52671 + Interview Questions in Chemistry in Class 11 Page 1043 InterviewSolution

52101.	0.15 g of an organic compound gave 0.12 g of silver bromide by carius by carius method. Find out the percentage of bromine in the compound.
Answer» SOLUTION :PERCENTAGE of BROMINE `=80/188xx ("Mass of AgBr")/("Mass of COMPOUND")xx100=80/188xx0.12/0.15xx100=34%`

Discussion

52102.	0.15 g of iodoform gave 0.2682 g of AgI. Calculate the percentage of iodine.
Answer» Solution :Weight of compound = 0.15 G Weight of silver IODIDE = 0.2682 g Molecular weight of silver iodide `= 108 + 127 = 235` 235 g of silver iodide contains 127 g of iodine 0.2682 g of AgI contains `=(127xx0.2682)/(235)` `=0.144 g` iodine 0.15 g of compound contains 0.144 g of iodine 100 g of compound contains `=(100xx0.1449)/(0.15)=96.6g` `therefore"Percentage of iodine"=96.6`

Discussion

52103.	0.14kg of nitrogen at 300K is expanded isothermally and reversibly until its volume becomes doubled. Calculate the work done by the gas.
Answer» SOLUTION :`1.729 KJ`

Discussion

52104.	0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element ?
Answer» Nitrogen Carbon Fluorine Sulphur Answer :C

Discussion

52105.	0.14 g of an element A combines with 0.16 g of the elements B. Also 0.05 g of another element C combines with 0.35 g of A. It was also found that 0.2857 g of C combines with 2.2857 g of B. Show that the data illutrates the law of reciprocal proportions.
Answer» SOLUTION :In the compound AB 0.14 g of A are combined with `B = 0.16 g` 1.0 g of A are combined with `B=(0.16)/(0.14)g=1.143g` In the compound AC 0.35 of A are combined with C = 0.05 g 1.0 g of A are combined with `C=(0.05)/(0.35)g=0.143 g` RATIO by weight of B and C combined with 1.0 g of C in the two compounds `= (1.143)/(0.143)` In the compound BC Ratio by weight of B and C `= (2.2857)/(0.2857)` Comparing (i) and (II), `(1.143)/(0.143):(2.2857)/(0.2857)or 8:8or 1:1` This illustrates the Law of Reciprocal Proportions.

Discussion

52106.	0.132g of an organic compound gave 50ml of N_(2) at NTP. The weight percentage of nitrogen in the compound is close to
Answer» 15 20 48.9 47.43 Solution :WT% of `N_(2) = (28)/(22400) XX ("VOL of" N_(2)(ml) "at stp")/("wt. of compound") xx 100`

Discussion

52107.	0.13 g of Cu, when treated with AgNO_3 solution, displaced 0.433 g of Ag. 0.13 g of AI, when treated with CuSO_4 solution displaced 0.47g of Cu. 1.17 g of AI displaces 0.13 g of hydrogen from an acid. Find the equivalent weight of Ag if equivalent weights of Cu and Al are not known.
Answer» Solution :`underset(0.13g)(Cu)+ AgNO_3 to underset(0.433g)(Ag) ""...(1)` `underset(0.13g)(AI)+ CuSO_4 to underset(0.47g)(Cu)""...(2)` `underset(1.17g)(AI) + "Acid" to underset(0.13g)(H_2)""...(3)` For Eqn. (3) : EQ. of AL = eq. of hydrogen `therefore (1.17)/("eq. W. of AI")= (0.13)/(1),` eq. WT. of Al is 9. For Eqn. (2) : eq. of Al = eq. of Cu `therefore (0.13)/(9) = (0.47)/("eq. wt. of Cu"),` eq. wt. of Cu= 32.5. For Eqn. (1) : eq. of Cu = eq. of Ag `therefore (0.13)/( 32.5) = (0.433)/("eq. wt. of Ag")` eq. wt. of Ag= 108.25.

