94420 + Interview Questions in Chemistry in Class 12

1.	Solutions of two electrolytes 'A' and 'B' are diluted. The Lambda_(m)of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.
Answer» Solution : Electrolyte B is strong as on DILUTION, the NUMBER of ions remains the same. Only the interionic attraction decreases. THEREFORE the increase in `Lambda_(m)`is small. However in case of A, which is a weak electrolyte, dilution increases the ionization. Therefore `Lambda_(m)`increases much more.

Discussion

2.	Solutions of two electrolytes A and B each having a concentration of 0.2 M have conductivities 2xx10^(-2) and 4xx10^(-4)" S "cm^(-1) respectively. Which will offer greater resistance to the flow of current and why?
Answer» SOLUTION :`kappa=Gxx(1)/(a)=(1)/(R)(1)/(a),i.e., kappa PROP(1)/(R)`. Hence, B will OFFER GREATER resistance.

Discussion

3.	Solutions of two electrolytes 'A' and 'B' are diluted. The Lamda_(m) of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.
Answer» Solution :* Electrolyte .B. is strong as on dilution the number of ions remains the same, only interionic attraction decreases THEREFORE increase in `Lamda_(m)` is small. * Electrolyte B is strong electrolyte with respect to electrolyte A. * Electrolyte A is strong, so it gives complete ionization and no dilution, there no significant CHANGE is OBSERVED so, in value of `Lamda_(m)` there will be less difference. * Electrolyte B is weak, so it does not give complete ionization and on dilution, significant change is observed in number of ions. so, its `Lamda_(m)` value drastically increases (25 times) on dilution. These difference in the value of `Lamda_(m)` is due to increase in the distance between INTER ions which decrease in interionic attaction. "strong electrolyte gives less increase in `Lamda_(m)` on dilution."

Discussion

4.	Solutions in test tubes containing H_(2)O and aqueous NaOH can be differentiated with the help of
Answer» red LITMUS blue litmus `Na_(2)CO_(3)` HCL (aqueous) Solution :`H_(2)O` is neutral while aqueous NAOH is basic . Thus , `H_(2)O` has no effect no red litmus while aqueous NaOH TURNS it blue.

Discussion

5.	Solutions having same osmotic pressure are called _____
Answer» SOLUTION :INCREASES

Discussion

6.	Solutions distilled without change in composition at a temperature are called:
Answer» Atmorphous Azeotropic mixture Ideal solution Super SATURATED solution Answer :B

Discussion

7.	Solutions are of "______________" types
Answer» three six nine eleven SOLUTION :THEORY BASED "INTRODUCTION and DEFINITION of solution."

Discussion

8.	Solution "X" contains Na_(2)CO_(3) and NaHCO_(3) 20 mL of X when titrated using methyl orange indicator consumed 60 mL of 0.1 M of X solution, when titrated using phenolphthalein consumed 20 ml. of 0.1 M HCI solution. The concentrations (in mol L^(-1) ) of Na_(2)CO_(3) and NaHCO_(3), in X are respectively
Answer» 0.01,0.02 0.1,0.1 0.01,0.01 0.1,0.01 Solution :(i) If phenolphthalein is used as indicator for titration of `Na_(2)CO_(3)` with HCl ,the END point is reached only when half of the `Na_(2)CO_(3)` neutralised. `Na_(2)CO_(3) + HCI rarr NAHCO_(3) + NaCl` At the end point, there is no reaction between `NaHCO_(3)` and HCI. Volume of `Na_(2)CO_(3)` = Volume of HCl CONSUMED 20mL of 0.1 M=20mL of 0.1 M `:.`The concentration of `Na_(1)CO_(3)` in given solution 0.1M Since only half volume of HCl is consumed for titration of `Na_(2)CO_(3)` at phenolphthalein end point. The remaining half volume of HCI is used for complete NEUTRALIZATION of `Na_(2)CO_(3)` with methyl orange as indicator `NAHCO_(3) + HCI rarr NaCI + CO_(2)+ H_(2)O` `:.`Volume of HCI used for complete neutralization of `Na_(2)CO_(3)` in given solution = `2 xx20 ml = 40 ml` (ii) The remaining HCl used to neutralize `NaHCO_(3)` from the given X solution = 60– 40= 20 ml of 0.1 M From the equation Volume of `NaHCO_(3)`= Volume of HCl consumed 20 mL of 0.1M = 20 mL of 0.1 M `:.`The concentration of `NaHCO_(3)`in given X solution = 0.1M

