52671 + Interview Questions in Chemistry in Class 11

1.	The law of multiple proportions was proposed by
Answer» Lavoisier Dalton Proust Gay Lussac. Answer :B

Discussion

2.	The law of multiple proportions is ilustrated by the two compounds
Answer» SODIUM chloride and sodium bromide Ordinary WATER and heavy water CAUSTIC soda and caustic potash SULPHUR dioxide and sulphur trioxide. Answer :D

Discussion

3.	The law of conservation of mass is given by study of which reaction ...........
Answer» COMBUSTION fusion endothermic vaporization Answer :A::B::C

Discussion

4.	The law of conservation of mass holds good for all of the following except
Answer» CHEMICAL REACTIONS NUCLEAR reactions Endothermic reaction Exothermic reactions Solution :For nuclear reactions, MASS can be CONVERTED to energy.

Discussion

5.	The lattice site in a pure crystal cannot be occupied by _____
Answer» MOLECULE ion electron atom Answer :C

Discussion

6.	The lattice site in a pur crystal cannot be occupied by ………
Answer» MOLECULE ion electron atom Answer :D

Discussion

7.	The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCI_((s)).
Answer» Solution :The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its IONS in gaseous STATE. For the reaction, `Na^(+) CI_((s))^(-) to Na_((g))^(+) + CI_((g))^(-) , Delta_("lattice") H^( Theta ) = + 788 "kJ mol"^(-1)` It is IMPOSSIBLE to determine lattice enthalpies directly by experiment, we use an INDIRECT method where we construct an enthalpy diagram called a Born-Haber cycle. Let us now calculate the lattice enthalpy of `Na^(+) CI_((s))^(-)` by following steps given below (I) `Na_((s))^(+) to Na_((g))`, Sublimation of sodium metal, `Delta_("sub") H^( Theta ) = 108.4 "kJ mol"^(-1)` (ii) `Na_((g)) to Na_((g))^(+) +e_((g))^(-)`, The ionization of sodium atoms, ionization enthalpy `Delta_(i) H^( Theta ) = 497 "kJ mol"^(-1)` (iii) `(1)/(2) CI_(2(g)) to CI_((g)) ,` The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy `(1)/(2) Delta_("bond") H^( Theta ) = 121 "kJ mol"^(-1)` (iv) `CI_((g)) e_((g))^(-) to CI_((g))^(-)`, electron gained by chlorine atoms. The electron gain enthalpy, `Delta_(eg) H^( Theta ) = -348.6 "kJ mol"^(-1)` (v) `Na_((g))^(+) + CI_((g))^(-) to Na^(+) CI_((s))^(-)` The sequence of steps is shown in given figure and is known as Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero. Applying Hess.s law, we get `Delta_("lattice") H^( Theta ) = 411.2 + 1084 + 121 + 496 - 3486` `Delta_("lattice") H^( Theta ) = + 788 "kJ"`

Discussion

8.	The lattice enthalpy and hydration enthalpy of four compounds are given below : {:("Compuounds","Lattice enthalpy ( in "kJ mol^(-1)")","Hydration enthalpy ( in kJ " mol^(-1)")"),(P,+780,-920),(Q,+1012,-812),(R,+828,-878),(S,+632,-600):} The pair of compunds which is soluble in water is
Answer» P and Q Q and R R and S P and R SOLUTION :A compound is soluble if hydration ENTHALPY `GT` lattice enthalpy

Discussion

9.	The lattice energy of solid NaCl is 180K. Call per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1K, Cal per mol. If the solvation energies of Na^(+) and Cl^(-) ions are in ratio 6:5, what is the enthalpy of hydration of sodium ion?
Answer» `-85.6K.Cal//mol` `-97.5K.Cal//mol` `82.6K.Cal//mol` `+100K.Cal//mol` ANSWER :B

Discussion

10.	The lattice energy of NaCl is 788 kJ mol^(-1) . This means that 788 kJ of energy is required
Answer» to separate one mole of SOLID NaCl into one mole of `Na_((g))` and one mole of `Cl_((g))` to infinite distance to separate one mole of solid NaCl into one mole of `Na_((g))^(+)` and one mole of `Cl_((g))^(-)` to infinite distance to CONVERT one mole of solid NaCl into one mole of GASEOUS NaCl to convert one mole of gaseous NaCl into one mole of solid NaCl. Solution :Lattice energy is the energy REQUIRED to separate 1 mole of an ionic compound into its ions in gaseous state. `NaCl_((s)) to Na_((g))^(+) + Cl_((g))^(-) , Delta_("lattice")H^(@) = 788" kJ " mol^(-1)`