Discussion

52108.	0.12g organic compound gave 0.22g Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in compound? (P= 31) (Molar mass of Mg_(2)P_(2)O_(7)= 222g)
Answer» Solution :`2P rarr Mg_(2)P_(2)O_(7)` % `P =("MASS of 2P")/("MOLAR mass of" Mg_(2)P_(2)O_(7)) xx ("mass of "Mg_(2)P_(2)O_(7))/("molar mass of compound") xx 100` `=(62)/(222) xx (0.22)/(0.12) xx 100= 51.20%`

Discussion

52109.	0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_(2)P_(2)O_(7) by the usual analysis. Calculate the percentage of phosphorus in the compound.
Answer» Solution :Here, the mas of the compound taken = 0.13 G Mass of `Mg_(2)P_(2)O_(7) -= 2g` atoms of P or `(2 xx 24 + 2 xx 31 + 16 xx 7) = 222g` of `Mg_(2)P_(2)O_(7) -= 62 g` of P i.e., 222g of `Mg_(2)P_(2)O_(7)` CONTAIN PHOSPHORUS = 62 g `:.` 0.22 g of `Mg_(2)P_(2)O_(7)` will contain phosphorus `= (62)/(222) xx 0.22 g` But this is the amount of phosphorus present in 0.12 g of the organic compound. `:.` Percentage of phosphorus `= (62)/(222) xx (0.22)/(0.12) xx 100 = 51.20`

Discussion

52110.	0.12 g of an organic compound gave on combustion 0.18 g of water and 0.11 g of CO_(2). Calculate the percentage of C and H in the organic compound.
Answer» SOLUTION :`C=25%, H=16.66%`

Discussion

52111.	0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_2 P_2 O_7 by usual analysis. Calculate the percentage of phosphorus in the compound.
Answer» ANSWER :0.512

Discussion

52112.	0.11 g of colourless oxide is nitrogen occupies56 cc at 273^(@)C and 2 atm. Name the gas.
Answer» SOLUTION :NITROUS OXIDE

Discussion

52113.	0.108g of finely divided copper was treated with an excess of ferric sulphate solution until copper was comnpletely dissolved. The solutionafter the addition of excess dilute sulphuric acidrequired 33.7mL of 0.1N KMnO_(4) for complete oxidation. Find theequaction whichrepresentsthe reaction between metallic copper and ferric sulphate solution. At wt. of Cu = 63.6 and Fe = 56.
Answer» Answer :`Cu + Fe_(2)(SO_(4))_(3) RARR CuSO_(4) + 2FeSO_(4)`

Discussion

52114.	0.10 moole of AgCI_((s)) is added to 1 litre of H_(2)O. Next crystal of NaBr are added until 75% of the AgCI is concerted to AgBr_((s)), the less soluble silver halide. What is Br^(-) at this point? K_(SP) ofAgCI is 1.78xx10^(-10) and K_(SP) of AgBr is 5.25xx10^(-13).
Answer» ANSWER :`2.2xx10^(-4)M;`

Discussion

52115.	0.1 N solution of Na_(2)CO_(3) is being titrated with 0.1 N HCl, the best indicator to be used is :
Answer» POTASSIUM FERRICYANIDE` phenolphthalein methyl orange litmus Solution :N//A

Discussion

52116.	0.1 N solution of a dibasic acid can be prepared by dissolving 0.45 g of the acid in water and diluting to 100 mL. The molecular mass of the acid is :
Answer» 45 90 135 180 Solution :N//A

Discussion

52117.	0.1 moles of Hydrocaron on complete combustion produced 17.6 gms of CO_(2).How many Carbon atoms are present in each molecule of the hydrocarbon.
Answer» SOLUTION :`C_(4)H_(8)+6O_(2)rarr4CO_(2)+4H_(2)O` 1 mole _________ `4xx44=176g` 0.1 mole _________ 17.6 G

Discussion

52118.	0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculate the concentration of Cl_(2) at equilibrium, if the equilibrium constant for the dissociation of PCl_(5) is 0.0414.
Answer» Solution : `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc. ",0.1 "mole",,0,,0),(" Conc. at EQM. (moles/litres)",(0.1 -x),,x,,x):}` Applying the law of chemical equilibrium, we get`K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]` Here `K_(c) = 0.0414 " (Given) "` `:. 0.0414 = (x XXX)/((0.1-x)) or x^(2)/(0.1 - x) = 0.0414 or x^(2) + 0.0414 x - 0.00414 = 0` `(x= (-0.0414 pm sqrt((0.0414)^(2) - 4 xx 1xx (-0.00414)))/2 ""[ "Using the formula " x = (-b pm sqrt(b^(2) - 4ac))/(2A)]` ( The negative value of x is meaningless and hence is rejected ) Thus, the concentration of `Cl_(2) " at equilibrium will be " 0.0468 mol L^(-1)`

