) ,p (34) ,n(36)` |
|