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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. |
If the solubility product of CuS is 6xx10^(-16), calculate the maximum molarity of CuS in aqueous solution. |
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Answer» Solution :Maximum molarity of CUS in aqueous solution = Solubility of Cus in mol `L^(-1)` If S is the solubility of CuS in mol `L^(-1)`, then `CuShArrunderset("S")(CU)^(2+)+underset(S)(S)^(2-),K_(SP)=[Cu^(2+)][S^(2-)]=SxxS=S^(2)` `therefore""S^(2)=6XX10^(-6)"or"S=sqrt(6xx10^(-16))=2.45xx10^(-8)" mol L"^(-1)` |
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| 2. |
If the solubility product of lead iodide is (PbI_2) is 3.2 times 10^-8. Then its solubility in moles/litre will be…….. |
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Answer» `2 times 10^-3 M` `4s^3=3.2 times 10^-8 IMPLIES s=2 times 10^-3 M` |
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| 3. |
If the solubility product of lead iodide is 3.2 times 10^-8, its solubility will be…….. |
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Answer» `2 times 10^-3 M` `K_(SP)=(s) (2s)^2` `3. 2 times 10^-8=4s^3` `s=((3.2 times 10^-8)/4)^(1//3)=(8 times 10^-9)^(1//3)=2 times 10^-3 M` |
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| 4. |
If the solubility product of CuS is 6xx10^(-15), calculate the maximum molarity of CuS in aqueous solution. |
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Answer» Solution :SOLUBILITY product of CuS, `K_(SP)=6xx10^(-15)` Let us be the solubility of CuS in mol `L^(-1)` `CuS harr underset(S)(Cu^(2+))+underset(S)(S^(2+))` Now,`K_(SP)=[Cu^(2+)][S^(2-)]` `= S XX S` `= S^(2)` Then we have, If `S^(2)=6xx10^(-15)` `S = sqrt(6xx10^(-15))` thus,`S=sqrt(60XX10^(-16))` `= sqrt(60)xx10^(-8)` `S=2.45xx10^(-8)M`. |
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| 5. |
If the solubility product of CuS is 6 xx 10^(-16) , calculate the maximum molarity of CuS in aqueoussolution. |
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Answer» SOLUTION :MAXIMUM molarity of CuS in aqueous solution = Solubility of CuS in mol `L^(-1)` If S is the solubility of CuS in mol `L^(-1)`, then `CuS IFF Cu^(2+)(s) + S^(2-)(s)` ` therefore S^2=6 xx 10^(-16) " or" S = sqrt(6 xx 10^(-16))` Hence, solubility = `2.45 xx 10^(-8 )"mo "L^(-1)` |
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| 6. |
If the solubility product of CuS is 6 xx 10^(-16), calculate the maximum molarity of CuS in aqueous solution . |
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Answer» Solution :`K_(sp)` of CuS = `6 XX 10^(-16)` If S is the solubility , then `CuS to Cu^(2+) + S^(2-)` `[Cu^(2+)] = S , [S^(2-)]= S` `K_(sp) = [Cu^(2+)] [S^(2-)]` `= S xx S= S^(2)` Solubility S = `sqrt(K_(sp)) = sqrt(6 xx 10^(-6))` `= 2.45 xx 10^(-8) M` Highest molarity = `2.45 xx 10^(-8) M` |
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| 7. |
If the solubility product of AgBrO_(3) and Ag_(2)SO_(4) are 5.5 xx 10^(-5) and 2 xx 10^(-5) respectively, the relationship between the solubilities of these can be correctly represented as |
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Answer» `S_(AgBrO_(3)) gt S_(Ag_(2)SO_(4))` `K_(sp) = 4S^(3), K_(sp) = 2 xx 10^(-5)` `S = sqrt((2 xx 10^(-5))/(4)) = 0.017 m//l = 1.7 xx 10^(-2)` `{:(AgBrO_(3), rarr, Ag^(+) + ,BrO_(3)^(-)),(,,S,S):}` `K_(sp) = S^(2), K_(sp) = 5.5 xx 10^(-5)` `S = sqrt(5.5 xx 10^(-5)) = 7.4 xx 10^(-3) m//l`. |
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| 8. |
IF the solubility product K_(sp) of a sparingly soluble salt MX_(2) at 25^(@)C is 1.0 xx 10^(-11), the solubility of the salt in mole "litre"^(-1) at this temperature will be |
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Answer» `2.46 xx 10^(14)` `S = 3sqrt((K_(sp))/(4)) = 3sqrt((1 xx 10^(-11))/(4)) = 1.35 xx 10^(-4)` |
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| 9. |
If the solubility product of AgBrO_3 and Ag_2 SO_4 are 5.5 xx 10^(-5) and 2 xx 10^(-5) respectively, the relationship between the solubilities of these have correctly represented as |
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Answer» A) `s AgBrO_(3) -= Ag_(2) SO_(4)` |
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| 10. |
if the solubility of product of lead iodide(PbI_2) is 3.2xx10^(-8) its solubility will be: |
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Answer» `2XX10^(-3)` M |
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| 11. |
If the solubility of Pbbr_(2) is S g-mole per litre, its solubility product, considering it to be 80% ionized, is |
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Answer» `2.048 S^(2)` `S(S xx 80)/(100) (2S xx 80)/(100)` [`because PbBr_(2)` ionises to 80%] `K_(SP) = [(S xx 80)/(100)][(2S xx 80)/(100)]^(2)` `= 0.8 S xx 2.56 S^(2) = 2.048 S^(3)` |
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| 12. |
If the solubility of Pb3 (PO_4)_2 is mol per litre, then the solubility product of Pb_3(PO_4)_2 will be : |
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Answer» `6s^2` |
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| 14. |
