52671 + Interview Questions in Chemistry in Class 11 Page 1040 InterviewSolution

51951.	0.5 mole each of two ideal gases A(C_(v,m)=(5)/(2)R) and B(C_(v,m)=3R) are taken in a container and expanded reversibely and adibatically , during this process, temperature of gaseous mixture decreased from 350 K and 250 K. Find DeltaH (in "cal"//"mol") for the process :
Answer» `-100R` `-137.5R` `-375 R` NONE of these Answer :C

Discussion

51952.	0.5 mole each of H_2SO_4and CH_4 are kept in a container. A hole was made in the container. After 3 hours the order of partial pressures in the container will be:
Answer» <P>`P_(SO_2) gt P_(CH_4) gt P_(H_2)` `P_(CH_4) gt P_(SO_2) gt P_(H_2)` `P_(H_2) gt P_(SO_2) gt P_(CH_4)` `P_(H_2) gt P_(CH_4) gt P_(SO_2)` SOLUTION :`H_2 ` is LOST FASTER `:. P_(H_2)` is LEAST .

Discussion

51953.	0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 400 K reaction CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) equilibrium constant of K_c = 0.9 "mol L"^(-1). Calculate mol of CO_2 at equilibrium how much percentage of reaction completed ?
Answer» SOLUTION :`[CO_2]= 0.45 "MOL L"^(-1)`, 90% decomposition `[CO_2]`= 0.3 mol and 60% REACTION if `CaCO_3` and CaO used in calculation

Discussion

51954.	0.5 litres of an ideal gas is present at a pressure of 700 mm of Hg and 30^@C. Find its volume at standard temperature and pressure.
Answer» SOLUTION :0.415 It

Discussion

51955.	0.5 litres of an ideal gas is present at a pressure.
Answer» ANSWER :`0.415 I t`

Discussion

51956.	0.5 g of organic compound in Kjeldhal's method liberated ammonia, which nutralised 60 ml of 0.1 N H_(2)SO_(4) solution. The percentage of nitrogen in the compound is
Answer» `1.68` `16.8` `33.6` `8.4` Solution :% of `N=("No. of m.eq. of " NH_(3)" liberated"xx1.4)/("WEIGHT of organic compound")` `(60xx0.1xx1.4)/(0.5)=16.8 %`

Discussion

51957.	0.5 g of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in 150 mL of N//5 H_(2)SO_(4) solution. Excess sulphuric acid required 20 mL of 1N NaOH for complete neutralization. The percentage of NH_(3) in the ammonium chloride is :
Answer» 0.68 0.34 0.48 0.17 Answer :B

Discussion

51958.	0.5 g of bleaching powder was suspended in water an excess of KI added. On acidifying with dilute H_(2)SO_(4) iodine was liberated which required 50 mL of N//10 hypo solution. Calculate the percentage of available chlorine in bleaching powder.
Answer» SOLUTION :N//A

Discussion

51959.	0.5 g of an organic compound cantaining N on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralization of NH3 . The percentage of N in the compound is
Answer» 34.3 16.2 21.6 14.8 Solution :`% ofN_2 = (28xx0.129xx100)/(1000xx 0.5)=16.2`

Discussion

51960.	0.5 g of an organic compound containing nitrogen on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralisation of ammonia. The percentage of nitrogen in the compound is
Answer» 34.34 40 21.64 14.84 Solution :`%N = (1.4 XX N xx V)/(W)`

Discussion

51961.	0.5 g of a sample of bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4). I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O). Calculate the percentage of available Cl_(2) in bleaching powder.
Answer» SOLUTION :N//A