Discussion

11.	Thelattice energy of NaCl is 180 kcal // mol . The dissociation of the solid in water in the form of ioins is endothermicto the extent of 1 kcal // mol.If the solvation energies ofNa^(+) and Cl^(-) ions are in theratesof6:5 , calculate the enthalpy of hydration of Na^(+) ions.
Answer» Solution :(i) `NaCl(s) rarr Na^(+) (g) + Cl^(-) (g), DeltaH =180kcal mol^(-1)` (ii) `Na^(+) Cl^(-)(s) overset( + aq)(rarr) Na^(+)(aq), Cl^(-)(aq) , DELTA H = 1 kcalmol^(-1)` (iii) `Na^(+)(g)+aq rarr Na^(+)(aq) , Delta H = 6 x ( `Aim) (iv) `Cl^(-)(g) +aq rarrCl^(-) (aq), DeltaH = 5X` Eqn. (ii) - Eqn. (iv) GIVES Enq. (i) `:. 180 = 1-5x ` or ` - 11X = 179 `or ` x= ( -179)/( 11)` `:. ` Enthalpy of hydration of`Na^(+)`ion `= 6X = ( -179)/( 11) xx 6 =-97.6 kcal mol^(-1)`

Discussion

12.	The Lassaigne's extract is boiled with conc. HNO_(3) while testing for halogens. By doing so it
Answer» decomposes `Na_(2)S` and NACN, if formed helps in the precipitation of AgCl increases the SOLUBILITY PRODUCT of AgCl increases the CONCENTRATION of `NO_(3)^(-)` ions Solution :Conc. `HNO_(3)` decomposes `Na_(2)S` and NaCN, if formed.

Discussion

13.	The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
Answer» `3d^(1)4s^(2)` `3d^(3)4s^(2)` `3d^(5)4s^(1)` `3d^(5)4s^(2)` Solution :Maximum oxidation no. of transition ELEMENT = `(n-1)d" "e^(-)+NS" "e^(-)` SO, (A) `3d^(1)4s^(2)=1+2=3` (B) `3d^(3)4s^(2)=3+2=5` ( C) `3d^(5)4s^(1)=5+1=6` (D) `3d^(5)4s^(2)=5+2=7`

Discussion

14.	The largest stable nucleus is
Answer» U-238 U-235 Pb-206 Bi-209 Solution :U-235

Discussion

15.	The largest oxidation number exhibited by an element depend on its outer electronic configuratoin withwhich of the following outer eletronic configuration the element will exhibit largest oxidation number ?
Answer» Solution :Hishest O.N of any transition ELEMENT =(n-1) d electrons + ns electrons therefore larger the number of electrons in the 3d orbitals higher is the maximum O.N (a) `4d^(1) 4S^(2)=3 , (b) 3d^(3) 4s^(2)=3+2=5 , (c ) 3d^(5)4s^(1)=5+1=6 and (d) 3d^(5)=5+2=7` THUS option (d) is correct

Discussion

16.	The large number of organic compounds is due to ............. of carbon.
Answer» SOLUTION :CATENATION

Discussion

17.	Lanthanide contraction occurs because
Answer» Zr and Hf have same RADIUS Zr and ZN have the same OXIDATION STATE Zr and Y have same radius Zr and Nb have similar oxidation state ANSWER :A

Discussion

18.	The lanthanide compound which is used as a most powerful liquid laser after dissolving in selenium oxychloride is:
Answer» CERIUM oxide Neodynium oxide Promethium sulphate Cerium sulphate Answer :B

Discussion

19.	The K_(w) of water at two temperature 25^(@)C and 50^(@)C are 1.08xx10^(-14), 5.474xx10^(-14) respectively. Assuming DeltaH of any reaction is neutralisation of a strong acid with strong base.
Answer» ANSWER :`-12.5xx10^(3)CAL;`

