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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. |
At 27^(@)C temperature, 36 gm glucose is in 1 litre aqueous solution has pi = 4.98 bar. Find out concentration if pi = 1.52 bar at same temperature. |
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Answer» Solution :`pi = iCRT = CRT (because i=1" for GLUCOSE")` `4.98 =(36)/(180)xx RT ""` ….(1) `pi = C xx RT ""` …..(2) Now,`("equation (1)")/("equation (2)")=(4.98)/(pi)=(36)/(180)xx(1)/(C )` `therefore C = 0.061 MOL L^(-1)` |
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| 2. |
At 27^(@)C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The values of DeltaE andq are (R=2) |
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Answer» `0, -965.84 CAL` `DeltaE=q+w=40-8=32 J` |
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| 3. |
At 27^(@)C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of DeltaE and q are (R=2 cal) |
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Answer» <P>0,-965.84 CAL `W=2.303" NRT LOG"(p_(2))/(p_(1))=2.303xx1xx300xx"log"(10)/(2)` `=2.303xx600xxlog5=965.84` At constant temperature, `DeltaE=0` `DeltaE=q+W,q=-W=-965.84` cal |
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| 4. |
At 27^(@)C, latent heat of fusion of a compound is 2930 J/mol. Entropy change during fusion is |
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Answer» 9.77 J/mol K `=(2930)/(300)=9.766JK^(-1)mol^(-1)` `=9.77JK^(-1)mol^(-1)` |
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| 5. |
At 27^(@)C in the presence of a catalyst, the activation energy of a reaction is lowered by 2 kcal. Calculate by how much the rate of reaction will increase ? |
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Answer» Solution :In the ABSENCE of catalyst, SUPPOSE rate constant = k. Then `logk=logA-(E_(a))/(2.303"RT")""…(i)` In the presence of catalyst, suppose rate constant = k'. Now, activation energy `=E_(a)-2("if "E_(a)" is in KCAL "mol^(-1))` `:.logk'=logA-(E_(a)-2)/(2.303"RT")" or "logk'=logA-(E_(a))/(2.303"RT")+(2"kcal mol"^(-1))/(2.303"RT")""...(ii)` Subtracting eqn (i) from eqn (ii), we get `logk'-logk=(2" kcal mol"^(-1))/(2.303"RT")` `log""(k')/(k)=(2" kcal mol"^(-1))/(2.303(2xx10^(-3)"kcal K"^(-1)mol^(-1))(300" K"))=1.4474""(R=2xx10^(-3)"kcal K"^(-1)mol^(-1))` or `(k')/(k)="Antilog "1.4474=28" or "k'=28" k`, i.e., the rate of REACTION will INCREASE 28 times. |
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| 6. |
At 27^@C , hydrogen leaks through a tiny hole in a vessel for 20 minutes. Anotherunknown gas at the same temperature and pressure as that of H_2leaks through the same hole for 20 minutes. After the effusion of the gases, the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, what is the molecular weight of the unknown gas? |
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Answer» SOLUTION :`(r_X)/(r_(H_2)) = SQRT(2/M) = (n_X)/(n_(H_2)) , pV = (n_X + n_(H_2) ) RT` 1032 |
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| 7. |
At 27^(@)C, hydrogen is leaked through a tiny hole into a vessel for 20 minute. Another unknown gas at the same temperature and pressure as that of hydrogen leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre, what is the molecular mass of the unknown gas ? |
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Answer» Solution :Let `P_(H_(2)) and P_(x)` be the partial pressures of hydrogen and unknown gas respectively and n be the number of moles of unknown gas. `P_(H_(2)) =(0.7)/(3)xx0.821xx300` `P_(x) =(n)/(3) xx0.0821xx300` Adding both `P_(H_(2))+P_(x) =6=(1)/(3) xx0.0821xx300(0.7+n)` `n=0.0308` mole Applying LAW of diffusion `(0.7//20)/(0.0308//20)=sqrt((M)/(2))` or, M=1033 |
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| 8. |
At 27^(@)C and 37^(@)C , the rates ofa reactionare given as 1.6 xx 10^(-2) mol L^(-1)s^(-1)and 3.2 xx 10^(-2)mol L^(-1)s^(-1) . Calculate the energy ofactivationfor thegivenreaction . |
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Answer» SOLUTION :The ration of specific rates at two DIFFERENT temperatures are gives as, `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the VALUES `k_(1),k_(2),T_(1),T_(2)` and molar GAS constant R , `0.301=(E_(a)xx1000)/(2.303xx8.314)xx(10)/(300xx310)` Energy of activation `=E_(a)=53" kJ mol"^(-1)`. |
