52671 + Interview Questions in Chemistry in Class 11

1.	Following three gaseous equilibrium reactions are occurring at 27^(@)C. (A) 2CO+O_(2)hArr2CO_(2) (B) PCl_(5)hArrPCl_(3)+Cl_(2) (C) 2HIhArrH_(2)+I_(2) The correct order of K_(p)//K_(c) for the following reaction is
Answer» <P>`AltBltC` `CltBltA` `AltCltB` `BltAltC` SOLUTION :(A) `Deltan=2-3=-1,K_(p)=K_(C)(RT)^(-1),K_(p)//K_(c)=(1)/(RT)` (B) `Deltan=2-1=1,K_(p)=K_(c)(RT),K_(p)//K_(c)=RT` (C) `Deltan=2-2=0,K_(p)=K_(c)," "K_(p)//K_(c)=0`

Discussion

2.	Following two eqilibrium is simultaneously establised in a containerPCl_(s(g)) hArr PCl_((3)(g)) + Cl_((2)(g)), CO_((g)) + Cl_(2(g)) hArr COCl_((2)(g)). If some Ni_((g)) is introduced in the container forming Ni(CO)_(4)then at new equilibrium
Answer» `PCI_3`. concentration will increase `PCl_3` concentration will decrease `CI_2` concentration will remain same `CO` concentration will remain same Solution :CO is removed through COMPLEX formation `:. Cl_(2)` is formed through 2nd REACTION `:. 1^(ST)` reaction SHIFTS in the backward direction

Discussion

3.	Following statements regarding the periodic trends of chemical reactivity of alkali metals and halogens are given. Which of these statements give the correct picture?
Answer» In both, the alkali metals and the HALOGENS, the chemical reactivity DECREASES with INCREASE in atomic number down the group Chemical reactivity increases with increase in atomic number in both alkali metals and halogens In alkali metals the reactivity increases but in the halogens, it decreases with increase in atomic number down the group The reactivity decreases in the alkali metals but increases in the halgons with increase in atomic number down the group Solution :Reactivity of alkali metals`LiltNaltKltRbltCs` Reactivity of halogens :`FgtCIgtBrgtI`

Discussion

4.	Followingstatementregardingthe periodictrends of chemicalreactivity of thealkalimetalsand thehalogensare given. Whichof thesestatement gives the correctpicture ?
Answer» Chemicalreactivityincreaseswith INCREASES in atomicnumberdown thegroup in BOTHTHE alkalimetalsand haloges In alkalimetals the reactivity increasesbut inthe HALOGENS itdecreaseswith INCREASESIN atomicnumberdown thegroup . Thereactivitydecreases in thealkalimetalsbut increasesin the halogens withincreasesin atomicnumber . Inboth thealkalimetalsand thehalogensthe chemicalreactivitydecreaseswithincreasesin atomicnumber down thegroup. Answer :B

Discussion

5.	Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : (1) 60 mL (M)/(10) HCl + 40 mL (M)/(10) NaOH 2. 55 mL (M)/(10) HCl + 45 mL (M)/(10) NaOH 75 mL (M)/(5) HCl + 25 mL (M)/(5) NaOH (4) 100 mL (M)/(10) HCl + 100 mL(M)/(10) NaOH pH of which one of them will be equal to 1 ?
Answer» `(2)` `(1)` `(4)` `(3)` Solution :(1) 4 mL `(M)/(10)` NAOH will neutralize 40 mL of `(M)/(10)` HCl . `(M)/(10)` HCl left unneutralised = 20 mL. TOTAL volume = 100 mL . Dilution = 5 TIMES. In finalsolution, `[HCl]=(M)/(50). pH != 1`. (2) `(M)/(10)`HCl left unneutralised = 10 mL . Total volume = 100 mL . Dilution = 10 times. In final solution, `[HCl]=(M)/(100)=10^(-2)M` `:. pH = 2`. `(3) (M)/(5) ` HCl left unneutralized = 50 mL . Total volume = 100 mL . Dilution = 2 times. In final solution, `[HCl]=(M)/(10)=10^(-1)M`. Hence, pH=1. (4) There is exact NEUTRALISATION. Hence, pH = 7.

Discussion

6.	Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:(a) 60 ml, (M)/(10) HCl + 40 ml (M)/(10) NaOH (b) 55ml, (M)/(10) HCl +45 ml (M)/(10) NaOH (c) 75ml , M/5 HCl + 25ml M/5 NaOH (d) 100 ml, M/5 HCl + 100 ml (M)/(10) NaOH pH of which one of them will be equal to 1?
Answer» c b d a Solution :`75ml M/5 HCl + 25 ml M/5 NAOH` ` THEREFORE (75-25) = 50ml M/5 + HCl` REMAIN Total volume=100ml x m `1/5 M xx 50 ml = 100 ml xx xm` ` therefore x= 1/5 xx 50 xx 1/100 = 1/10 M HCl =[H^+]` ` therefore PH = -log (1/10) = 1.0`