Discussion

20.	The lable is .volume. is stiched on one bottle of H_2O_2 then …….... is its strength.
Answer» `10%` `40%` `8%` `4.55%` SOLUTION :`M=V/11.2 =15/11.2`=1.3392 % w/v = `(Mxx34)/10 =(1.3392xx34)/10` =4.55% w/v

Discussion

21.	The K_(sp) values of Al(OH)_(3) and 1.8 xx 10^(-14) respectively at room temp. If a salt contains equal concentration of Al^(+3) and Zn^(+2) ions, the ion first precipitated by adding NH_(4)OH is ?
Answer» `Zn^(+2)` `Al^(+3)` both starts precipitation at same time no ion precipitates Solution :`K_(SP)` of `Al(OH)_(3) = 8.5 xx 10^(-23)` `K_(sp)` of `Zn(OH)_(2) = 1.8 xx 10^(-4)` `Al(OH)_(3) hArr Al^(+3) + 3OH^(-)` `Zn(OH)_(2) hArr Zn^(+2)+2OH^(-)` `K_(IP) = [Al^(+3)][OH^(-)]^3` `K_(IP) = [Zn^(+2)][OH^(-)]^2` If `[Al^(+3)]=[Zn^(+2)] = 10^(-1)` `[Zn^(+2)][OH^(-)]^(2) gt 1.8 xx 10^(-14)` `[Al^(+3)][OH^(-)]^(3) gt 8.5 xx 10^(-23)` `[OH^(-)]^(2) gt 18xx10^(-14),[OH^(-)]^(3)gt9.4xx10^(-8)` `[OH^(-)]gt4.2xx10^(-7)`

Discussion

22.	The K_(sp) of PbSO_4 is 1.44xx10^(-8) then calculate concentration of Pb^(2+).
Answer» Solution :`S=sqrtK_(sp)` `=SQRT(1.44xx10^(-8))=sqrt(144xx10^(-6))=12XX10^(-3)` `=1.2xx10^(-2) M`=0.012 M =`[Pb^(2+)]`

Discussion

23.	The K_(sp)of PbCO_(3) and MgCO_(3) are 1.5xx10^(-15) and 1xx10^(-15) respectively at 298 K.The concentration of Pb^(2+) ions in a saturated solution containing MgCO_(3) and PbCO_(3) is
Answer» `1.5 xx 10^(-8)M` `3xx10^(-8) M` `2xx10^(-8) M` `2.5xx10^(-8)M` Solution :When both `PbCO_(3) and MgCO_(3)` are present in the solution, suppose solubility of `PbCO_(3)` is x mol `L^(-1)` and that of `MgCO_(3)` is y mol `L^(-1)` . Then `{:(PbCO_(3),HARR,Pb^(2),+,CO_(3)^(2-)),(,,x,,x+y):}` `{:(MgCO_(3),hArr,"Mg"^(2),+,CO_(3)^(2-)),(,,x,,x+y):}` `(K_(sp)(PbCO_(3)))/(K_(sp)(MgCO_(3)))=(x(x+y))/(y(x+y))=(x)/(y)` `=(1.5xx10^(-15))/(1.0xx10^(-15))=1.5` Thus, x = 1.5 y Also `K_(sp)(PbCO_(3))=x(x+y)=1.5xx10^(-15)` `:. 1.5 y (1.5 y +y)= 1.5xx10^(-15)` or `3.75 y^(2)=1.5xx10^(-15)` or, `y=((1.5xx10^(-15))/(3.75))^(1//2) = 2 xx 10^(-8)` `x=1.5 y = 1.5 xx (2xx10^(-8))=3xx10^(-8)M`

Discussion

24.	The K_(sp) of PbCrO_(4) is 1.0xx10^(-16). Then the molar solubility of PbCrO_(4)is
Answer» `1.0xx10^(-6)` `1.0xx10^(-4)` `1.0xx10^(-16)` `1.0xx10^(-8)` Answer :B::C::D

Discussion

25.	The K_(sp) of Mg(OH)_2 is 1.0xx10^(-12) . At which pH the 0.01 M Mg(OH)_2 begins to precipitate ?Calculate solubility.
Answer» SOLUTION :SOLUBILITY is `6.29xx10^(-5) "MOL L"^(-1)` , pH= 10.1 So, the precipitation occurs when the PHIS less than 10.1