Discussion

9.	Solution with reserve acidity and lakalinity are called :-
Answer» Isohydric solution True solution Normal soltion Buffer solution Answer :4

Discussion

10.	Solution S_1 contains 3g of urea per litre and solution S_2 contains 9g glucose per litre. At 298 K, the osmotic pressure of
Answer» `S_1` is GREATER than that of `S_2` `S_1`is LESS than that of`S_2` Both the SOLUTIONS is same Both the solutions is 1 atm. Answer :C

Discussion

11.	Solution (s) containing 40 gm NaOH is/are
Answer» 50GM of 80% (w/w) NAOH 50gm of 80% (w/v) NaOH `[d_("SOLN.") = 1.2 gm//m]` 50 gm of 20 NaOH `[d_("soln.") = 1 gm//ml]` 50 gm of 5 m NaOH Answer :A::C

Discussion

12.	Solution prepared by dissolving equal number of moles of HOCl (K_(a) = 3.2 xx 10^(-8)) and NaOCl in a buffer of pH
Answer» 3.2 G `8.0 g` 4.8 g 7.5 g Answer :D

Discussion

13.	Solution prepared by dissolving equal number of mole of HOCl ( K_s = 3.2 xx 10^-8) and NaOCl is a buffer of pH :
Answer» 8 3.2 7.5 4.8 Answer :C

Discussion

14.	Solution of ZnS in dil HCl , when treated with NaOH solution , a white ppt.is formed which dissolves in excess of NaOH due to the formation of :
Answer» `ZN(OH)_2` `Zn(OH)Cl` `Na_2ZnO_2` `Na_2ZnO_3`. Solution :`ZnS + 2HCL to ZnCl_2 + H_2S ` `ZnCl_2 + 2NaOH to Zn(OH)_2 + 2NaCl` `Zn(OH)_2 + underset("EXCESS")(2NaOH) to underset("Sod zincate (soluble)")+ 2H_2O`

Discussion

15.	Solution of (X) in dil HCI + H_(2)O rarrwhite turbidily (X) overset(H_(2)S//HCI) toback ppt (Y),(Y) is solution in
Answer» `NAOH` YAS `NHO_(3)` HCI Answer :C

Discussion

16.	Solution of sodium metal in liquid NH_3 is strongly reducing due to the presence of :
Answer» SODIUM atoms Solvated electrons NaOH Spdoum amide Answer :B

Discussion

17.	Solution of SO_2 in water is known as:
Answer» Hydrosulphuric acid Sulphurous acid Thiosulphuric acid Thiosulphurous acid Answer :B

Discussion

18.	Solution of following salt on heating with solid K_2Cr_2O_7 and conc. H_2SO_4 orange redvapours are evolved and resultant solution turn aquous NaOH solution yellow
Answer» NaBr NaCl `NaNO_3` NAI ANSWER :B

Discussion

19.	Solution of a salt in sulphanilicacid a naphithylamine give red ppt ,due to
Answer» `BR^(THETA)` `I^(Theta)` `NO_(2)^(Theta)` `NO_(3)^(Theta)` ANSWER :b

Discussion

20.	Solution of a salt in dilute H_(2)SO_(4) or acetic acid produces deep blue colour with starch iodide solution.The salt contains:
Answer» `Br^(-)` `I^(-)` `NO_(3)^(-)` `NO_(2)^(-)` Solution :NITRITE ion liberates `I_(2)` from potassium iodide TURNING starch blue. `2NO_(2)^(-)+3l^(-)+4CH_(3)COOH to I_(3)^(-)+2NOuarr+4CH_(3)COO^(-)+2H_(2)O` `I_(2)`+starch `to` blue colour

Discussion

21.	Solution of a salt in dil H_(2)SO_(4) produces deepblue colour with starchiodide solution .The salt contains
Answer» `BR^(THETA)` `I^(Theta)` `NO_(2)^(Theta)` `NO_(3)^(Theta)` ANSWER :C

Discussion

22.	Solution of a salt in dilute H_2 SO_4 produces deep blue colour with starch iodine solution. The salt contains
Answer» `Br^(-)` `I^(-)` `NO_(3)^(-)` `NO_(2)` Solution :NITRITE ion LIBERATES `I_(2)` from POTASIUM iodide TURNING starch blue.