Discussion

52119.	01 mole of N_(2)O_(4) (g) was sealed in a tube under atmospheric conditions at 25^(@)C. Calculate the number of moles of NO_(2) (g) present if the equilibrium N_(2)O_(4) (g) hArr 2 NO_(2)is reached after some time (K_(p) = 0 14).
Answer» Solution :` {: (,N_(2)O_(4)(G),hArr,2NO_(2)(g),) ,("Intial amounts",0.1 "mole",,0,), ("At equilibrium " ,(01 -x),,2x," TOTAL " 01 +x "moles"):}` As P=1atm, ` p_(N_(2)O_(4)) = (01 -x)/(01 +x), p_(NO_(2))= 2/(01 + x) atm ` ` K_(p) - (p_(N_(O_(4)))^(2))/p_(N_(2)O_(4)) = (2x // (01 + x))^(2)/((01-x) //( 01 +x))= (4x^(2))/((01- x) ( 01 +x) )= (4x^(2))/(001 - x^(2) )` ` :. (4x^(2))/(001 - x^(2) = 014 )or 414x^(2) = 00014or x = 0 018 ` ` :. " No. of moles of " NO_(2) " at equilibrium " = 2x = 2 xx 0* 018= 0* 036 "mole" `

Discussion

52120.	0.1 mole of KMnO_(4) present in 500 ml solution got converted into K_(2)MnO_(4). The concentration of KMnO_(4) is
Answer» 0.2 M 0.2 N 1 N 0.04 N Solution :n-factor = `1,M=(0.1)/(0.5)=0.2M=0.2N`

Discussion

52121.	0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and the solution diluted to one litre. The H^(+) ion concentration in the solution will be
Answer» `1.6xx10^(-11)` `8xx10^(-11)` `5XX10^(-5)` `8xx10^(-2)` Solution :`{:(,CH_(3)NH_(2),+,HCl,rarr,CH_(3)N overset(+)H_(3)Cl^(-),,,),("Initial conc.",0.1,,0.08,,0 "mol" L^(-1),,,),("Conc. in solution after mixing",0.1-0.08 ,,0,,0.08 "mol " L^(-1),,,):}` `=0.02` (HCl being limiting REACTANT) Thus, the final solution is abasic buffer of `CH_(3)NH_(2) and CH_(3)NH_(3)^(+)Cl` `:. pOH = pK_(b) + LOG. (["Salt"])/(["Base"])` i.e., `-log [OH^(-)]=-log K_(b) + log. (["Salt"])/(["Base"])` or `[OH^(-)]=K_(b) (["Salt"])/(["Base"]) = 5xx10^(-4)xx(0.02)/(0.08) ` `=(5)/(4) xx 10^(-4)M = 1.25 xx 10^(-4)M` `:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)`

Discussion

52122.	0.1 mol acetic acid and 0.1 mol sodium acetate in 500 mL solution pH is 4.74. Find ionization constant.
Answer» SOLUTION :`1.82xx10^(-5)`

Discussion

52123.	0.1 M-KMnO_(4) is used for the following titration. How much volume of the solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3)? S_(2)O_(3)^(2-)+MnO_(4)^(-)+H_(2)OtoMnO_(2)(s)+SO_(4)^(2-)+OH^(-)
Answer» 80 ML 26.67 ml 13.33 ml 16 ml Solution :eq. `KMnO_(4)` = eq. `Na_(2)S_(2)O_(3)` `(0.1xxV_(mL))/(1000)xx3=(0.158)/(158)xx8impliesV_(mL)=26.67mL`

Discussion

52124.	0.1 M HQ acid has pH=3 then findits ionization constant.
Answer» `3xx10^(-1)` `1xx10^(-3)` `1xx10^(-5)` `1xx10^(-7)` Solution :`{:(,HQoversettolarr, H^+ +, Q^-),("Initial " , 0.1 M, O,O),("Ionized MOLE",X,x,x),(,darr,darr,darr),("Mole at EQUILIBRIUM",0.1-x,x,x):}` pH=3 `THEREFORE [H^+]=10^(-3)` `therefore x=10^(-3)` `therefore K_a=((x)xx(x))/(0.1-x)` `=(10^(-3))^2/(0.1-10^(-3)) approx 10^(-6)/0.1 = 1xx10^(-5)`

Discussion

52125.	0.1 MHCl and 0.1 M H_(2)SO_(4), each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 N NaCl solution. The pH of the resulting mixture is
Answer» `1.17` `1.0` `0.3` log 2 - log 3 Solution :2 MLOF 0.1 M HCl =0.2 MILLIMOLE of HCl = 0.2 millimole of `H^(+)` ion 2 ml of 0.1 M `H_(2)SO_(4)= 0 .2` millimole of `H_(2)SO_(4)` = 0.4 millimole of `H^(+)` ion Total `H^(+)` ions = 0.6 millimole Total volume = 6 ml `:. [H^(+)]=(0.6)/(6)M=0.1 M =10^(-1)M:. pH = 1`.