Discussion

26.	The K_(sp) of Mg(OH)_2 is 1.2 xx 10^(-11) calculate its solubility in pure water.
Answer» SOLUTION :`1.442xx10^(-4)`

Discussion

27.	The K_(sp)of Ca(OH)_(2)is 4.42xx10^(-5)at 25^(@)C. 500 mL of saturated solution of Ca(OH)_(2) is mixed with equal volume of 0.4 M NaOH . How much Ca(OH)_(2)in mg will be precipitated ? Also calculate percentage precipitation.
Answer» Solution :Suppose the solubility of `Ca(OH)_(2)` in saturated solution is S mol `L^(-1)`. `{:(Ca(OH)_(2),hArr,Ca^(2+),+,2OH^(-),),(,,S,,2S,):}` `K_(sp) ` of `Ca(OH)_(2) hArr [Ca^(2+) ] [ OH^(-)]^(2)` `4.42xx10^(-5) = S xx (2S)^(2) = 4S^(3)` or`S=((4.42xx10^(-5))/(4))^(1//3) = 0.223 ` mol `L^(-1)` After mixing the two solutions, total volume becomes = 500 + 500 = 100 mL Now, concentration will be `[Ca^(2+) ] = (0.0223)/(1000) xx 500 = 0.01115 ` mol `L^(-1)` `{:([OH^(-)]=(0.0223xx2xx500)/(1000)+(0.4xx500)/(1000)=0.2223 " mol "L^(-1)),(""["From"Ca(OH)_(2)]"" ["From" NAOH]):}` On ADDING NaOH solution to saturated solution of `Ca(OH)_(2)` some `Ca(OH)_(2)` will be precipitated (DUE to common ion effect). Now, as `Ca^(2+)` ion left are still present in the left out saturated solution, `[ Ca^(2+) ] _("left") [OH^(-)]^(2) = K_(sp)` `:. [Ca^(2+) ]_("left") = (K_(sp))/([OH^(-)]^(2))=(4.42xx10^(-5))/((0.2223)^(2))=8.94xx10^(-4)` mol `L^(-1)` Moles of `Ca(OH)_(2)` precipitated = Moles of `Ca^(2+)` ions precipitated `= [Ca^(2+) ] _("initial")-[Ca^(2+)]_("left")` `=0.01115-8.94xx10^(-4) = 111.5xx10^(-4) - 8.94xx10^(-4) = 102.56xx10^(-4)` moles `=102.56xx10^(-4) xx 74 g = 7589.44 xx 10^(-4) g = 758.944 ` mg Initial `[Ca(OH)_(2)] = 0.01115 ` mol `L^(-1) = 111.5xx10^(-4) ` mol `L^(-1)` `Ca(OH)_(2)` precipitated `= 102.56xx10^(-4) ` mol `L^(-1)` `:. ` % `Ca(OH)_(2)` precipitated `= (Ca(OH)_(2)"precipitated")/("Initial conc. ")xx100=(102.56xx10^(-4))/(111.5xx10^(-4))xx100=91.98% = 92 %` Note. The above example illustrates the method of calculation of amount precipitated and amount left after precipitation when to a saturated solution of a sparingly soluble salt present in excess, some solution containing a common ion is added .

Discussion

28.	The K_(sp) of CaF_2 is 1.7 xx 10^(-10). Then what is the volume in mililitre of saturated solution of 10 miligram CaF_2 ? (Molecular mass of Ca (40), F(19)).
Answer» SOLUTION :367.45 ML

Discussion

29.	The K_(sp) of BaSO_4 is 1.1 xx 10^(-10), Will a precipitate form when equal volume of 2 xx 10^(-4) BaCl_2 and 5.0 xx 10^(-3) M H_2SO_4 solution are mixed ? Explain by calculation.
Answer» SOLUTION :`Q_(SP)` of `BaSO_4 =2.5xx10^(-7)`So no PPt. will be FORMED

Discussion

30.	The K_(sp) of BaSO_4 is 1.0xx10^(-9) .Then What is the solubility of it in 0.1 M MgSO_4 and 0.01 M BaCl_2 ?
Answer» Solution :`1.0xx10^(-8)` and `1.0xx10^(-7) "MOL L"^(-1)` RESPECTIVELY.