Discussion

23.	Solution of a chemical compound 'X reacts with AgNO_3 solution to form a white ppt. 'Y which dissolves in excess of NH4OH to give complex 'Z'. When Z' is treated with dil HNO_3. Y reappears. The chemical compound 'X' may be :
Answer» NaBr NaI KCL `CH_3Cl` ANSWER :C

Discussion

24.	Solution of a mono basic acid has a pH =5 . If1 mL of it is diluted to 1 litre , what will be the pH of the resulting solution ?
Answer» 3.45 6.96 8.58 10.25 Answer :2

Discussion

25.	Solution having the same osmotic pressure are called ……………..and they have same…………….. .
Answer» SOLUTION :ISOTONIC, MOLAR CONCENTRATION

Discussion

26.	Solution A + solution B rarr Fehling,s reagent. What are A and B ?
Answer» aq. Copper sulphate, aq. SODIUM potassium tartarate aq. `AgNO_(3)`, aq. `NH_(4)OH` aq. `AgNO_(3)`, alkaline sodium POTASIUM tartarate aq. Copper sulphate, alkaline Sodium potassium tartarate Answer :D

Discussion

27.	Solution (A) is ideal and solution (B) is non-ideal with -ve deviation, which of the following are correct?
Answer» `DeltaH_("mix")=0, DeltaV_("mix")=0` for A `DeltaS_("mix")` is positive for both A, B `Delta H_("mix")` is positive, `DEltaV_("mix")` is NEGATIVE for B `DeltaH_("mix")` is negative, `DeltaV_("mix")` is positive for B Answer :A::B::C

Discussion

28.	Solution A contains 7g//L MgCl_(2) and solution B sontains 7g//L of NaCl . At room temperature, the osmotic pressure of
Answer» SOLUTION A is grater that B both have same osmotic PRESSURE solution B is greater than A Can't determine Answer :C

Discussion

29.	Solution 'A' contains acetone dissolved in chloroform and solution B' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions A and B, respectively are
Answer» POSITIVE and positive positive and NEGATIVE negative and negative negative and positive. Solution :(d): Solution of acetone in CHLOROFORM shows negative deviation and in `CS_(2)` it shows positive deviation from Raoult's law.

Discussion

30.	Solution A contains 7g/L MgCl_2 and solution B contains 7g/L of NaCl. At room temperature, the osmotic pressure of
Answer» Solution A is GREATER than B Both have same OSMOTIC pressure Solution B is greater than A Can't determine. Answer :C

Discussion

31.	Solution A contains 7g/L of MgCl_(2) and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of :
Answer» SOLUTION A is greater than B Both have same osmotic pressure Solution B is greater than A Can't determine Solution :Concentration of NaCl solution `=(7)/(58.5)=0.119 M` Concentration of `MgCl_(2)` solution `=(7)/(95)=0.074 M` HENCE, particles (IONS) in NaCl solution `(B) = 2xx0.119 = 0.238 M` Hence, particles (ions) in `MgCl_(2)` solution `(A)=3xx0.074 = 0.222 M` As the particles in solution B (NaCl) is more than in solution A, the osmotic pressure of solution B will be greater than that of solution A.

Discussion

32.	Solution (A) containing FeCl_(3) is separated from solution (B) containing K_(4)FE(CN)_(6) by a semipermeable membrane as shown below : {:("Solution (A)","Solution (B)"),(FeCl_(3),K_(4)Fe(CN)_(6)):} In FeCl_(3) on reaction with K_(4)[Fe(CN)_(6)] produces blue colour of Fe_(4)[Fe(CN)_(6)]_(3), the blue colour will appear in
Answer» A B In both A and B Neither in A nor in B Solution :As there is only a NET FLOW of solvent from one soltuion to the other through the semi-permeable membrane and there is no flow of solute, the two solute will not come in CONTACT with each other. Hence, no colour will APPEAR in A or B.