Discussion

7.	Following solutions were prepared by mixing different volume of NaOH and HC1 of different concetrations: (1) 60 mL (M)/(10) HC1 + 40 mL (M)/(10)NaOH (2) 55 mL (M)/(10) HC1 + 45 mL (M)/(10) NaOH (3) 75 mL (M)/(5) HC1 + 25 mL (M)/(5) NaOH 100 mL (M)/(10) HC1 + 100 mL (M)/(10) NaOH pH of which one of them will be equal to 1 ?
Answer» 1 2 4 3 Solution :Where `N_(1) V_(1) gt N_(2)V_(2)` Acid is left at the end of REACTION and solutuion will be acidic `N_(FINAL solution) = [H^(+)] = (N_(1)V_(1) - N_(2)V_(2))/(V_(1) + V_(2))` , as basicity and acidity of acid and base is 1 so M = N (1) `[H^(+)] = (6-4)/(100) = (2)/(100)``pH /+ 1` (2) `[H^(+)] = (5.5 - 4.5)/(100) = (1)/(100) = 10^(-2)``pH = 2` (3)`[H^(+)] = (15 - 5)/(100) = (10)/(100) = 10^(-1)``pH = 1` (4) `[H^(+)] = (10 - 10)/(200) = 0`Neutral

Discussion

8.	Following solutions were prepared by mixing different volumes of NaOH and HCI of different concentrations: (i) 60 mL M/10 HCl + 40 mL M/10 NaOH (ii) 55 mL M/10 HCl + 45 mL M/10 NaOH, (iii) 75 mL M/5 HCl + 25 mL M/5 NaOH (iv) 100 mL M/10 HCl + 100 mL M/10 NaOH pH of which one of them will be equal to 1?
Answer» ii i iv iii Answer :D

Discussion

9.	Following results were observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant. {:(lamda (nm) ,500,450,400),(v xx 10^(-6) (ms^(-1)),2.55,4.35,5.20):}
Answer» Solution :Suppose threshold wavelength `= lamda_(0) nm = lamda_(0) xx 10^(-9)m` Then `h (v - v_(0)) = (1)/(2) mv^(2) or hc ((1)/(lamda) - (1)/(lamda_(0))) = (1)/(2) mv^(2)` Substituting the GIVEN results of the three EXPERIMENTS, we get `(hc)/(10^(-9)) ((1)/(500) - (1)/(lamda_(0))) = (1)/(2) m (2.55 xx 10^(6))^(2)`....(i) `(hc)/(10^(-9)) ((1)/(450) - (1)/(lamda_(0))) = (1)/(2) m (4.35 xx 10^(6))^(2)`...(II) `(hc)/(10^(-9)) ((1)/(400) - (1)/(lamda_(0))) = (1)/(2) m (5.20 xx 10^(6))^(2)` Dividing eqn. by eqn. (i), we get `(lamda_(0) - 450)/(450 lamda_(0)) xx (500 lamda_(0))/(lamda_(0) - 500) =((4.35)/(2.55))^(2)` or `(lamda_(0) - 450)/(lamda_(0) - 500) xx (450)/(500) = ((4.35)/(2.55))^(2) = 2.619 or lamda_(0) - 450 = 2.619 lamda_(0) - 1309.5` or `1.619 lamda_(0) = 859.5 :. lamda_(0) = 531nm` Substituting this value in eqn. (iii), we get `(h xx (3 xx 10^(8)))/(10^(-9)) ((1)/(400) - (1)/(531)) = (1)/(2) (9.11 xx 10^(-31)) (5.20 xx 10^(6))^(2) or h = 6.66 xx 10^(-34) Js`

Discussion

10.	Followingresultsare observedwhensodiummetal isirradiated withdifferent wavelengthCalculate( a)thresholdwavelengthand( b)planks constant
Answer» Solution : Takewavelength`(lambda_(N))`in nmabd`v_(n) ` in`mn^(_1)` of1,2,3 observation Thresholdwavelength`=(lambda_(0)nm =10 ^(9) lambda _(0) M` `= (hc )/( lambda_(0)xx 10^(9))` = workfunction Emitted energy(E ) = `(hc )/(10^(9)xx lambda _(n))` `lambda_(0)`= 530.89 nm =531thresholdwavelength Calculationof h planksconstant Putvalueof `lambda `and `lambda_(0)` in observation: `(hxx 3 xx 10^(8) (531 - 500))/(10^(9)(531) (500)) = (1)/(2) (9.1 xx 10^(31) kg) (2.55 xx 10^(5))^(2)` `=84464 xx 10^(-36)` `8.4464 xx 10^(34)` plankconstant

Discussion

11.	Following redox in balancing state has how many P, H, O and electric charge on left side respectively. P_(4)+OH^(1-)toPH_(3)+H_(2)PO_(2)^(-)
Answer» <P>`4,1,1,-1` `2,9,6,0` `4,6,9,-3` `4,9,6,-3` SOLUTION :BALANCE reaction : `P_(4)+3OH^(-)+3H_(2)OtoPH_(3)+3H_(2)PO_(2)^(-)`