Discussion

31.	The K_(sp) of AgCl is 1.0xx10^(-10) calculate solubility of AgCl in 0.2 M AgNO_3
Answer» SOLUTION :`5XX10^(-10)` M

Discussion

32.	The K_(sp) of AgBr is 1 xx 10^(-10) at 25^@C temperature then what is the value of [Ag^1]In saturated solution ?
Answer» `10^(-4)` `10^(-5)` `10^(-6)` `10^(-10)` ANSWER :B

Discussion

33.	The K_(sp) of Ag_2CrO_4, AgCl, AgBr and Agl are respectively, 1.1 xx 10^(-12), 1.8 xx 10^(-10), 5.0 xx 10^(-13), 8.3 xx 10^(-17), Which one of the following salts will precipitate last if AgNO_3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal andNa_2CrO_4 ?
Answer» AgI AgCl AgBr `Ag_2CrO_4` Solution :`Ag_2CrO_4 to K_(sp)=[Ag^+]^2 [CrO_4^(-2)]` `rArr [Ag^+]=SQRT(K_(sp)/[CrO_4^(-2)])=sqrt((1.1xx10^(-12))/[CrO_4^(-2)])`=MAXIMUM

Discussion

34.	The K_(sp) of Ag_(2)CrO_(4), AgCl, A gBr and AgIare respectively 1.1xx10^(-12), 1.8xx10^(-6), 5.0xx10^(-13) and 8.3xx10^(-17). Which one of the following salts will precipitate last if AgNO_(3)solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na_(2)CrO_(4)?
Answer» AgBr<BR>`Ag_(2)CrO_(4)` AgI AgCl Solution :`Ag_(2)CrO_(4)hArr2Ag+CrO_(4)^(2-), K_(sp)=(2s)^(2)xxs=4s^(3), s=((K_(sp))/(4))^(1//3)=((1.1xx10^(-12))/(4))^(1//3)` `=0.65xx10^(-4)` `AgCl HARR Ag^(+)+Cl^(-),K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(1.8xx10^(-10))=1.34xx10^(-5)` `AgBr hArr Ag^(+)+Br^(-), K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(5.0xx10^(-13))=0.71xx10^(-6)` `AgI hArr Ag^(+)+I, K_(sp)=sxxs=s^(2)`, `s=sqrt(K_(sp))=sqrt(8.3xx10^(-17))=0.9xx10^(-8)` As solubility of `Ag_(2)CrO_(4)` is highest, it will be PRECIPITATED last of all.

Discussion

35.	The K_(sp) of Ag_(2) CrO_(4) is 1.1xx10^(-12)at 298 K. Thesolubility (in mol/L) of Ag_(2)CrO_(4) in 0.1 MAgNO_(3) solution is
Answer» `1.1xx10^(-11)` `1.1xx10^(-10)` `1.1xx10^(-12)` `1.1xx10^(-9)` Solution :`{:(Ag_(2)CrO_(4),hArr,2"Ag"^(+),+,CrO_(4)^(-)),(s,,2s+0.1,,s),(,,(0.1 "from AgNO"_(3)),,),(,,~= 0.1 M,,),(,,"(as sis negligible incomparison to 0.1)",,),(,,,,):}` `K_(SP)=[Ag^(+)]^(2)[CrO_(4)^(2-)]` `1.1xx10^(-12)= (0.1)^(2)XXS or s=1.1xx10^(-10)M`

Discussion

36.	The K_(p) " value for reaction " H_(2) + I_(2) hArr 2 HI " at "460^(@)C " is 49. If intial pressure of " H_(2) and I_(2)" is "0*5 atm respectively , determine the partial pressure of each gas at equilibrium.
Answer» <P> Solution :`{:(,H_(2),+,I_(2),hArr,2 HI),("Intial ",05,,05,,0),("At. EQM.",05-x,,05-x,,2" "x):}` `K_(p) = (2x)^(2)/(05-x)^(2) = 49 or (2 x)/(05 - x)=7 or 2 x = 35- 7 x` or ` 9 x = 35 or x= (35)/9=039` `:. " Pressure of " H_(2) and I_(2) " at eqm."= 05 - 039= 011 " atm"` Pressure of HI at eqm. `= 2 xx 039 = 0*78 " atm "`