Discussion

33.	Solution A contains 7 g/L MgCl_(2) and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of
Answer» solution A is greater than B both have same osmotic pressure solution B is greater than A can't determine. Solution :`7gL^(-1)MgCl_(2)=(7)/(24+71)"mol L"^(-1)` `=(7)/(95)" mol L"^(-1)=(7xx3)/(95)"mol L"^(-1) " of ions "=0.22M` `7gL^(-1)NaCl=(7)/(23+35.5)M` `=(7)/(58.5)M=(7xx2)/(58.5)"mol L"^(-1)" of ions = 0.24 M"` As concentration of ions in NaCl solution is greater, NaCl soltuion (solution B) will have greater osmotic pressure.

Discussion

34.	Solution A, B, C and D are respectively 0.1M Glucose, 0.05M NaCl, 0.05M BaCl_2 and 0.1M AlF_3. Which one of the following pairs is isotonic ?
Answer» A & C B & C A & B A & D ANSWER :C

Discussion

35.	Solute when dissolved in water:
Answer» Decreases the vapour PRESSURE of WATER Increases the boiling POINT of water Decreases the FREEZING point of water All of the above Answer :D

Discussion

36.	Solute when dissolve in water
Answer» increases the VAPOUR PRESSURE of water decreases the BOILING point of water decrease the freezing point of water All of the above ANSWER :D

Discussion

37.	Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P, then 0.05 M solution of A at the same temperature will produce an osmatic pressure equal to:
Answer» P 1.5P 2P 3P Answer :D

Discussion

38.	Solubilty product of Mg(OH)_(2),Cd(OH)_(2),Al(OH)_(3) and Zn(OH)_(2) are 4xx10^(-11),8xx10^(-6),8.5xx10^(-23) and 1.8xx10^(-14) resectively. The cation, that will precipitate first as hydroxide, on adding limited quantity of NH_(4)OH in a solution containing equimolar amount of metal cation, is :
Answer» `Al^(3+)` `ZN^(2+)` `MG^(2+)` `Cd^(2+)` ANSWER :a

Discussion

39.	Solubility (s) of CaF_(2) in terms of its solubility product is given as
Answer» `s = (K_(sp))^(1//3)` `s = (K_(sp)//2)^(1//3)` `s = (K_(sp)//4)^(1//3)` `s = (K_(sp)//2)^(1//2)` SOLUTION :It is TERNARY electrolyte, `CaF_(2) harr Ca^(2+) + 2F^(-)` Let solubility of `CaF_(2)` be s mol `"litre"^(-1)` Then `{:(CaF_(2) hArr,Ca^(2+),+,2F^(-)),(,s,,2s):}` Let `K_(sp) = [Ca^(2+)] [F^(-)]^(2) = (s) xx (2s)^(2) = 4S^(3)` `s = sqrt((K_(sp))/(4))`.

Discussion

40.	Solubility products of AI(OH)_(3) and An(OH)_(2) are 8.5 xx 10^(-23) and 1.8 xx 10^(-14) respectively. If both AI^(3+) and Zn^(2+) ions are present in a solution, which one will be precipitated first on addition of NH_(4)OH ?
Answer» `AI(OH)_(3)` `ZN(OH)_(2)` Both 1 and 2 None of these Answer :A

Discussion

41.	Solubility product of silver bromide is 5.0xx10^(-13). The quantity of potassium bromide (molar mass taken as 120 g " mol"^(-1)) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is
Answer» `1.2xx10^(-10)g`<BR>`1.2xx10^(-9)g` `6.2xx10^(-5)g` `5.0xx10^(-8)g` Solution :`AgBr HARR Ag^() Br^()` `K_(sp) =([Ag^(+)][Br^(-)]` For precipitation to occur Ionic product `gt` Solubility product `[Br^()] (K_(sp))/([Ag^()])(5 10^(-13))/(0.05) 10^(-11)` i.e. precipitation just STARTS when `10^(-11)` moles of KBR is added to `1//AgNO_(3)` solution `therefore` Number of moles of `Br^(-)` needed from `KBr=10^(-11)` `therefore` Mass of `KBr=10^(-11)xx120=1.2xx10^(-9) g`