Discussion

52126.	0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point. Dissociation constant for the acid HA is 5xx10^(-6) and the degree of hydrolysis, h lt 1.
Answer» Solution :`UNDERSET("weak")HA+NaOH rarr NAA+H_(2)O` At the END POINT, theirequivalent AMOUNTS react together `:.` In thefinal solution. `[NaA]=(0.1)/(2)=0.05M` AsNaA is a salt of weak acid and strong base, it hydrolyses as `a^(-) + H_(2)O hArr HA+ OH^(-)` For such a salt, `pH = 7 + (1)/(2) [pK_(a) + log c]=7+(1)/(2) [-log(5xx10^(-6))+log 0.05]` `=7+(1)/(2) [ 6-0.6990+0.6990-2]=9`

Discussion

52127.	0.1 M CH_(3)CO OH (pH = 3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized.(ii) 3/4th of the acid has been neutralized.
Answer» Solution :Calculation of dissociation constant of the acid `CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)` As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M` `K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)` (i) When 1/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial CONC.",0.1 M,,,,,,),("After 1/4th NEUTRALIZATION",0.1xx(3)/(4),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}` `:. pH = pK_(a) + LOG. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229` (II) When 3/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa , +,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 3/4th",0.1xx1/4M,,,,0.1xx3/4M,,),("neutralization",=0.025 M,,,,=0.075 M,,):}` `:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771`

Discussion

52128.	0.1 litre of 0.01 M KMnO_(4) is used by 100 mL of H_(2)O_(2) in acidic medium. Volume of same KMnO_(4) required in alkaline medium to oxidise 0.1 litre of some H_(2)O_(2) will be :
Answer» `(100)/(3) ML` `(500)/(3) mL` `(300)/(5) mL` `(400)/(3) mL` Answer :B

Discussion

52129.	0.1 gram mole of urea is dissolved in 100g. of water. The molality of the solution is
Answer» 1 m 0.01 M 0.01 M 1.0 M Answer :A

Discussion

52130.	0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metal is
Answer» 12 24 18 36 Solution :a) 1 mole of `O_(2)` = 4 equivalent of OXYGEN 22400 ML of `O_(2)` = 4 equivalent of oxygen 46.6 mL of `O_(2)` = (4/22400) xx 46.6 = 0.00832 eq. Equivalent of metal = Equivalent of oxygen (WEIGHT)/(Equivalent) = 0.00832 0.1/E = 0.00832 `rARr` E= (0.1)/(0.00832) = 12.0

Discussion

52131.	0.1 g of an organic compound gave 0.222g of Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in the compound ?
Answer» <P>0.32 0.42 0.52 0.62 Solution :Amountof `P=(0.222 XX 62)/(222) = 0.062 G ` `therefore " % of" P= (0.062)/(0.1) xx 100 =62.`

Discussion

52132.	0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :
Answer» 0.252 M 0.126 M 0.0313 M 0.0623 M Solution :The reaction involved are : `IO_(3)^(-)+5I^(-)+6H^(+) to 3I_(2)+H_(2)O` `2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(6)` Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)` `=(3xx0.1)/(214)` Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)` `(Mxx45)/(1000)=(2xx3xx0.1)/(241)` M=0.623

Discussion

52133.	0.1g of carbon dioxide occupies a volume of 320cc at certain conditions. Under similar conditions 0.2g of 3 dioxide of element 'X' occupies 440cc. Calculate the atomic weight of 'X'.
Answer» Solution :Applying Avogadro.slaw,` (n_(1))/(n_(2)) = (V_(1))/(V_(2))` But number of moles ` =n = w//M` `(V_(1))/(V_(2)) = (w_(1)M_(2))/(w_(2)M_(1))` Molecular weight of dioxide of X ` = M_(2) = (V_(1))/(V_(2)) XX (w_(2)m_(1))/(w_(1)) = (320)/(440) xx (0.2)/(0.1) xx 44 = 64` Molecular weight of `XO_(2) = 64` Therefore, ATOMIC weight of X = 32