Discussion

37.	The K_(P) values for three reactions are 10^(-5), 20 and 300 then what will be the correct order of the percentage composition of the products.
Answer» Solution :Since `K_(P)` order is `10^(-5)lt20lt300` so the PERCENTAGE COMPOSITION of products will be greatest for `K_(P)=300.`

Discussion

38.	The K_p/K_c is equal to which of the following in given reaction ? Reaction CO_((g)) + Cl_(2(g)) hArr COCl_(2(g))
Answer» `1/(RT)` RT `sqrt(RT)` `(RT)^2` Solution :`{:(CO_((g))+Cl_(2(g)) hArr, COCl_(2(g))),(ubrace(1"+"1)_("2 mol"),"1 mol"):}` `THEREFORE Deltan_((g)) =(1-2)=-1` `K_p=K_c(RT)^(DELTAN)` `therefore K_p/K_c=(RT)^(Deltan)` `=(RT)^(-1) =1/(RT)`

Discussion

39.	The K_(p) for the reaction, N_(2)O_(4)(g) hArrNO_(2)(g) " is " 640 " mm at " 775 K. Calculate the percentage dissociation of N_(2)O_(4)at equilibrium pressure of 160mm. At what pressure the dissociation will be 50%.
Answer» <P> ANSWER :`70.7%, P = 480 ` MM

Discussion

40.	The K_(p) " for the reaction," N_(2) O_(4) hArr 2NO_(2)is 640 mm at 775 K. Calculate the percentage dissociation of N_(2)O_(4) at equilibrium pressure of 160 mm. At what pressure the dissociation will be 50% ?
Answer» Solution : Supposeintially `N_(2)O_(4)` taken = 1 mole and its degree of dissociation = `ALPHA` ` {:(,N_(2)O_(4),hArr,2NO_(2),),("Intial ",1 "mole",,,),("At eqm.",1-alpha,2alpha,"Total " =1-alpha + 2 alpha = 1+alpha,):}` If P is the total pressure at EQUILIBRIUM, then ` p_(N_(2)O_(4))= (1-alpha)/(1+alpha) xxP and p_(NO_(2)) = (2 alpha)/(1+alpha) xxP ` Now, ` K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= ((2 alpha)/(1+alpha).P)^(2)/((1-alpha)/(1+alpha) P)= (4 a^(2))/((1+alpha)(1-alpha)) = (4a^(2))/(1-alpha^(2)) xx P` Putting` K_(p) = 640 " mm (Given ) and equilibrium pressure ,P+ 160 mm, we get "` ` 640 = (4a^(2))/(1-alpha)^(2) xx 160 or (alpha^(2))/(1-alpha)^(2) = or alpha^(2) = 1 - alpha^(2) or 2 alpha^(2)= 1 or alpha^(2) = 05 or alpha = 0* 707 = 70* 7 % ` For dissociation to be ` 50 % , alpha = 050 , K_(p)= 640` mm ( constant ) ` :. 640 = (4 ( 05)^(2))/(1-(0*5)^(2)) xx P or 640 = 1/(1-1/4) P= 4/3 P or P =480 "mm"` .

Discussion

41.	The Kolbe's electrolysis proceeds via
Answer» NUCLEOPHILIC substitution MECHANISM ELECTROPHILIC addition mechanism Free radical mechanism Electrophilic sustitution mechanism Answer :C

Discussion

42.	The kinetic theory of gases assumes all of the following except:
Answer» Gases are COMPOSED of particles in RANDOM ceaseless motion. The sizes of gas particles are negligible COMPARED to the size of the container. Gas particles do not attract or rapel each other. When gas particles COLLIDE, KINETIC energy is lost. Answer :d

Discussion

43.	The kinetic gas equation is applicable when the gas is present in a
Answer» CUBIC vessel Spherical vessel Vessel of any shape Cylindrical vessel Solution :KINETIC THEORY is APPLICABLE to any vessel.