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| 9. |
At 277 K, degree of dissociation water is1xx10^(-7)%. The value of ionic product of water is |
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Answer» `3.0xx10^(-14)` |
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| 10. |
At 273K, 100cm^(3) of a solution contaning 3gm of an unidentified solute exhibits an osmotic pressure of 2.24 atm , them molar mass of the solute is |
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Answer» `88g MOL^(-1)` |
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| 11. |
At 273K, 100"Cm"^3 of a solution containing 3gm of an unidentified solute exhibits an osmotic pressure of 2.24 atm, then molar mass of the solute is |
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Answer» 88 `"gmol"^(-1)` |
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| 12. |
At 273 K temperature, if 10 gm glucose (p_(1)), 10 gm urea (p_(2)) and 10 gm sucrose (p_(3)) is dissolved in 250 mL of water, then what is the correct relation of osmotic pressure for them ? |
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Answer» `p_(1)gt p_(2)gt p_(3)` So, `p_(2)gt p_(1)gt p_(3)`. |
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| 13. |
At 273 K, Pd vs P is plotted for various gases 1,2,3,4 assuming ideal behaviour for gases N_(2), He,CO_(2) and H_(2). The correct combination is [P denotes combination in atmosphere and d denotes density in gm/L: |
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Answer» `2 -N_(2),1 - He, 3 -CO_(2), 4-H_(2)` |
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| 14. |
At 273 K, ice and water are in equilibrium and enthalpy of fusion of ice is [6 x 10^3 J mol^-1]. The values of deltaS and deltaG for conversion of one mole of ice into water respectively are |
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Answer» `[21.98 J mol^-1 K^-1, 5227.17 J]` |
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| 15. |
At 27^@ C the ratio of root square speeds of ozone to oxygen is : |
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Answer» `SQRT (3/5)` |
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| 16. |
At 27 ^(@) C , one mole of an ideal gas is compressed isothermallty and reversiblty and reversibly from a pressure of 2 atm to 10 atm . The value of DeltaE and q are ( R = 2 cal ) |
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Answer» `- 965 . 84 cal ` ` = 2.303 xx1 xx 2 xx 300 LG (10//2) = 965.84 ` at constant temperature ,` DeltaE=0,q=-W=-965.84 cal` |
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| 17. |
At 25^oc the equilibrium constant K_1 and K_2 of two reaction are :2NH_3 hArr N_2 +3H_2 : 1/2N_2 +3/2H_2 hArr NH_3 the relation between two equilibrium constant is : |
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Answer» `K_1=K_2` |
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| 18. |
At 25^(@)C,lamda_(oo)(H^+)=3.4982xx10^(-2)S m^2mol^(-1) and lamda_(oo)(OH^-)=1.98xx10^(-2)S m^2 mol^(-1) Given : Sp. Conductance =5.7xx10^(-6)Sm^(-1) for H_2O determine pH and K_w |
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Answer» |
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| 19. |
At 25^(@)C, when 0.5 mol of HCl reacts completely with 0.5 mol of NaOH in a dilute solution, 28.65 kJ of heat is liberated. If at 25^(@)C" "DeltaH_(f)^(0)[H_(2)O(l)]=-285.8kJ*mol^(-1), then DeltaH_(f)^(0)OH^(-)(aq)is- |
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Answer» `-314.45kJ*MOL^(-1)` |
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| 20. |
At 25^(@)C, the vapour presure of pure water is 23.76 mm of Hg and that of an aueous dilute solution of urea is 22.98 mm of Hg. What is the molalitty of the solution ? |
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Answer» `22.98 mm Hg=23.76 mm HgxxX_(A) or X_(A)=((22.98 mm))/(23.76 mm)=0.967` `"Mole FRACTION of solute"(X_(B))=1-X_(A)=1-0.967=0.033` `"MASS of water"=(0.976 MOL)xx(18 g mol^(-1))=17.406 g` 17.405 g of water contain solute = 0.033 mole `"1000 g of water contain solute "=((0.033mol))/((17.406 g))xx1000 g = 1.896 mol` `therefore "Molality of solution"=1.896 m.` |