Discussion

12.	Following reaction in equilibrium at 25^(@)C: 2NO(g,1 xx 10^(-5) "atm")+Cl_2(g,1 xx 10^(-2)"atm") harr 2NOCl(g, 1 xx 10^(-2) "atm") DeltaG^@ is
Answer» `-45.65 KJ` `-28.53 KJ` `-22.82 KJ` `-57.06 KJ` Solution :`K_(P)=(P_(NOCL)^(2))/(P_(NO)^(2) xx P_(Cl_(2)))=((10^(-2))^(2))/((10^(-5))^(2) xx 10^(-2))=10^(8)` `DeltaG^(@)= -RT` 1 N k = -45.65 kJ

Discussion

13.	Following reaction describes the rusting of iron 4Fe +3O_(2) rarr 4Fe^(3+) +6O^(2-) .Which one of the following statement is incorrect .
Answer» This is an example of a redox reaction METALLIC iron is reduced to `FE^(3+)` `Fe^(3+)` is an oxidising agent Metallic iron is a REDUCING agent Solution :In RESTING of Fe, Fe gets oxidized and oxygen undergoes reduction .

Discussion

14.	Following reaction describes the rusting of iron 4Fe+3O_(2) rarr 4Fe^(3+)+6O^(2-) Which one of the following statements is incorrect?
Answer» This is an example of a redox reaction. Metallic IRON is reduced to `Fe^(3+)` `Fe^(3+)` is an oxidizing agent. Metallic iron is a reducing agent. Answer :B

Discussion

15.	Following reaction describes the rusting of 4Fe+3O_(2) to 4Fe^(3+) +6O^(2-). Which one of the following statement is incorrect
Answer» This is an EXAMPLE of a redox reaction Metallic iron is REDUCED to `FE^(3+)` `Fe^(3+)` is an oxidising agent Metallic iron is a REDUCING agent Answer :B

Discussion

16.	following is the fact about the newly discover superconductor fo C_(60) (fullerence).the alkali mental fulleride superconductor M_(3)C_(60)has a cubic closesr - packed (face0centered cubic) arrangement of nearly spherical C_(2-)^(60) anions with M^(+) cations in the holes between the larger C_(3-)^(60) ions, the holes are of two types - octahedral hoels , which are surrounded octachedralby six C_(3-)^(60) ions. tetarahedral holes , which are surrounded tetrahedrally by for C_(3-) ^(60) ions . how many C_(3-)^(60) ions , octahedral holes , and tetrachedral holesare prsent pre unit cell
Answer» 5,4,4 3,8,4 4,4,8 2,1,2 Answer :C

Discussion

17.	following is the fact about the newly discover superconductor fo C_(60) (fullerence).the alkali mental fulleride superconductor M_(3)C_(60)has a cubic closesr - packed (face0centered cubic) arrangement of nearly spherical C_(2-)^(60) anions with M^(+) cations in the holes between the larger C_(3-)^(60) ions, the holes are of two types - octahedral hoels , which are surrounded octachedralby six C_(3-)^(60) ions. tetarahedral holes , which are surrounded tetrahedrally by for C_(3-) ^(60) ions . the ionic radii of Na^(+), K^(+) and Rb^(+) are 97 , 133 and 147pm , respectivley . which of these ions will fit into the octanhedrral holes ? (radius of C_(3-)^(60) is about 350 pm )
Answer» `NA^(+)` `K^(+)` `RB^(+)` all of these Answer :C

Discussion

18.	Following frightening diagrams show identical cubes such that edge of cube 2 lies exactly in the middle of one of the faces of Cube 1 and Cube 4 has a corner at the body centre of the cube 3 . Find the contribution (in fraction) of the spheres shown to each of the cube 1 ,2 , 3 and 4 respectively is
Answer» `(1)/(2) , (1)/(4) , 1 , (1)/(8)` `(1)/(2) , (1)/(4) , (1)/(8) , 1` `(1)/(8) , (1)/(4) , 1 , (1)/(2)` `(1)/(8) , (1)/(4), (1)/(2) , 1` ANSWER :A

Discussion

19.	Following graph between DeltaG and reaction progress in for/can be :
Answer» `S_N1` reaction `E_1` reaction Aromatic electrophilic substitution Electrophilic addition reaction Answer :A::B::C::D

Discussion

20.	Following figure shows an FCC unit cell with atoms of radius r market 1(corner), 2(face centre), 3(face centre). A quadrilateral is also shown by joining the centers of 4 face centered atoms. Find : (i) The distances between atoms 1& 2, 2 &3, and 1 &3 (ii) The shape and dimensions of the quadrilateral
Answer» Answer :(i) 2R, 2r, `2SQRT3 R`; (II) square, SIDE = 2r

Discussion

21.	Following diagrams show identical cubes such that edge of cube 2 lies exactly in the middle of one of the faces of Cube1 and Cube 4 has a corner at the body centre of the Cube3. Find the contributions (in fraction) of the spheres shown to each of the cubes
Answer» Answer :CUBE1 - 1/2; Cube 2 - 1/4; CUBE3 - 1; Cube4 - 1/8