Discussion

44.	The kinetic enerrgy of 4 mole sof nitrogen gas at 127^(@)C is (R=2" cal "mol^(-1)K^(-1))
Answer» 4400 cal 3200 cal 4800 cal 1524 cal Answer :C

Discussion

45.	The kinetic energy of two moles of CO_(3) at a certain temperature is 1800cal. The temperature of the gas is :
Answer» 300K 150K 200K 400K Answer :a

Discussion

46.	The kinetic energy of the molecules in a sample of H_(2)O in its stable state at -10^(@)C and 1 atm is doubled. What are the initial and final phases?
Answer» SOLID `RARR` liquid liquid `rarr` GAS solid `rarr` gas solid `rarr` solid Answer :C

Discussion

47.	the kineticenergyof protonwaveis 500 eV.Calculatedebrogliewavelength(mass ofproton =1.67 xx 10^(-27) kg
Answer» SOLUTION :`4.953 XX 10^(10) m`

Discussion

48.	The kinetic energy of n moles of an ideal gas is given by the expression
Answer» `3/2 RT ` `3/2 N RT` `2/3 RT ` `2/3 NRT ` ANSWER :B

Discussion

49.	The kinetic energy of 'N' molecules of H_(2)is 3J at -73°C. The kinetic energy of the same sample of H_(2) at 127°C is
Answer» `12 J` `6 J` `9 J` `3 J` Answer :B

Discussion

50.	An electron and a proton are detected in a cosmic ray experiment , the first with kinetic energy 10 keV and the second with 100 keV.Which is faster , the electron or the proton ? Obtain the ratio of their speeds . (electron mass = 9.11 xx10^(-31) kg , proton mass = 1.67 xx10^(-27)kg , 1 eV = 1.60 xx10^(-19)J)
Answer» SOLUTION :`1.7 xx10^(9) HZ`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The law of multiple proportions was proposed by

The law of multiple proportions is ilustrated by the two compounds

The law of conservation of mass is given by study of which reaction ...........

The law of conservation of mass holds good for all of the following except

The lattice site in a pure crystal cannot be occupied by _____

The lattice site in a pur crystal cannot be occupied by ………

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCI_((s)).

The lattice enthalpy and hydration enthalpy of four compounds are given below : {:("Compuounds","Lattice enthalpy ( in "kJ mol^(-1)")","Hydration enthalpy ( in kJ " mol^(-1)")"),(P,+780,-920),(Q,+1012,-812),(R,+828,-878),(S,+632,-600):} The pair of compunds which is soluble in water is

The lattice energy of solid NaCl is 180K. Call per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1K, Cal per mol. If the solvation energies of Na^(+) and Cl^(-) ions are in ratio 6:5, what is the enthalpy of hydration of sodium ion?

The lattice energy of NaCl is 788 kJ mol^(-1) . This means that 788 kJ of energy is required

Thelattice energy of NaCl is 180 kcal // mol . The dissociation of the solid in water in the form of ioins is endothermicto the extent of 1 kcal // mol.If the solvation energies ofNa^(+) and Cl^(-) ions are in theratesof6:5 , calculate the enthalpy of hydration of Na^(+) ions.

The Lassaigne's extract is boiled with conc. HNO_(3) while testing for halogens. By doing so it

The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?

The largest stable nucleus is

The largest oxidation number exhibited by an element depend on its outer electronic configuratoin withwhich of the following outer eletronic configuration the element will exhibit largest oxidation number ?

The large number of organic compounds is due to ............. of carbon.

Lanthanide contraction occurs because

The lanthanide compound which is used as a most powerful liquid laser after dissolving in selenium oxychloride is:

The K_(w) of water at two temperature 25^(@)C and 50^(@)C are 1.08xx10^(-14), 5.474xx10^(-14) respectively. Assuming DeltaH of any reaction is neutralisation of a strong acid with strong base.

The lable is .volume. is stiched on one bottle of H_2O_2 then …….... is its strength.

The K_(sp) values of Al(OH)_(3) and 1.8 xx 10^(-14) respectively at room temp. If a salt contains equal concentration of Al^(+3) and Zn^(+2) ions, the ion first precipitated by adding NH_(4)OH is ?

The K_(sp) of PbSO_4 is 1.44xx10^(-8) then calculate concentration of Pb^(2+).

The K_(sp)of PbCO_(3) and MgCO_(3) are 1.5xx10^(-15) and 1xx10^(-15) respectively at 298 K.The concentration of Pb^(2+) ions in a saturated solution containing MgCO_(3) and PbCO_(3) is

The K_(sp) of PbCrO_(4) is 1.0xx10^(-16). Then the molar solubility of PbCrO_(4)is

The K_(sp) of Mg(OH)_2 is 1.0xx10^(-12) . At which pH the 0.01 M Mg(OH)_2 begins to precipitate ?Calculate solubility.