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| 21. |
At 25^(@)C Vapour Pressureof Pure benzene and pure toluene are 93.4 and 26.9 torr respectively . Asolution is prepared by mixing 60g benzene and 40g of toluene . What pressure should be maintained in the flask containing this solution so that it start boiling at 25^(@)C. |
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Answer» `0.693mm` |
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| 23. |
At 25^(@)C , the vapour pressure of pure water is23.76mm of Hgand thatof an aqueousdilutesolutionof urea is 22.98 mm of Hg . Calculate themolalityof the solution . |
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Answer» or `"" x_(B) = 0.0033` Now for thedilutesolution , `x_(B) = (n_(B))/(n_(A))""….(i)` Molalitym ` =(n_(B))/(w_(A)) xx 1000""….(II)` Dividing eq.(ii) by eq.(i) . `(m)/(x_(B)) = (1000)/(w_(A))xx n_(A)= (1000)/(w_(A)) xx (w_(A))/(18)` ` m= x_(B) xx (1000)/(18) = (0.033 xx 1000)/(18)` ` = 1.83m` |
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| 24. |
At 25^(@)C, the vapour pressure of pure water is 23.76 mm of Hg and that of an aqueous dilute solution of urea is 22.98 mm of Hg. Calculate the molality of this solution? |
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Answer» |
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| 25. |
At 25^@C the vapour pressure of methyl alcohol is 96 torr. What is the mole fraction of CH_3OHin a solution in which the (partial) vapour pressure of CH_3OH is 23 torr at 25^@C ? |
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Answer» |
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| 26. |
At 25^(@)C, the standard heats of formation of H_(2)O(g),H_(2)O_(g),H(g) and O(g) are -241.8, -135.66, 218 and 248.17kJ*mol^(-1) respectively. The bond energy (in kJ*mol^(-1)) of O-O bond in H_(2)O_(2)(g)molecule is- |
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Answer» 179.23 |
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| 27. |
At25^@C, the specific conductance of 0.01M alkaline earth metal chloride is 0.000158 "ohm"^(-1) cm^(-1) Calculate the equivalent conductance. |
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Answer» |
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| 28. |
At 25^@C the specific conductance of a saturated solution of AgCl after substracting the specific conductance of water is 1.82 xx 10^(-4)Sm^(-1). The molar conductance at Infinite dilution of AgNO_3, HNO_3, and HCI are respectively, 133.0 xx 10^(-4), 421.0 xx 10^(-4) and 426.0 xx 10^(-4)S m^2 mol^(-1). Write down half cell reaction and calculate standard reduction potential for the half cell: Pt|50 mL 0.5 M KCI solution, in which few drops of 0.1 M solution of AgNO_3 is added to precipitate out some AgCl. E_(Ag^(+),Ag)^(0)=0.80V |
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Answer» |
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| 29. |
At 25^@C, the standard emf of cell having reactions involving two electron change is found to be 0.295V. The equilibrium constant of the reaction is : |
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Answer» `29.5xx 10^(-2)` |
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| 30. |
At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0 xx 10^(-11). At which pH, will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.001 M Mg^(2+) ions |
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Answer» 8 `1 xx 10^(-11) = 10^(-3)xx [OH^(-)]^(2)` `[OH^(-)]^(2) =10^(-8)` `OH^(-) = 10^(-4)` `POH = 4[pH + pOH + pOH = 14]` pH = 10. |
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| 31. |
At 25^(@)C the solubility product of Mg(OH)_(2) is 1.0xx10^(-11) . At which pH, will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.001 M Mg^(2+) ions ? |
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Answer» 1 `Mg(OH)_(2)HARR Mg^(2+)+2OH^(-)` `K_(SP)=[Mg^(2+)][OH^(-)]^(2)` `[OH^(-)]=sqrt((K_(sp))/([Mg^(2+)]))` `[OH^(-)]=sqrt((1xx10^(-11))/(0.001))=1xx10^(-4)` `THEREFORE pOH = 4`andpH = 10. |