Discussion

22.	Following equilibrium is established to decomposing of Ammonium carbonate NH_4COONH_2 in closed vessel at 700 K temperature. NH_4COONH_(2(s)) hArr 2NH_(3(g)) + CO_2 At initial if there is vaccum and at equilibrium total pressure is P bar than derive the value of K_p with respect to P.
Answer» Solution :At EQUILIBRIUM the RATIO of `NH_(3(g))` and `CO_(2(g))` is 2:1. If total pressure = P, So, `p_(NH_3)=2/3P` and `p_(CO_2)=1/3 P` `{:("Reaction:",NH_4COONH_(2(s)) hArr 2NH_(3(g)) , + CO_(2(g))),("Pressure at equilibrium :", p_(NH_3)=2/3P , p_(CO_2)=1/3P):}` `therefore K_p=(p_(NH_3))^2(p_(CO_2))=(2/3P)^2 (1/3P)` `therefore K_p=4/27P^3` `P^3=27/4K_p` So, `P=3(K_p/4)^(1/3)`

Discussion

23.	Following data is given for the reacson, CaCO_(3(s))toCaO_((s))+CO_(2(s)) Delta_(f)H^(@)[CaO_((s))]=-650.0" kJ mol"^(-1) Delta_(f)H^(@)[CO_(2(g))]=-395.9" kJ mol"^(-1) Delta_(f)H^(@)[CaCO_(3(s))]=-1206.9" kJ mol"^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.
Answer» Solution :`CaCO_(3(s))toCaO_((s))+CO_(2(s))` `Delta_(f)H^(@)=Delta_(f)H^(@)[CaO_((s))]+Delta_(f)H^(@)[CO_(2(G))]-Delta_(f)H^(@)[CaCO_(3(s))]` `Delta_(f)H^(@)=-650+(395.9)-(-1206.9)` `=+161" kJ mol"^(-1)` Because `DELTAH` value is positive so the reaction is ENDOTHERMIC. Hence, according to Le-Chatelier's PRINCIPLE, reaction will proceed in forward direction on increasing temperature.

Discussion

24.	Following data is given for the reaction : CaCO_(3) (s) rarr CaO(s) + CO_(2) (g) Delta_(f) H^( Θ) [CaO(s) ] = - 635.1 kJ "mol" ^(-1) Delta_(f) H^(Θ) [ CO_(2) (g) ] = - 393.5 kJ"mol" ^(-1) Delta_(f) H^(Θ)[CaCO_(3)(s) ] = - 1206.9 kJ "mol" ^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.
Answer» SOLUTION :`delta_(r) H^(@) = Delta_(f) H^(@) [CaO (s) ] + Delta_(f) H^(@)[CO_(2)(g) ] -Delta_(f) H^(@)[CaCO_(3)(s)]` `=- 635.1 + (-393.5) - (-1206.9) = 178.3 kJ"mol"^(-1)` Thus, the reaction is endothermic. Hence, according to Le Chatelier'sprinciple, onincreasingthe temperature, the equilibrium will PROCEED in the FORWARD direction .

Discussion

25.	Following data is given for the reaction: CaCO_(3(s)) to CaO_((s)) + CO_(2(g)) Delta_f H^ө [CaO_((s))]=-"635.1 kJ mol"^(-1) Delta_f H^ө [CO_(2(g))]=-393.5 "kJ mol"^(-1) Delta_f H^ө [CaCO_(3(s))]=-1206.9 "kJ mol"^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.
Answer» Solution :GIVEN that, `Delta_f H^ө [CaO_((s))]=-"635.1 kJ mol"^(-1)` `Delta_f H^ө [CO_(2(g))]=-393.5 "kJ mol"^(-1)` `Delta_f H^ө [CaCO_(3(s))]=-1206.9 "kJ mol"^(-1)` In the REACTION `CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))` `Delta_f H^ө=Delta_f H^ө [CaO_((s))]+ Delta_f H^ө [CO_(2(g))]-Delta_f H^ө [CaCO_(3(s))]` `THEREFORE Delta_f H^ө`=-635.1+(-393.5)-(-1206.9) =178.3 kJ `"mol"^(-1)` As `DeltaH` value is positive, hence the reaction is ENDOTHERMIC. So, according to Le-Chatelier.s principle, reaction will proceed in FORWARD direction on increasing temperature.

Discussion

26.	Following criteria should be taken in consideration for resonance. The major contributor is the one with the lower energy. Good contributors generally have all octets satisfied, as many bonds as possible and as little charge separation as possible. Negative charges are more stable on the more electronegative atoms. Resonance stabilization is most important when it serves to delocalize a charge over two or more atoms Observe the following structures H_(2)C=""^(+)underset(I)(N)H_(2) harr""^(+) underset(II)(CH_(2))-NH_(2)
Answer» STRUCTURE - I is major contributor in the two resonating structures Structure - II is major contributor in the two resonating structures Both are EQUAL stable Resonance is not possible Answer :B

Discussion

27.	Following criteria should be taken in consideration for resonance. The major contributor is the one with the lower energy good contributors generally have all octets satisfied, as many bonds are possible andas little charge separation as possible negative charges are more stable on the more electronegative atoms. Resonance stabilization is most important when it serves to delocalize a charge over two or more atoms How many resonating structures can be drawn for 2, 4-pentadienyl radical
Answer» 1 2 3 4 Answer :C