Discussion

42.	Solubility product of silver bromide is 5.0 xx 10^(-13). The quantity of potassium bromide (molar mass taken as 120 g mol^(-1)) to be added to 1 litre of 0.5 M solution of silver nitrate to start the precipitation of AgBr is
Answer» `5.0 xx 10^(-8) G` `1.2 xx 10^(-10) g` `1.2 xx 10^(-9) g` `6.2 xx 10^(-5) g` Solution :`Br^(-) = (K_(sp)(AGBR))/(C_(AG^(+)))` `Br^(-) = (5 xx 10^(-13))/(0.05) = 10^(-11)` Conc. `= [KBr] = 10^(-11)` ltbgt Moles of KBr `= 10^(-11)` Weight of KBr `= 10^(-11) xx 120 xx 1.2 xx 10^(-9)g`.

Discussion

43.	Solubility product of Mg(OH)at ordinary temperature is 1.96 xx10^(-11). pH of a saturated soln. of Mg(OH)_2 will be
Answer» `10.53 ` `8.47` `6.94` `3.47` Solution :`MG(OH)_(2(s)) hArrMg^(2+) +2OH^(-)` ` K_(sp ) =([Mg]^(2+)[OH^(-)]^2)/( [Mg (OH)_2])= 1.96 xx 10^(-11)` ` x xx (2x)^2= 1.96 xx 10^(-11)` ( concentrationofsolid is UNITY ) `4x^3= 1.96 xx 10^(-11)` ` x=((1.96 xx 10^(-11))/(4) )^(1//3)` ` x= ( 4.9 xx 10^(-12))^(1//3)= 1.7 xx10^(-4)` So ` OH^(-) ` concentration`= 2XX 1.7 xx 10^(-4)` ` i.e.,[OH^-] = 3.4 xx 10^(-4)` Now `pOH=- log [OH^-]` `=- log [3.4 xx 10^(-4)]=4- 0.531= 3.469 ` ` therefore pH= 14 - 3.469= 10.531`

Discussion

44.	Solubility product of PbCl_(2) at 298 K is 1.0xx10^(-6). At this temperature solubility of PbCl_(2) in moles per litre is :
Answer» `(1.0xx10^(-6))^(1//2)` `(1.0xx10^(-6))^(1//3)` `(0.25xx10^(-6))^(1//3)` `(0.25xx10^(-6))^(1//2)` Solution :Let SOLUBILITY of `PbCl_(2)=x` `K_(sp)=[Pb^(2+)][Cl^(-)]^(2)=(x)(2x)^(2)=4x^(3)` `THEREFORE x = ((1.0xx10^(-8))/(4))^(1//3)` `=(0.25xx10^(-6))1//3`

Discussion

45.	Solubility product of M(OH), is 10^(-14). What should be the concentration of M^(2+) in 0.1 M solution of NH_(4)OH, if NH_(4)OH gets 10% ionised?
Answer» `10^(-10)` `10^(-5)` `10^(-12)` `10^(-4)` Solution :`K_(sp)=[M^(+2)][OH^(-)]^(2)` `[OH^(-)]=(10)/(100)xx0.1=0.01` `[M^(+2)]=(K_(sp))/([OH^(-)]^(2))=(10^(-14))/([0.01]^(2))=10^(-10)`

Discussion

46.	Solubility product of Mg(OH)_(2) at ordinary temp. is 1.96 xx 10^(-11). pH of a saturated of Mg(OH)_(2) will be
Answer» 10.5 8.47 6.94 3.47 Solution :`{:("MG(OH)"_(2(s)),"Mg"^(2+),+,2OH^(-)),(,x,,2x):}` `K_(sp) = [Mg]^(2+) [ OH^(-)]^(2) = 1.96 xx 10^(-11)` `x xx (2 x)^(2) = 1.96 xx 10^(-11)` (concentration of solid is unity) `4x^(3) = 1.96 xx 10^(-11)` `x = ((1.96 xx 10^(-11))/(4))^(1//3)` `x = (4.9 xx 10^(-12))^(1//3) = 1.6 xx 10^(-4)` So, `OH^(-)` concentration `= 2 xx 1.6 xx 10^(-4)` i.e. `[OH^(-)] = 3.2 xx 10^(-4)` `= -log [3.2 xx 10^(-4)] = 4- 0.505 = 3.495` `:. pH = 14 - 3.495 = 10.505`.