Discussion

52134.	0.093 g of Na_2H_2 EDTA.2H_2O is dissolved in in 250 " mL of " aqueous solution. A sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with the above EDTA solution using a buffer of NH_4OH+NH_4Cl using erochrome black-T as indicator. 10 " mL of " hard water at equivalence point. Another sample of hard water is titrated with 10 " mL of " above EDTA solution using KOH solution (pH=12). Using murexide indicator, it requires 40 " mL of " hard water at equivalence point. (a). Calculate the amount of Ca^(2+) and Mg^(2+) present in 1 L of hard water. (b). Calculate the hardness due to Ca^(2+),Mg^(2+) ions and the total hardness of water in ppm of CaCO_3 (Given: Mw(EDTA sal t)=372gmol^(-1) Mw(CaCO_3)=100gmol^(-1))
Answer» Solution :Case I (Using eriochrome black-T indicator) (a). M of EDTA solution`=(0.093xx1000)/(372xx250)=0.001M` Volume of EDTA used `=10ML` Volume of water sample `=40mL` `M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)` in hard water) `0.001xx10=M_2xx10` `M_2=0.001` Molarities of `(Ca^(2+)+Mg^(2+))` ions `=0.001M` `=1.0 m` moles/L Case II: (Using murexide indicator) `M_1V_1(EDTA)=M_2V_2` (Hard water) `0.001xx10=M_2xx40` `M_2=0.25xx10^(-3)=0.25mmolL^(-1)` Total m mol `L^(-1) of Ca^(2+) and Mg^(2+)=1.0` MMOL `L^(-1) of Mg^(2+)=1.0-0.25=0.75mmolL^(-1)` mmoles `L^(-1) of Ca^(2+)=0.25 mmol L^(-1)` Amount of `Ca^(2+)L^(-1)implies0.25xx40xx10^(-3)=0.01gL^(-1)` Amount of `Mg^(2+)L^(-1)implies0.75xx24xx10^(-3)=0.018gL^(-1)` (B). `[MW(CaCO_3)=100gmol^(-1))` Total m" mol of "`Ca^(2+)` and `Mg^(2+)` ions `L^(-1)` `=1.0=0.001 (mol)/(L)` `=0.001M` 0.1 " mol of "`Ca^(2+) and Mg^(2+)-=0.1mol CaCO_3L^(-1)` `-=(0.001xx100xx10^(6))/(10^(3))` `-=100ppm` `{:[("Total hardness due to Ca^(2+) and Mg^(2+) ions"),("of the sample in gram of CaCO_3in 10^(6) " mL of " H_2O"),(=("Total MxxMw(CaCO_3)xx10^(6))/(10^(3))]:}` Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O=(0.25xx10^(-3)xx100xx10^(5))/(10^(3))=25ppm` Hardness due to `Mg^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O=100-25=75ppm`

Discussion

52135.	0.096kg of oxygen is compressed iso thermally and reversibly from 10^5 N.M^(-2) to 2 xx 10^(5) N.M^(2) at 300 K. Calculate the work done during the impression process.
Answer» ANSWER :12.078 KJ

Discussion

52136.	0.092 g of an organic compound containing phosphorus gave 0.111 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentage of phosphorus in the organic compound.
Answer» ANSWER :`33.7%`

Discussion

52137.	0.08 g/2.901 L is a saturated solution of CaF_2 at 298 K temp. calculate K_(sp) of CaF_2.
Answer» SOLUTION :`1.767xx10^(-10)`

Discussion

52138.	0.0755 g of selenium vapours occupying a volume of 114.2 mL at 700^(@)C and 185 mm of Hg. The vapours are in equilibrium as: Se_(6(g))hArr3Se_(2(g)) Calculate: (i) Degree of dissociation of Se, (ii) K_(p), (iii) K_(c). Atomic weight of Se is 79.
Answer» Solution :(`i`) `59%`, (`II`) `0.168 ATM`, (`III`) `2.633xx10^(-5) mol^(2) litre^(-2)` ,

Discussion

52139.	0.075 g of a monobasic acid required 10 mL of N//12 NaOH solution for complete neutralisation. Calculate the molecular mass of the acid
Answer» SOLUTION :MASS of the organic acid taken = 0.075 g Volume of `N//12 NaOH` required for neutralisation = 10 mL STEP I. Calculation of equivalent mass of the acid. `10 mL` of `N//12 NaOH -= 0.075 g` of acid `:.` 1000 mL of `1N NaOH = ((0.075 g) xx (1000 mL) xx (1N))/((10 mL) xx (N//12)) = 90 g` 1000 mL of 1 N NaOH solution contain one gram equivalent of base and it reacts with one gram equivalent of the acid. `:.` Equivalent mass of the acid = 90 step II. Calculation of the molecular mass of the acid. Molecular mass = Equivalent mass `xx` Basicity `= 90 xx 1 = 90` [Foe a monobasic acid, basicity = 1]