Discussion

28.	Following cell is set up between copper and silver electrodes : Cu\|Cu^(2)(aq)\|\|Ag^(+)(aq)\|(Ag) if its two half cells work under standard condition calculate the e.m.f of the cell ["Given" E_(Cu^(2+)/Cu)^(@)=0.34 "volt" E_(Ag^(+)/Ag=+0.80 "volt")^(@)]
Answer» ANSWER :`E^(2)_("CELL")=E_("CATHODE")^(@)-E^(2)_("anode")=E_(Ag^(+)//Ag)^(@)-E_(Cu^(2+)//Cu)^(@)=+0.80-(+0.34)=0.46V`

Discussion

29.	Following are the structures of four isomer of hexane. Among the names given below, whichcorrectly identifies the fifth isomer? CH_3 CH_2 CH_2 CH_2CH_2 ""CH_3(CH_3)_3 C CH_2CH_3 (CH_3)_2CHCH_2CH_2CH_3 ""(CH_3)_2CHCH(CH_3)_2
Answer» 2-methyl pentane 2-ethyl butane 2, 3-dimethyl butane 3-methyl pentane Solution :ISOMERS

Discussion

30.	Following are some statements about modern periodic table i) It consists of s, p, d and f blocks ii) The energy levels filling order in 6th period is 6s, 4f, 5d and 6p iii) IIIA group contains maximum number of elements
Answer» only i & II are CORRECT only i is correct only ii & III are correct all are correct Answer :A

Discussion

31.	Following are some gases with their compressibility factor values at S.T.P. The gas supposed to deviateleast from ideal nature is
Answer» GAS A( Z = 1.15) Gas B ( Z= 0.98) Gas C ( Z = 0.86) Gas D ( Z = 0.92) ANSWER :B

Discussion

32.	Following are explain by reasons : The solution of strong base and weak acid salt is basic
Answer» Solution :The solution of `NaCH_3COO` (Sodium Acetate ) is basic : Sodium acetate is a salt of strong base NaOH and weak acid acetic acid `(CH_3COOH)` `CH_3COONa_((aq))hArr Na_((aq))^(+) +CH_3COO_((aq))^(-)` …. (a) CATION `Na^+` is not hydrolysed and REMAIN in solution as `Na_((aq))^(+)` . The acetate ion formed by ionization of sodium acetate is interactionwith water and dehydrolysis and produce weak acid `CH_3COOH` and `OH^-` ion. `CH_3COO_((aq))^(-) + H_2O_((l)) hArr CH_3COOH_((aq))+ OH_((aq))^(-)` ...(b) This `CH_3COOH` is a weak acid so remain as a dissociate molecule in solution. Thus `[OH^-]` increase in solution. Total REACTION of (a) and (b) : `CH_3COONa_((aq)) + H_2O_((l)) hArr ubrace(Na_((aq))^(+)+ OH_((aq))^(-))_(NaOH)+ CH_3COOH_((aq))` Thus concentration of `OH^-` in solution increase and solution BECOME basic . The pH of solution is more than `(pH gt 7)`. Similarly , all strong base and weak acid salts are basic.

Discussion

33.	Following are explain by reasons : The solutionof weak base and strong acid salt is acidic.
Answer» Solution :The solution of `NH_4Cl` (Ammonium chloride ) is acidic . The ammonium chloride `(NH_4Cl)` being a salt of weak base `NH_4OH` and STRONG acid HCl. It dissociate completely in water `NH_4Cl_((aq)) to NH_(4(aq))^(+) + Cl_((aq))^(-)` ….(a) Ammonium ions undergo hydrolysis with water to from `NH_4OH` and `H^+` ion. `NH_(4(aq))^(+) + H_2O_((l)) HARR NH_4OH_((aq)) + H_((aq))^(+)` ...(b) Ammonium hydroxide is a weak base and therefore remaining almost unionised in solution.This results in increased of `H^+` ions CONCENTRATION in solution, If total reaction =(a)+(b) `NH_4Cl_((aq))+ H_2O_((l)) hArr NH_(4(aq))^(+) +[H_((aq))^(+) + Cl_((aq))^-]` Thus , `[H^+]` is solutionincrease and solution become acidic and the value of pH is less than 7.0 Similarly to understand EVERY weak base and strong acid salts solution in acidic.

Discussion

34.	Following are explain by reasons : The solution of strong acid and strong base salts is neutral.
Answer» Solution :The solution of NaCl is NEUTRAL : Sodium chloride (NaCl) is a salt of strong ACID HCl and strong base NaOH . In NaCl solution `NA^+` and `Cl^-` is a hydrated ions. `NaCl to Na_((AQ))^(+) + Cl_((aq))^(-)` `Na^+` and `Cl^-` ions are not hydrolyse by water . So, pH of NaCl solution is equal to pH of water (7) . So , this solution is neutral. Similarly all strong base and strong acid salts solution are neutral.