If the solubility of lithium sodium hexafluoro aluminate Li_(3)Na_(3)(AlF_(6))_(2) is x mol L^(-1), then its solubility product is equal to : |
| Answer» Answer :D | |
| 15. |
If the solubility of lithium sodium hexafluro-aluminate, Li_3Na_3(AlF_6)_2 is 'a' mol/litre, its solubility product is equal to: |
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Answer» `a^2` |
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| 16. |
If the solubility of Ca(OH)_2 is sqrt3, the solubility product of Ca(OH)_2 is : |
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Answer» 3 |
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| 17. |
If the solubility of an aqueous solution of Mg(OH)_(2) be Xmoll t^(-1) than k_(sp) of Mg(OH)_(2)is: |
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Answer» Solution :`Mg(OH)_(2) to underset(s)(Mg^(2+)) + underset(2S)2OH^(-)` `K_(sp)=[Mg^(2+)][OH^(-)] = Sxx(2S)^(2) = 4S^(3 = 4X^(3)` |
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| 18. |
If the solubility of a sparingly soluble saltof the type BA_(2) (giving three ions on dissociation of a molecule) is x moles per litre, then its solubility product is given by |
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Answer» `x^(2)` `K_(SP) = 4x^(3)` |
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| 19. |
If the solubility of a sparingly soluble salt of the type BA_2 (giving three ions on dissociation of a molecule) is x mole per litre, Then its solubility product is given by : |
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Answer» `x^2` |
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| 20. |
If the shortest wavelength of H atom in Layman series is x, then the longest wavelength in Balmer series of He^(+) is |
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Answer» `(36x)/5` `1/(lamda)=R(1/(1^(2))-1/(prop^(2)))=R` or `lamda=1/R=x`………..(i) For largest wavelength of Balmer series. `1/(lamda.)=Rxx2^(2)xx(1/4-1/9)=4Rxx5/36` or `1/(lamda.)=(5R)/9`…………(ii) or `lamda.=9/(5R)=(9x)/5""( :. 1/R=x)` |
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| 21. |
If the shortest wave length of Lyman series of H atom is x, then the wave length of the first line of Balmer series of H atom will be- |
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Answer» `9x//5` |
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| 22. |
If the series limit of wavelength of the Lyman series for the hydrogen atom is 912 overset@A, then the series limit of wavelength for the Balmer series of the hydrogen atom is: |
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Answer» 912 `OVERSET@A` |
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| 23. |
If the shortest wavelength of H-atom in Lyman series is x, then longest wavelenght in Balmer series of He^(2+) is : |
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Answer» `(X)/(4)` |
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| 24. |
If the % -s character in one Sb -H bond in SbH_(3) is 1.0 % . What is % p-character is the orbital occupied by its lone pair. |
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Answer» 99 |
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| 25. |
If the rms speed of the NH_3molecule is found to be 0.51 km/s, what is the temperature? |
| Answer» SOLUTION :`-95^@C` | |
| 26. |
If the root mean square speed of helium is 4.75 m s^(-1) at 25^(@)C, then its speed will become 9.50 ms^(-1) at |
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Answer» `100^(@)C` `( 4.75 )/( 9.50) = sqrt(( 298)/( T))` or `( 4.75^(2))/(9.50) = ( 298)/( T )` or `T =1192 K` or `T = 1192 - 273 = 919^(@)C` |
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| 27. |
If the relation between time of decay (t) and half life period (t_(1//2)) is (t = 4 t_(1//2)), the relation between r and mean life (T) is: |
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Answer» `(In 2)/(T^(2))` |
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| 28. |
If the reduction potential is more, then |
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Answer» it is EASILY oxidised |
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| 29. |
if the reaction is reversible all is at equilibrium (DeltaS_universe = 0), then the entropy of the system |
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Answer» will CHANGE abruptly |
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| 31. |
If the reaction between A and B to give C shows first order kinetics in A and Second order in B, the rate equation can be writen as : |
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Answer» `rate=k[A][B]^(1/2)` |
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| 32. |
If the ratio of the rates of diffusion of two gases A and B is 4 : 1, then what is the ratio of densitiesof B and A ? |
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Answer» `16 : 1 ` `4/1=sqrt(d_2/d_1)` `16/1=d_2/d_1` |
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| 33. |
If the ratio of PV_(m) & RT for a real gas is (x)/(24) at a temp where ((delP)/(delV_(m))) =0. The find value of 10x. |
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Answer» |