Discussion

52140.	0.06mole of KNO_(3) is added to 100cm^(3) of water at 298K . The enthalpy of KNO_(3) aq. Solution is 35.8kJ mol^(-1). After the solute is dissolved , the temperature of the solution will be
Answer» 293K 298K 301K 304 K Solution :Dissoluation of `KNO_(3)` in WATER is endothermic, `DeltaH _(sol) = 35.8 kJ mol^(-1)` `:.` Heat absorbed when 0.06 mole of `KNO_(3)` is DISSOLVED `= 35.8 xx 0.06 kJ = 2148 J` `q= m xx C xx DeltaT` `:. 2148 = 100 xx 4.184 xx DeltaT` or `DeltaT = 5K` `:. `TEMPERATUREOF the solution `= 298 -5= 293K`.

Discussion

52141.	0.05g of a commercial sample of KClO_(3) on decomposition liberated just sufficient oxygen for complete oxidation of 20 mL CO at 27^(circ)C and 750 mm pressure. Calculate % of KClO_(3) in sample.
Answer» ANSWER :`65.4%` ;

Discussion

52142.	0.05 g of magnesium when treated with dilute HCI gave 51 mL of hydrogen at 27°C and 780 mm pressure. Calculate the equivalent mass of magnesium.
Answer» ANSWER :11.76

Discussion

52143.	0.05 g of a gas at 750 mm pressure and 25°C occupy a volume of 46.5 mL. Calculate the molecular mass of the gas.
Answer» Solution :In the present case, `P = 750 MM= 750/760 atm, V = 46.5 mL = (46.5)/1000 L`, `T = 25^(@) C = 298 K, w=0.05 g` and `R = 0.0821 L atm K^(-1) "mol"^(-1)` Using the ideal gas equation, `PV = w/M RT` and substituting the values, we have `750/760 xx (46.5)/1000 = 0.05/Mxx 0.0821 xx 298 = 26.66`

Discussion

52144.	0.049 g of H_(2)SO_(4) is dissolved per litre of thegiven solution . Calculate the pH of the solution.
Answer» SOLUTION :`[H_(2)SO_(4)]=(0.049)/(98) "mol" L^(-1)=5xx10^(-4)M`. As `H_(2)SO_(4)rarr2H^(+)+SO_(4)^(2-), [H^(+)]=2XX(5xx10^(-4))=10^(-3)M :. pH = 3`

Discussion

52145.	0.03 mole of Ca^(2+) ions is added to a litre of 0.01 M SO_(4)^(2-) solution. Will it cause precipitation of CaSO_(4) ?K_(sp) "for" CaSO_(4)=2.4xx10^(-5).
Answer» Solution :In the final solution, `[Ca^(2+)]=0.03 M and [SO_(4)^(2-)]=0.01 M`. Hence, ionic PRODUCT of `CaSO_(4) = (0.03) (0.01)=3xx10^(-4)` which is greater than `K_(SP)`. Hence, precipitation will OCCUR.

Discussion

52146.	0.023 g of sodium metal is reacted with 100 cm^(3) of water. The pH of the resulting solution is
Answer» 10 11 9 12 Solution :`Na+H_(2)OrarrNaOH=(1)/(2)H_(2)` 1 mole of Na produces 1 mole of NAOH `0.023 gNa = (0.023)/(23) "mole" = 10^(-3) "mole"` `:. ` NaOH PRODUCED `= 10^(-3)` mole `:. [NaOH]=(10^(-3))/(100)xx1000=10^(-2) M` `:. pOH=2 ` or pH `= 14-2=12`.