Discussion

35.	Following are explain by reasons : The solution of weak acid-weak base salts is almost neutral.
Answer» Solution :The solution of AMMONIUM cynate is neutral . Ammonium cynate being a salt of weak base `NH_4OH` and weak acid `CH_3COOH` . It undergo dissociated in water to give ammonium ion and acetate ion. `NH_4COOCH_(3(aq)) to NH_(4(aq))^(+) + CH_3COO_((aq))^(-)` The produced positive and negative ions hydrolysis with water. `CH_3COO_((aq))^(-) + NH_(4(aq))^(+) + H_2OhArr CH_3COOH_((aq)) + NH_4OH_((aq))` `CH_3COOH` and `NH_4OH` , ALSO remain into partially dissociated FORM : `CH_3COOH_((aq)) hArr CH_3COO_((aq))^(-) + H_((aq))` `NH_4OH_((aq)) hArr NH_(4(aq))^(+) + OH_((aq))^(-)` `H_((aq))^(+) + OH_((aq))^(-) hArr H_2O_((l))` Thus , the EXTEND of `[H^+]` and `[OH^-]` depends on nature of weak acid and weak base. If `[H^+] = [OH^-]` , So this solution is neutral. `therefore pH=7 + 1/2 (pK_a - pK_b)` Similarly , to understand every weak base and weak acid salts solution are neutral.

Discussion

36.	Following are explain by reasons : The solution of NH_4CH_3COO is (almost) neutral.
Answer» Solution :The solution of ammonium cynate is neutral . Ammonium cynate being a salt of weak base `NH_4OH` and weak acid `CH_3COOH` . It undergo dissociated in water to give ammonium ion and acetate ion. `NH_4COOCH_(3(aq)) to NH_(4(aq))^(+) + CH_3COO_((aq))^(-)` The produced positive and negative ions hydrolysis with water. `CH_3COO_((aq))^(-) + NH_(4(aq))^(+) + H_2OhArr CH_3COOH_((aq)) + NH_4OH_((aq))` `CH_3COOH` and `NH_4OH` , also REMAIN into partially dissociated form : `CH_3COOH_((aq)) hArr CH_3COO_((aq))^(-) + H_((aq))` `NH_4OH_((aq)) hArr NH_(4(aq))^(+) + OH_((aq))^(-)` `H_((aq))^(+) + OH_((aq))^(-) hArr H_2O_((L))` Thus , the extend of `[H^+]` and `[OH^-]` depends on nature of weak acid and weak base. If `[H^+] = [OH^-]` , So this solution is neutral. `THEREFORE pH=7 + 1/2 (pK_a - pK_b)` Similarly , to understand every weak base and weak acid salts solution are neutral.

Discussion

37.	Following are explain by reasons : NH_4Cl solution is acidic.
Answer» Solution :The solution of `NH_4Cl` (Ammonium chloride ) is acidic . The ammonium chloride `(NH_4Cl)` being a salt of WEAK base `NH_4OH` and strong acid HCl. It dissociate completely in water `NH_4Cl_((aq)) to NH_(4(aq))^(+) + Cl_((aq))^(-)` ….(a) Ammonium ions undergo hydrolysis with water to from `NH_4OH` and `H^+` ion. `NH_(4(aq))^(+) + H_2O_((l)) hArr NH_4OH_((aq)) + H_((aq))^(+)` ...(b) Ammonium hydroxide is a weak base and THEREFORE REMAINING almost unionised in solution.This RESULTS in increased of `H^+` ions concentration in solution, If total reaction =(a)+(b) `NH_4Cl_((aq))+ H_2O_((l)) hArr NH_(4(aq))^(+) +[H_((aq))^(+) + Cl_((aq))^-]` Thus , `[H^+]` is solutionincrease and solution become acidic and the value of pH is less than 7.0 Similarly to understand every weak base and strong acid salts solution in acidic.

Discussion

38.	Following are explain by reasons : NaCl solution is neutral.
Answer» Solution :The solution of NACL is neutral : Sodium chloride (NaCl) is a salt of strong acid HCl and strong base NaOH . In NaCl solution `Na^+` and `Cl^-` is a HYDRATED ions. `NaCl to Na_((aq))^(+) + Cl_((aq))^(-)` `Na^+` and `Cl^-` ions are not HYDROLYSE by WATER . So, pH of NaCl solution is equal to pH of water (7) . So , this solution is neutral. Similarly all strong base and strong acid SALTS solution are neutral.