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| 34. |
If the rate of reaction increases by two times with every 10^@C rise in temperature, then if we raisethe temperature by 40^@C, the new rate of the reaction will be :- |
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Answer» 4 TIMES |
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| 35. |
If the rate of the reaction is equal to the rate constant , the order of the reaction is |
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Answer» 3 |
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| 36. |
If therate ofreactionincreases by a factorof 2.36 , when thetemperatureis raised from300 K to 313 K .Whatis theenergyactivation? R =8.314 JK^(-1) mol^(-1) |
| Answer» Solution :`E_(a) =67 .6 KJ "MOL"^(-1)` | |
| 37. |
If the rate of reaction increases by 27 times, when temperature is increased by 30 K, then temperature coefficient of the reaction is___________________ |
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Answer» |
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| 38. |
If the rate of reaction between A and B is given by rate =k[A][B]^(2), then the reaction is |
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Answer» FIRST order in A |
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| 39. |
If the rate of reaction between A and B is. given by,rate =K[A][B]^n, then the reaction is : |
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Answer» FIRST ORDER in A |
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| 40. |
If the rate of reaction, 2SO_2(g)+O_2(g)overset(Pt)to2SO_3(g) is given by : Rate=K([SO_2])/([SO_3]^(1//2)) which statement are correct : |
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Answer» The OVERALL order of reaction is `-1//2` |
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| 41. |
If the rate of gaseous reaction is independent of pressure, the order of raction is |
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Answer» 0 |
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| 42. |
If the rate of a reaction is 5 at 10^(@)Cthen onincreasing the temperature to 30^(@)C , new rate is |
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Answer» 20 For `20^(@)` rise of temperature , rate constant becomes 4 TIMES , i.e., `k_(2) = 4k_(1) = 4 xx 5 = 20.` |
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| 43. |
If the rate of formation of NH_3 for the reaction N_2+3H_2rarr2NH_3 is 2xx10^-4 Msec^-1, the rate of disappearance of hydrogen is |
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Answer» `1xx10^-4Msec^-1` |
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| 44. |
What would be the activation energy of a reaction when the temperature is increased from 27^(@)C to 37^(@)C ? |
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Answer» SOLUTION :`ln. (k_(2))/(k_(1))=(E_(a))/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]` `ln 2 = (E_(a))/(R )[(10)/(300xx310)]` `E_(a)=9300` R ln 2 `= 53.4 "KJ mol"^(-1)` |
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| 45. |
If the rate law for a reaction is A+B rarr is C(d[A])/(dt)xx(1)/(2)=k[A]^(n)[B]^(m) then the order of a reaction is __________ |
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Answer» N |
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| 46. |
If the rate expresson for a chemical reaction is Rate= k [A] [B]^n then |
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Answer» the ORDER of reaction is 1 `(because m = 1)` |
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| 47. |
If the rate for the chemical reaction is expresssed at Rate =K[A][B]" then |
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Answer» one |
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| 48. |
If the rate constant of a first order reaction at a certain temperature is 1.5xx10^(-1)s^(-1) and t_(1) and t_(2) are the respective times for 50% and 75% completion of the reaction, determine the ratio of t_(2)" to "t_(1). |
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Answer» For `3//4` of the REACTION to OCCUR, `t=t_(3//4),(a-x)=a-3a//4=a//4:.t_(3//4)=(2.303)/(k)log4` For HALF of a reaction to occur, `t=t_(1//2),(a-x)=a-a//2=a//2.:.t_(1//2)=(2.303)/(k)log""(a)/(a//2)=(2.303)/(k)LOG2` Hence, `(t_(3//4))/(t_(1//4))=(log4)/(log2)=(0.6021)/(0.3010)=2.` |
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| 49. |
If the rate constant of a reaction is k = 3 xx 10^(-4) s^(-1), then identify the order of the reaction. |
| Answer» Solution :On the BASIS of UNIT of rate constant `(s^(-1))`, the ORDER of reaction is FIRST order. | |
| 50. |
If the rate constant for a first order reactions K ,the time (t) required for the completion of 99% of the reaction is given by : |
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Answer» t=2.303 /k `[R]_(t)=((100-99)/(100))[R]_(0)` `[R]_(t)=((100-99)/(100))[R]_(0)` `t=(2.303)/(K)` log `([R]_(o))/([R]_(t))` `therefore t=(2xx303)/(K)` log` ([R]_(o))/((100-99)/(100))[R]_(o)` `t=(2.303)/(K)` log `100=(2.303)/(K)`log`10^(2)` `t=(2.303)/(K).2"log"10` `t=(2.303xx2)/(K)[therefore log 10=1]` `t=(4.606)/(K)` |
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