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| 32. |
At 25^(@)C, the solubility product of Hg_(2)Cl_(2) in water is 3.2 xx 10^(-17) mol^(3) dm^(-9). What is the solubility of Hg_(2)Cl_(2) in water 25^(@)C |
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Answer» `1.2 XX 10^(-12)M` Let the concentration of `Hg_(2)^(2+)` be = x Now, for each `Hg_(2)^(2+)` ion, two `Cl^(-)` ions are produced `:.` Concentration of `Cl^(-)` ions = 2X `K_(sp) = [Hg_(2)^(2+)][Cl^(-)]^(2)` `x(2x)^(2) = 3.2 xx 10^(-17)rArr 4x^(3) = 32 xx 10^(-18)` `rArr x^(3) = (32)/(4) xx 10^(-18) :. x = 2 xx 10^(-6) M` |
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| 33. |
At 25^(@)C , the saturated vapour pressure of water is 3.165 k Pa (23.75 mm Hg). Find the saturated vapour pressure of a 5% aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea = 60.05 g mol^(-1) ) |
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Answer» Solution :`(P^(0)- P_(S))/(P^(0)) = n_(2) = (n_(2))/(n)` (for DILUTE solution`n_(2) ltlt n_(1)`) `(P^(0) - P_(S))/(P^(0)) = (W_(2) xx M_(1))/(W_(1) xx M_(2))` GIVEN `P^(0)` = 3.165 kPa, `W_(2) = 5 g, W_(1) = 95 ` g `M_(2) = 60.05 g "mol"^(-1), M_(1) = 18 g mol^(-1)` `(3.165 - P_(s))/(3.165) = (5 xx 18)/(60.05 xx 95) = 0.0158` 3.165` - P_(s) ` = 0.049 `P_(s) = ` 3.116 kPa |
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| 34. |
At 25^@C, the reduction potential of hydrogen electrode is -0.118 V at 1 atm. What is the pH of acid solution used for the construction of the electrode? |
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Answer» SOLUTION :`E = E^(@) + (0.059)/(N) log C " (or) " E = E^@ + (0.059)/(2) log [H^+]` `-0.118 = 0.059 log [H^+] " (or) " - log[H^+] = 2, pH = 2`. |
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| 35. |
At 25^(@)C, the reduction potential of hydrogen electrode is -0.118V at 1 atm. What is he pH of acid solution used for the construction of the electrode ? |
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Answer» Solution :`E=E^(@)+(0.059)/(n) LOG C` (or) `E =E^(@)+(0.059)/(2) log [H^(+)]` `-0.118=0.059 log [H^(+)]` (or) `-log[H^(+)] =2 , pH=2`. |
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| 36. |
At 25^(@)C the pH value of a solution is 6. The solution is |
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Answer» BASIC |
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| 37. |
At 25^(@)C, the pH of 0.1 (M) aqueous solution of NH_3 is 11.13. At the same temperature, the pH of a solution containing 0.1 (M) of NH_4Cl and 0.01 (M) of NH_3 is |
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Answer» 4.74 |
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| 38. |
At 25^(@)C, the molar conductance at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl_(2) are 248 xx 10^(-4) , 126 xx 10^(-4) and 280 xx 10^(-4) S m2 mol^(-1) respectively. The value of wedge_(m)^(infty) Ba(OH)_(2) in S m^(2) mol^(-1) will be: |
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Answer» `52.xx10^(-4)` |
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| 39. |
At 25°C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl_(2) are 248 xx 10^(-4) ,126 xx 10^(-4) and 280xx10^(-4)Sm^(2)mol^(-1) respectively, lambda_(m)^(o)Ba(OH)_(2) inSm^(2)mol^(-1)is |
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Answer» `52.4 xx 10^(-4)` |
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| 40. |
At 25^(@)C, the following heat of formation is given: Compound : SO_(2)(g), H_(2)O(l) DeltaH_(1)^(@) kJ//"mole" : -296.81, -285.83 For the reactions at 25^(@)C 2H_(2)S(g) + Fe(s) to FeS_(2)(s) + 2H_(2)(g), DeltaH^(@) = -137 kJ/mole. H_(2)S(g) + 3//2O_(2)(g) to H_(2)O(l) + SO_(2)(g), DeltaH^(@) = -562 kJ/mole. Calculate heat of formation of H_(2)S(g) and FeS_(2)(s) at 25^(@) C. |
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Answer» |
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| 41. |
At 25^(@)C the emf of the cell Pb_(2)|PbCl_(2) HCl (0.5 M) || HCl (0.5 M) |AgCl_((s))|Ag is 0.49 volts and its temperature coefficient (dE)/(dt)=-1.8 xx 10^(-4) volt/degree. Calculate (a) The empty change when 1 gm mol of silver is deposited and (b) The heat of formation of AgCl, if the heat of formation of lead chloride is -8600 cal. |