Discussion

47.	Solubility product of Ba(OH)_2 is 4xx10^-9 its solubbility in water is
Answer» `1XX10^(-3)` M `1xx10^(-9)` M `4XX10^(-27)` M `1xx10^(-27)` M Answer :A

Discussion

48.	Solubility product of BaCl_(2) is 4 xx 10^(-9). Its solubility in moles/litre would be
Answer» `1 XX 10^(-3)` `1 xx 10^(-9)` `4 xx 10^(-27)` `1 xx 10^(-27)` Solution :`K_(sp) = 4S^(3), S^(3) = (4 xx 10^(-9))/(4) = 10^(-9)` `:. S = 10^(-3) M`.

Discussion

49.	Solubility product of AgI at 25^@C is 4xx10^(-18)"mole"^2 //L^2 . The solubility of AgI in presence of 2xx10^(-4) M Kl solution at 25^@Cis approximately equal to : (in mol/L)
Answer» `8xx10^(-8)` `2XX10^(-16)` `2xx10^(-14)` `0.5xx10^(-12)` Solution :`K_"sp"=[Ag^+][I^-]` `4xx10^(-18)=[Ag^+]xx2xx10^(-4)` `[Ag^+] = 2 XX 10^(-14)` M

Discussion

50.	Solubility product of AgCl is 1 xx 10^(-6) at 298 K. Its solubility in mole "litre"^(-1) would be
Answer» `1 xx 10^(-6)` mol/ litre `1 xx 10^(-3)` mol/litre `1 xx 10^(-12)` mol/litre None of these Solution :`{:("For",AgClrarr,Ag^(+)+,Cl^(-)),(,,X,x):}` `K_(sp) = x^(2), x = sqrt(K_(sp)), sqrt(1 xx 10^(-6)) = 1 xx 10^(-3)` mole/litre.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Solutions of two electrolytes 'A' and 'B' are diluted. The Lambda_(m)of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.

Solutions of two electrolytes A and B each having a concentration of 0.2 M have conductivities 2xx10^(-2) and 4xx10^(-4)" S "cm^(-1) respectively. Which will offer greater resistance to the flow of current and why?

Solutions of two electrolytes 'A' and 'B' are diluted. The Lamda_(m) of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte ? Justify your answer.

Solutions in test tubes containing H_(2)O and aqueous NaOH can be differentiated with the help of

Solutions having same osmotic pressure are called _____

Solutions distilled without change in composition at a temperature are called:

Solutions are of "______________" types

Solution with reserve acidity and lakalinity are called :-

Solution S_1 contains 3g of urea per litre and solution S_2 contains 9g glucose per litre. At 298 K, the osmotic pressure of

Solution (s) containing 40 gm NaOH is/are

Solution prepared by dissolving equal number of moles of HOCl (K_(a) = 3.2 xx 10^(-8)) and NaOCl in a buffer of pH

Solution prepared by dissolving equal number of mole of HOCl ( K_s = 3.2 xx 10^-8) and NaOCl is a buffer of pH :

Solution of ZnS in dil HCl , when treated with NaOH solution , a white ppt.is formed which dissolves in excess of NaOH due to the formation of :

Solution of (X) in dil HCI + H_(2)O rarrwhite turbidily (X) overset(H_(2)S//HCI) toback ppt (Y),(Y) is solution in

Solution of sodium metal in liquid NH_3 is strongly reducing due to the presence of :

Solution of SO_2 in water is known as:

Solution of following salt on heating with solid K_2Cr_2O_7 and conc. H​_2​SO_​4 orange redvapours are evolved and resultant solution turn aquous NaOH solution yellow

Solution of a salt in sulphanilicacid a naphithylamine give red ppt ,due to

Solution of a salt in dilute H_(2)SO_(4) or acetic acid produces deep blue colour with starch iodide solution.The salt contains:

Solution of a salt in dil H_(2)SO_(4) produces deepblue colour with starchiodide solution .The salt contains

Solution of a salt in dilute H_2 SO_4 produces deep blue colour with starch iodine solution. The salt contains

Solution of a chemical compound 'X reacts with AgNO_3 solution to form a white ppt. 'Y which dissolves in excess of NH4OH to give complex 'Z'. When Z' is treated with dil HNO_3. Y reappears. The chemical compound 'X' may be :

Solution of a mono basic acid has a pH =5 . If1 mL of it is diluted to 1 litre , what will be the pH of the resulting solution ?