Discussion

39.	Following are explain by reasons : NaCH_3COO solution is basic
Answer» Solution :The solution of `NaCH_3COO` (Sodium Acetate ) is basic : Sodium acetate is a salt of strong BASE NaOH and weak acid acetic acid `(CH_3COOH)` `CH_3COONa_((aq))hArr Na_((aq))^(+) +CH_3COO_((aq))^(-)` …. (a) Cation `Na^+` is not hydrolysed and remain in solution as `Na_((aq))^(+)` . The acetate ion formed by ionization of sodium acetate is interactionwith water and dehydrolysis and produce weak acid `CH_3COOH` and `OH^-` ion. `CH_3COO_((aq))^(-) + H_2O_((l)) hArr CH_3COOH_((aq))+ OH_((aq))^(-)` ...(B) This `CH_3COOH` is a weak acid so remain as a dissociate molecule in solution. Thus `[OH^-]` increase in solution. TOTAL reaction of (a) and (b) : `CH_3COONa_((aq)) + H_2O_((l)) hArr ubrace(Na_((aq))^(+)+ OH_((aq))^(-))_(NaOH)+ CH_3COOH_((aq))` Thus concentration of `OH^-` in solution increase and solution become basic . The PH of solution is more than `(pH GT 7)`. Similarly , all strong base and weak acid salts are basic.

Discussion

40.	Fog is colloidal solution of:-
Answer» Liquid is GAS Gas in liquid Solid in gas Gas in gas Solution :FOG is a colloidal solution of liquid in gas.

Discussion

41.	Fog is an example of colloidal system of
Answer» liquid DISPERSED in a liquid liquid dispersed in a gas gas dispersed is a liquid solid dispersed in a liquid Answer :B

Discussion

42.	Fly ash is mainly the waste produced by which industry?
Answer» DAIRY INDUSTRY Detergent industry Petroleum industry THERMAL POWER industry SOLUTION :Thermal Power industry

Discussion

43.	Fluorocarbon polymers can be made by florinating polyethylene according to the reaction 2CoF_(2)+F_(2)rarr2CoF_(3) (CH_(2))_(n)+4nCoF_(3)rarr(CF_(2))_(n)+2nHF +4nCoF_(2) The CoF_(3) can be regenerated by the reaction 2CoF_(2)+F_(2)rarr2CoF_(3) Calculate kg of fluorine consumed per kg of fluorocarbon produced (CF_(2))_(n) [Write the answer excluding the decimal places]
Answer» ANSWER :1

Discussion

44.	Fluorobenzene is prepared from benzene diazonium chloride by ………………… .
Answer» SOLUTION :Balz-Scheimann REACTION

Discussion

45.	Fluorobenzene is prepared by treating benzene diazonium chloride with fluoroboric acid and heating the product obtained. This reaction is known as:
Answer» Schiemann reaction Sandmeyer reaction Gattermann reaction Ullmann reaction. Answer :A

Discussion

46.	Fluorine reacts with ice and results in the change: H_(2)O(s) + F_(2)(g) to HF(g) + HOF(g) Justify that this reaction is a redox reaction .
Answer» Solution :`H_(2)O(s)+F_(2)(G)toHF(g)+HOF(g)` In this reaction `F_(2)` is undergoing reduction as well as OXIDATION as it is ADDING H (an electropositive element) to form HF and 0 (an electronegative element) to form HOF. Hence, it is a redox reaction.

Discussion

47.	Fluorine reacts with ice and results in the change : H_(2)O_((s))+F_(2(g))toHF_((g))+HOF_((g)) Justify that this reaction is a redox reaction.
Answer» SOLUTION :`overset(+1)(H_(2))overset(-2)(O_((s)))+overset(0)(F_(2(G)))tooverset(+1)(H)overset(-1)(F_((g)))+overset(+1)(H)overset(-2)(O)overset(+1)(F_((g)))` Oxidation number of F in `F_(2)` reduces from 0 to -1 and in HOF is +1. Therefore F gets reduced as WELL as OXIDIZED. Therefore it is disproportionate redox reactions.

Discussion

48.	Fluorine reacts with ice and r esults in the change : H_(2)O(s)+F_(2)(g)rarrHF(g)+HoF(g) justify that this reaction is a redox reaction
Answer» Solution :Writing the O.N of reach atom aboe its SYMBOL we have `H_(2)+F_(2)(g)rarrHF (g)+HOF` here the O.N of F decreases form 0 in `F_(2)` to -1 in HF and increases form 0 in `F_(2)` to +1 in HOF therefore `F_(2)` is both reduced as well as oxidised thus it isredox and more specifically it is a DISPROPORTIONATION reaction

Discussion

49.	Fluorine is very reactive because its bond dissociation energy is less. Why bond energy is less in F_2?
Answer» Solution :FLUORINE ATOM has seven valence electrons. One electron each of two F ATOMS are shared to form the single bond. Each F atom has thre IONE pairs. The Ione pair Ione pair repulsion are predominent in `F_(2)` molecule. Hence bond energy in `F_(2)` is less.

Discussion

50.	Fluorine is the best oxidizing agent because it has _____
Answer» HIGHEST electron affinity highest `E^(@)` (REDUCTION) highest `E^(@)` (oxidation) LOWER electron affinity Answer :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Following three gaseous equilibrium reactions are occurring at 27^(@)C. (A) 2CO+O_(2)hArr2CO_(2) (B) PCl_(5)hArrPCl_(3)+Cl_(2) (C) 2HIhArrH_(2)+I_(2) The correct order of K_(p)//K_(c) for the following reaction is

Following two eqilibrium is simultaneously establised in a containerPCl_(s(g)) hArr PCl_((3)(g)) + Cl_((2)(g)), CO_((g)) + Cl_(2(g)) hArr COCl_((2)(g)). If some Ni_((g)) is introduced in the container forming Ni(CO)_(4)then at new equilibrium

Following statements regarding the periodic trends of chemical reactivity of alkali metals and halogens are given. Which of these statements give the correct picture?