Discussion

52147.	0.01 mole sucrose dissolve in .......... Litre water so it becomes 0.01solution.
Answer» 1 0.0001 1000 0.01 Answer :A

Discussion

52148.	0.01 mole solution of .10 volume. H_2O_2 is required to convert 0.01 mol PbS into PbSO_4 ?
Answer» 11.2 22.4 33.6 44.8 Solution :`PBS+4H_2O_2 to PbSO_4 + 4H_2O` `4H_2O_2 to 4H_2O + 2O_2` To convert 1 mole PbS 1 `PbSO_4` required = 4 mole `H_2O_2` `therefore` to convert 0.01 MOL PbS to `PbSO_4`= (?) `=(0.01xx4)/1`= 0.04 mol `H_2O_2` required STRENGTH of volume=10 `M=V/11.2 =10/11.2 `=0.8928 mol/lit `H_2O_2` 0.8928 mol `H_2O_2 to` 1 LITRE `H_2O_2` `therefore` 0.04 mol `H_2O_2 to` (?) `=(0.04xx1)/0.8928` =0.04480 litre =44.80ml `H_2O_2`

Discussion

52149.	0.01 mole of iodoform (CHI_(3)) reacts with Ag powder to produce a gas whose volume at NTP is
Answer» 224 ml 112 ml 336 ml None Answer :B

Discussion

52150.	0.01 mole of FeS_n(iron (II) sulphide) required 0.06 mole of AO_(4)^(3-)for complete oxidation. The species formed are FeO, SO_2 and A^(2+) . Calculate the value of n.
Answer» SOLUTION :`0.01xxn-f =0.06xx3`, `N-f =18" "18 =(2/n+4)XXN, n =4`

Discussion

Explore topic-wise InterviewSolutions in .

0.15 g of an organic compound gave 0.12 g of silver bromide by carius by carius method. Find out the percentage of bromine in the compound.

0.15 g of iodoform gave 0.2682 g of AgI. Calculate the percentage of iodine.

0.14kg of nitrogen at 300K is expanded isothermally and reversibly until its volume becomes doubled. Calculate the work done by the gas.

0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element ?

0.14 g of an element A combines with 0.16 g of the elements B. Also 0.05 g of another element C combines with 0.35 g of A. It was also found that 0.2857 g of C combines with 2.2857 g of B. Show that the data illutrates the law of reciprocal proportions.

0.132g of an organic compound gave 50ml of N_(2) at NTP. The weight percentage of nitrogen in the compound is close to

0.13 g of Cu, when treated with AgNO_3 solution, displaced 0.433 g of Ag. 0.13 g of AI, when treated with CuSO_4 solution displaced 0.47g of Cu. 1.17 g of AI displaces 0.13 g of hydrogen from an acid. Find the equivalent weight of Ag if equivalent weights of Cu and Al are not known.

0.12g organic compound gave 0.22g Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in compound? (P= 31) (Molar mass of Mg_(2)P_(2)O_(7)= 222g)

0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_(2)P_(2)O_(7) by the usual analysis. Calculate the percentage of phosphorus in the compound.

0.12 g of an organic compound gave on combustion 0.18 g of water and 0.11 g of CO_(2). Calculate the percentage of C and H in the organic compound.

0.12 g of an organic compound containing phosphorus gave 0.22 g of Mg_2 P_2 O_7 by usual analysis. Calculate the percentage of phosphorus in the compound.

0.11 g of colourless oxide is nitrogen occupies56 cc at 273^(@)C and 2 atm. Name the gas.

0.10 moole of AgCI_((s)) is added to 1 litre of H_(2)O. Next crystal of NaBr are added until 75% of the AgCI is concerted to AgBr_((s)), the less soluble silver halide. What is Br^(-) at this point? K_(SP) ofAgCI is 1.78xx10^(-10) and K_(SP) of AgBr is 5.25xx10^(-13).

0.1 N solution of Na_(2)CO_(3) is being titrated with 0.1 N HCl, the best indicator to be used is :

0.1 N solution of a dibasic acid can be prepared by dissolving 0.45 g of the acid in water and diluting to 100 mL. The molecular mass of the acid is :

0.1 moles of Hydrocaron on complete combustion produced 17.6 gms of CO_(2).How many Carbon atoms are present in each molecule of the hydrocarbon.

0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculate the concentration of Cl_(2) at equilibrium, if the equilibrium constant for the dissociation of PCl_(5) is 0.0414.

0*1 mole of N_(2)O_(4) (g) was sealed in a tube under atmospheric conditions at 25^(@)C. Calculate the number of moles of NO_(2) (g) present if the equilibrium N_(2)O_(4) (g) hArr 2 NO_(2)is reached after some time (K_(p) = 0 *14).