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Answer» (B) -30506 cal/mole |
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| 42. |
At 25^(@)C the highest osmotic pressure is exhibited by 0.1 M solution of |
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Answer» `CaCl_(2)` |
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| 43. |
At 25^@C the half life of decomposition of H_2O_2 is 50 min. If initially 4M H_2O_2 is present, amount of H_2O_2 left after 200 min is |
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Answer» 2M |
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| 44. |
At 25^(@)C, the dissociation constant of a base BOH is 1.0xx10^(-12). The concentration of hydroxyl ions in 0.01 M aqueous solution of base would be : |
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Answer» `1.0xx10^(-5) MOL L^(-1)` `[BOH]=0.01 M` `{:(,BOH,hArr,B^(+),+,OH^(-)),("Initial",1,,0,,0),("conc.",,,,,),("At equal.",(1-X),,CX,,Cx):}` `K_(b)=((Cx)xx(Cx))/(C(1-x))=(Cx^(2))/(1-x)=1.0xx10^(-12)` or `(0.01x^(2))/(1)=1.0xx10^(-12)(1-x ~~ 1)` `x = 1.0xx10^(-7)mol L^(-1)` |
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| 45. |
At 25^(@)C, the dissociation constant of a base BOH is 1.0 xx 10^(-12). The concentration of Hydroxyl ion in 0.01 M aqueous solution of the base would be |
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Answer» `2.0 xx 10^(-6) mol L^(-1)` `K_(b) = (C^(2)alpha^(2))/(C(1-alpha)) = Calpha^(2)` assuming `alpha lt lt 1, 1 - alpha ~= 1` `10^(-12) = 10^(-2) xx alpha^(2) , alpha^(2) = 10^(-10) , alpha = 10^(-5)` `[OH^(-)] = C alpha = .01 xx 10^(-5) = 10^(-7)`. |
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| 46. |
At 25^(@)C , the activation energies for a reaction in absence and presence of a catalyst areE_(a) and (E_(a) -2) "kcal.mol"^(-1) , respectively . For the reaction , the reaction -rate and enthalpy change in absence of catalyst are r "mol.L"^(-1).s^(-1) and DeltaHk "cal.mol"^(-1) , respectively . which of the following would be true regarding the reaction when it is carried out in presence of a catalyst - |
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Answer» reaction -rate`= 14xxr "mol"^(-1).L.s^(-1)` |
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| 47. |
At 25^@C the degree of ionization of water was found to be 1.8 xx 10^(–9). Calculate the ionization constant and ionic product of water at this temperature |
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Answer» Solution : If x is the degree of ionization of water, then `{:(H_2O, hArr , H^(+), + , OH^(-)),(c(1-ALPHA), , CALPHA , , calpha):}` `c=[H_2O]=1000/18`=55.56 M `K_(eq)=([H^+][OH^+])/([H_2O])=((calpha)^2)/(c(1-alpha))=calpha^2` (since `alpha` is very muchless than 1) `K_(eq)=55.56xx(1.8xx10^(-9))^2=1.8xx10^(-16)` M `K_W=[H^+][OH^-]=(c alpha )^2 =(55.56xx1.8xx10^(-9))^2` `K_W=1.0xx10^(-14) M^2` |
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| 48. |
At 25^(@)C temperature, if for given unknown half cell has 0.34 Volt potential, then calculate standard reduction potential for copper : Pt|H_(2(g))(1" atm")|H_((1M))^(+)||Cu_((1M))^(2+)|Cu |
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Answer» `-0.34` Volt Oxidation : `H_(2(g)) to 2H_((aq))^(+)+2e^(-) , E_(RED)^(@)=`zero REDUCTION : `Cu_((aq))^(2+)+2e^(-) to Cu_((S)),""E_(red)^(@)=(?)` `E_(cell)^(@)=E_(red)^(@)("CATHODE")-E_(red)^(@)("ANODE")` `therefore 0.34=E_(Cu^(2+)//Cu)^(@)-E_(H^(+)//H_(2))^(@)` `therefore 0.34V=E_(Cu^(2+)//Cu)^(@)-0.0` `therefore E_(Cu^(2+)//Cu)^(@)=+0.34V`. |
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| 49. |
At 25^(@)C specific coductivity of a normal solution of KCl is 0.002765 mho. The resistance of cell is 400 ohms. The cell constant is:- |
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Answer» 0.815 `=(0.002765)/(1//R)=0.002765xx400=1.106` |
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| 50. |
At 25^(@)C, pK_b(NH)_3=4.74,pK_a(HF)=3.14 and pK_a(HCN)=9.4 Hence- |
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Answer» AQUEOUS SOLUTION of `NH_4F` is acidic |
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