Followingstatementregardingthe periodictrends of chemicalreactivity of thealkalimetalsand thehalogensare given. Whichof thesestatement gives the correctpicture ?

Following results were observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant. {:(lamda (nm) ,500,450,400),(v xx 10^(-6) (ms^(-1)),2.55,4.35,5.20):}

Followingresultsare observedwhensodiummetal isirradiated withdifferent wavelengthCalculate( a)thresholdwavelengthand( b)planks constant

Following redox in balancing state has how many P, H, O and electric charge on left side respectively. P_(4)+OH^(1-)toPH_(3)+H_(2)PO_(2)^(-)

Following reaction in equilibrium at 25^(@)C: 2NO(g,1 xx 10^(-5) "atm")+Cl_2(g,1 xx 10^(-2)"atm") harr 2NOCl(g, 1 xx 10^(-2) "atm") DeltaG^@ is

Following reaction describes the rusting of iron 4Fe +3O_(2) rarr 4Fe^(3+) +6O^(2-) .Which one of the following statement is incorrect .

Following reaction describes the rusting of iron 4Fe+3O_(2) rarr 4Fe^(3+)+6O^(2-) Which one of the following statements is incorrect?

Following reaction describes the rusting of 4Fe+3O_(2) to 4Fe^(3+) +6O^(2-). Which one of the following statement is incorrect

Following frightening diagrams show identical cubes such that edge of cube 2 lies exactly in the middle of one of the faces of Cube 1 and Cube 4 has a corner at the body centre of the cube 3 . Find the contribution (in fraction) of the spheres shown to each of the cube 1 ,2 , 3 and 4 respectively is

Following graph between DeltaG and reaction progress in for/can be :

Following diagrams show identical cubes such that edge of cube 2 lies exactly in the middle of one of the faces of Cube1 and Cube 4 has a corner at the body centre of the Cube3. Find the contributions (in fraction) of the spheres shown to each of the cubes

Following equilibrium is established to decomposing of Ammonium carbonate NH_4COONH_2 in closed vessel at 700 K temperature. NH_4COONH_(2(s)) hArr 2NH_(3(g)) + CO_2 At initial if there is vaccum and at equilibrium total pressure is P bar than derive the value of K_p with respect to P.

Following cell is set up between copper and silver electrodes : Cu|Cu^(2)(aq)||Ag^(+)(aq)|(Ag) if its two half cells work under standard condition calculate the e.m.f of the cell ["Given" E_(Cu^(2+)/Cu)^(@)=0.34 "volt" E_(Ag^(+)/Ag=+0.80 "volt")^(@)]

Following are the structures of four isomer of hexane. Among the names given below, whichcorrectly identifies the fifth isomer? CH_3 CH_2 CH_2 CH_2CH_2 ""CH_3(CH_3)_3 C CH_2CH_3 (CH_3)_2CHCH_2CH_2CH_3 ""(CH_3)_2CHCH(CH_3)_2

Following are some statements about modern periodic table i) It consists of s, p, d and f blocks ii) The energy levels filling order in 6th period is 6s, 4f, 5d and 6p iii) IIIA group contains maximum number of elements

Following are some gases with their compressibility factor values at S.T.P. The gas supposed to deviateleast from ideal nature is

Following are explain by reasons : The solution of strong base and weak acid salt is basic

Following are explain by reasons : The solutionof weak base and strong acid salt is acidic.

Following are explain by reasons : The solution of strong acid and strong base salts is neutral.

Following are explain by reasons : The solution of weak acid-weak base salts is almost neutral.

Following are explain by reasons : The solution of NH_4CH_3COO is (almost) neutral.

Following are explain by reasons : NH_4Cl solution is acidic.

Following are explain by reasons : NaCl solution is neutral.

Following are explain by reasons : NaCH_3COO solution is basic

Fog is colloidal solution of:-

Fog is an example of colloidal system of

Fly ash is mainly the waste produced by which industry?

Fluorobenzene is prepared from benzene diazonium chloride by ………………… .

Fluorobenzene is prepared by treating benzene diazonium chloride with fluoroboric acid and heating the product obtained. This reaction is known as:

Fluorine reacts with ice and results in the change: H_(2)O(s) + F_(2)(g) to HF(g) + HOF(g) Justify that this reaction is a redox reaction .

Fluorine reacts with ice and results in the change : H_(2)O_((s))+F_(2(g))toHF_((g))+HOF_((g)) Justify that this reaction is a redox reaction.

Fluorine reacts with ice and r esults in the change : H_(2)O(s)+F_(2)(g)rarrHF(g)+HoF(g) justify that this reaction is a redox reaction

Fluorine is very reactive because its bond dissociation energy is less. Why bond energy is less in F_2?

Fluorine is the best oxidizing agent because it has _____