0.1 mole of KMnO_(4) present in 500 ml solution got converted into K_(2)MnO_(4). The concentration of KMnO_(4) is

0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and the solution diluted to one litre. The H^(+) ion concentration in the solution will be

0.1 mol acetic acid and 0.1 mol sodium acetate in 500 mL solution pH is 4.74. Find ionization constant.

0.1 M-KMnO_(4) is used for the following titration. How much volume of the solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3)? S_(2)O_(3)^(2-)+MnO_(4)^(-)+H_(2)OtoMnO_(2)(s)+SO_(4)^(2-)+OH^(-)

0.1 M HQ acid has pH=3 then findits ionization constant.

0.1 MHCl and 0.1 M H_(2)SO_(4), each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 N NaCl solution. The pH of the resulting mixture is

0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point. Dissociation constant for the acid HA is 5xx10^(-6) and the degree of hydrolysis, h lt 1.

0.1 M CH_(3)CO OH (pH = 3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized.(ii) 3/4th of the acid has been neutralized.

0.1 litre of 0.01 M KMnO_(4) is used by 100 mL of H_(2)O_(2) in acidic medium. Volume of same KMnO_(4) required in alkaline medium to oxidise 0.1 litre of some H_(2)O_(2) will be :

0.1 gram mole of urea is dissolved in 100g. of water. The molality of the solution is

0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metal is

0.1 g of an organic compound gave 0.222g of Mg_(2)P_(2)O_(7). What is the percentage of phosphorus in the compound ?

0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :

0.1g of carbon dioxide occupies a volume of 320cc at certain conditions. Under similar conditions 0.2g of 3 dioxide of element 'X' occupies 440cc. Calculate the atomic weight of 'X'.

0.096kg of oxygen is compressed iso thermally and reversibly from 10^5 N.M^(-2) to 2 xx 10^(5) N.M^(2) at 300 K. Calculate the work done during the impression process.

0.092 g of an organic compound containing phosphorus gave 0.111 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentage of phosphorus in the organic compound.

0.08 g/2.901 L is a saturated solution of CaF_2 at 298 K temp. calculate K_(sp) of CaF_2.

0.0755 g of selenium vapours occupying a volume of 114.2 mL at 700^(@)C and 185 mm of Hg. The vapours are in equilibrium as: Se_(6(g))hArr3Se_(2(g)) Calculate: (i) Degree of dissociation of Se, (ii) K_(p), (iii) K_(c). Atomic weight of Se is 79.

0.075 g of a monobasic acid required 10 mL of N//12 NaOH solution for complete neutralisation. Calculate the molecular mass of the acid

0.06mole of KNO_(3) is added to 100cm^(3) of water at 298K . The enthalpy of KNO_(3) aq. Solution is 35.8kJ mol^(-1). After the solute is dissolved , the temperature of the solution will be

0.05g of a commercial sample of KClO_(3) on decomposition liberated just sufficient oxygen for complete oxidation of 20 mL CO at 27^(circ)C and 750 mm pressure. Calculate % of KClO_(3) in sample.

0.05 g of magnesium when treated with dilute HCI gave 51 mL of hydrogen at 27°C and 780 mm pressure. Calculate the equivalent mass of magnesium.

0.05 g of a gas at 750 mm pressure and 25°C occupy a volume of 46.5 mL. Calculate the molecular mass of the gas.

0.049 g of H_(2)SO_(4) is dissolved per litre of thegiven solution . Calculate the pH of the solution.

0.03 mole of Ca^(2+) ions is added to a litre of 0.01 M SO_(4)^(2-) solution. Will it cause precipitation of CaSO_(4) ?K_(sp) "for" CaSO_(4)=2.4xx10^(-5).

0.023 g of sodium metal is reacted with 100 cm^(3) of water. The pH of the resulting solution is

0.01 mole sucrose dissolve in .......... Litre water so it becomes 0.01solution.

0.01 mole solution of .10 volume. H_2O_2 is required to convert 0.01 mol PbS into PbSO_4 ?

0.01 mole of iodoform (CHI_(3)) reacts with Ag powder to produce a gas whose volume at NTP is

0.01 mole of FeS_n(iron (II) sulphide) required 0.06 mole of AO_(4)^(3-)for complete oxidation. The species formed are FeO, SO_2 and A^(2+) . Calculate the value of n.

01 mole of N_(2)O_(4) (g) was sealed in a tube under atmospheric conditions at 25^(@)C. Calculate the number of moles of NO_(2) (g) present if the equilibrium N_(2)O_(4) (g) hArr 2 NO_(2)is reached after some time (K_(p) = 0 14).