Solution having the same osmotic pressure are called ……………..and they have same…………….. .

Solution A + solution B rarr Fehling,s reagent. What are A and B ?

Solution (A) is ideal and solution (B) is non-ideal with -ve deviation, which of the following are correct?

Solution A contains 7g//L MgCl_(2) and solution B sontains 7g//L of NaCl . At room temperature, the osmotic pressure of

Solution 'A' contains acetone dissolved in chloroform and solution B' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions A and B, respectively are

Solution A contains 7g/L MgCl_2 and solution B contains 7g/L of NaCl. At room temperature, the osmotic pressure of

Solution A contains 7g/L of MgCl_(2) and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of :

Solution A contains 7 g/L MgCl_(2) and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of

Solution A, B, C and D are respectively 0.1M Glucose, 0.05M NaCl, 0.05M BaCl_2 and 0.1M AlF_3. Which one of the following pairs is isotonic ?

Solute when dissolved in water:

Solute when dissolve in water

Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P, then 0.05 M solution of A at the same temperature will produce an osmatic pressure equal to:

Solubilty product of Mg(OH)_(2),Cd(OH)_(2),Al(OH)_(3) and Zn(OH)_(2) are 4xx10^(-11),8xx10^(-6),8.5xx10^(-23) and 1.8xx10^(-14) resectively. The cation, that will precipitate first as hydroxide, on adding limited quantity of NH_(4)OH in a solution containing equimolar amount of metal cation, is :

Solubility (s) of CaF_(2) in terms of its solubility product is given as

Solubility products of AI(OH)_(3) and An(OH)_(2) are 8.5 xx 10^(-23) and 1.8 xx 10^(-14) respectively. If both AI^(3+) and Zn^(2+) ions are present in a solution, which one will be precipitated first on addition of NH_(4)OH ?

Solubility product of silver bromide is 5.0xx10^(-13). The quantity of potassium bromide (molar mass taken as 120 g " mol"^(-1)) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

Solubility product of silver bromide is 5.0 xx 10^(-13). The quantity of potassium bromide (molar mass taken as 120 g mol^(-1)) to be added to 1 litre of 0.5 M solution of silver nitrate to start the precipitation of AgBr is

Solubility product of Mg(OH)at ordinary temperature is 1.96 xx10^(-11). pH of a saturated soln. of Mg(OH)_2 will be

Solubility product of PbCl_(2) at 298 K is 1.0xx10^(-6). At this temperature solubility of PbCl_(2) in moles per litre is :

Solubility product of M(OH), is 10^(-14). What should be the concentration of M^(2+) in 0.1 M solution of NH_(4)OH, if NH_(4)OH gets 10% ionised?

Solubility product of Mg(OH)_(2) at ordinary temp. is 1.96 xx 10^(-11). pH of a saturated of Mg(OH)_(2) will be

Solubility product of Ba(OH)_2 is 4xx10^-9 its solubbility in water is

Solubility product of BaCl_(2) is 4 xx 10^(-9). Its solubility in moles/litre would be

Solubility product of AgI at 25^@C is 4xx10^(-18)"mole"^2 //L^2 ​. The solubility of AgI in presence of 2xx10^(-4) M Kl solution at 25^@Cis approximately equal to : (in mol/L)

Solubility product of AgCl is 1 xx 10^(-6) at 298 K. Its solubility in mole "litre"^(-1) would be

Solution of following salt on heating with solid K_2Cr_2O_7 and conc. H_2SO_4 orange redvapours are evolved and resultant solution turn aquous NaOH solution yellow

Solubility product of AgI at 25^@C is 4xx10^(-18)"mole"^2 //L^2 . The solubility of AgI in presence of 2xx10^(-4) M Kl solution at 25^@Cis approximately equal to : (in mol/L)