The K_(sp) of Mg(OH)_2 is 1.2 xx 10^(-11) calculate its solubility in pure water.

The K_(sp)of Ca(OH)_(2)is 4.42xx10^(-5)at 25^(@)C. 500 mL of saturated solution of Ca(OH)_(2) is mixed with equal volume of 0.4 M NaOH . How much Ca(OH)_(2)in mg will be precipitated ? Also calculate percentage precipitation.

The K_(sp) of CaF_2 is 1.7 xx 10^(-10). Then what is the volume in mililitre of saturated solution of 10 miligram CaF_2 ? (Molecular mass of Ca (40), F(19)).

The K_(sp) of BaSO_4 is 1.1 xx 10^(-10), Will a precipitate form when equal volume of 2 xx 10^(-4) BaCl_2 and 5.0 xx 10^(-3) M H_2SO_4 solution are mixed ? Explain by calculation.

The K_(sp) of BaSO_4 is 1.0xx10^(-9) .Then What is the solubility of it in 0.1 M MgSO_4 and 0.01 M BaCl_2 ?

The K_(sp) of AgCl is 1.0xx10^(-10) calculate solubility of AgCl in 0.2 M AgNO_3

The K_(sp) of AgBr is 1 xx 10^(-10) at 25^@C temperature then what is the value of [Ag^1]In saturated solution ?

The K_(sp) of Ag_2CrO_4, AgCl, AgBr and Agl are respectively, 1.1 xx 10^(-12), 1.8 xx 10^(-10), 5.0 xx 10^(-13), 8.3 xx 10^(-17), Which one of the following salts will precipitate last if AgNO_3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal andNa_2CrO_4 ?

The K_(sp) of Ag_(2)CrO_(4), AgCl, A gBr and AgIare respectively 1.1xx10^(-12), 1.8xx10^(-6), 5.0xx10^(-13) and 8.3xx10^(-17). Which one of the following salts will precipitate last if AgNO_(3)solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na_(2)CrO_(4)?

The K_(sp) of Ag_(2) CrO_(4) is 1.1xx10^(-12)at 298 K. Thesolubility (in mol/L) of Ag_(2)CrO_(4) in 0.1 MAgNO_(3) solution is

The K_(p) " value for reaction " H_(2) + I_(2) hArr 2 HI " at "460^(@)C " is 49. If intial pressure of " H_(2) and I_(2)" is "0*5 atm respectively , determine the partial pressure of each gas at equilibrium.

The K_(P) values for three reactions are 10^(-5), 20 and 300 then what will be the correct order of the percentage composition of the products.

The K_p/K_c is equal to which of the following in given reaction ? Reaction CO_((g)) + Cl_(2(g)) hArr COCl_(2(g))

The K_(p) for the reaction, N_(2)O_(4)(g) hArrNO_(2)(g) " is " 640 " mm at " 775 K. Calculate the percentage dissociation of N_(2)O_(4)at equilibrium pressure of 160mm. At what pressure the dissociation will be 50%.

The K_(p) " for the reaction," N_(2) O_(4) hArr 2NO_(2)is 640 mm at 775 K. Calculate the percentage dissociation of N_(2)O_(4) at equilibrium pressure of 160 mm. At what pressure the dissociation will be 50% ?

The Kolbe's electrolysis proceeds via

The kinetic theory of gases assumes all of the following except:

The kinetic gas equation is applicable when the gas is present in a

The kinetic enerrgy of 4 mole sof nitrogen gas at 127^(@)C is (R=2" cal "mol^(-1)K^(-1))

The kinetic energy of two moles of CO_(3) at a certain temperature is 1800cal. The temperature of the gas is :

The kinetic energy of the molecules in a sample of H_(2)O in its stable state at -10^(@)C and 1 atm is doubled. What are the initial and final phases?

the kineticenergyof protonwaveis 500 eV.Calculatedebrogliewavelength(mass ofproton =1.67 xx 10^(-27) kg

The kinetic energy of n moles of an ideal gas is given by the expression

The kinetic energy of 'N' molecules of H_(2)is 3J at -73°C. The kinetic energy of the same sample of H_(2) at 127°C is