Discussion

51962.	0.5 g o fan impure sample of oxalate was dissolved in water and the solution made 100 mL on titration 10 mLof this solution requared 15 mL of N//20 KMnO_(4) solution . Calculate the percentage of pure oxalate in the sample
Answer» Solution :The reduction half reaction is `C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-)` `THEREFORE` Eq we of `C_(2)O_(4)^(2-)=(2xx12+4xx16)/(2)=44` Let `N_(1)` be the normality of the oxalte solution applying normality equation we have `15xx1//20(KmnO_(4))=N_(1)xx10(C_(2) O_(4))^(2-) therefore N_(1)=(15)/(10xx20)=3/40` N Amount of pure `C_(2)O_(4)^(2-)` present in 100 mL =`(33/1000)xx100=0.33 G` But amount of `C_(2)O_(4)^(2-)` present in impure sample =0.5 g (given ) PERCENTAGE of pure OXALATE `=(0.33)/(0.50)xx100=66`

Discussion

51963.	0.5 g mixture of oxalic acid (H_(2)C_(2)O_4)and some sodium oxalate (Na_(2)C_(2)O_4)with some impurities requires 40 ml of 0.1M NaOH for complete neutralization and 6ml of 0.2 M KMnO_4for complete oxidation. Calculate the % of Na_2C_(2)O_4in the mixture
Answer» `90%` ` 26.8%` `40%` `50%` Solution : No. of milli equivalents of `H_(2)C_(2)O_4` = No .of milli equivalents of NaOH = 4 No.of milli equivalents of `H_(2)C_(4)O_(4)+Na_(20)C_(2)O_(4)`= No. of milli equivalents of `KMnO_(4)= 6 xx 0.2 xx 5 = 6` Milli EQUIVALENT of `Na_(2)C_(2)O_(4)= 2` Weight of `Na_(2)C_(2)O_(4)=2xx10^(-3)xx67 =0.134` g % `Na_(2)C_2O_4=(0.134)/(0.5)xx100=26.8`

Discussion

51964.	0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In the reaction (I), which one is reduced ?
Answer» `CaOCl_(2)` `H_(2)SO_(4)` Both none of these Solution :N//A

Discussion

51965.	0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In reaction (II), Cl_(2) acts as :
Answer» REDUCING AGENT OXIDISING agent indicator both oxidising agent and indicator Solution :N//A

Discussion

51966.	0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI Starch forms iodo-starch complex in the given titration. The colour of the complex will be :
Answer» GREEN blue pale yellow milky white Solution :N//A

Discussion

51967.	0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI Percentage of available chlorine in bleaching powder is :
Answer» `35.5%` `71%` `17.25%` `50%` SOLUTION :N//A

Discussion

51968.	0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In the given titration, starch acts as :
Answer» OXIDISING agent indicator reducing agent catalyst Solution :N//A

Discussion

51969.	0.482 mol N_2 and 0.933 mol O_2 are placed in a 10 L vessel and allowed to form N_2O at constant temperature. Numerical valueof K_c for the reaction 2N_2(g)+O_2(g) leftrightarrow 2N_2O(g) is 2 times 10^-37 K mol^-1. Calculate the equilibrium concentration of nirous oxide.
Answer» SOLUTION :`6.6 TIMES 10^-21 MOL L^-1`

Discussion

51970.	0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 g of oxygen. What is the percentage composition of the compound ?
Answer» SOLUTION :MASS of compound TAKEN = 0.48 G Mass of boron present = 0.192 g Mass of oxygen present = 0.288 g Percent of B `= (("0.192 g"))/(("0.48g"))xx100=40%` Percentage of O `=(("0.288 g"))/(("0.48 g"))xx100=60%`.

Discussion

51971.	0.46g of an organic compound was analysed. The increase in mass of CaCl_(2) U-tube was 0.54g and potash bulb was 0.88g. The percentage composition of the compound is
Answer» `C=52.17%,H=13.04%,O=34.79%` `C=50%,H=50%` `C=32.19%,H=18.01%,O=49.8%` `C=72%,H=28%` Solution :MASS of `H_(2)O=0.54g` Mass of `CO_(2)=0.88g` % of `C=(12)/(44)xx(0.88)/(0.46)xx100=52.17%` % of `H=(2)/(18)xx(0.54)/(0.46)xx100=13.04%` % of `O=100-(52.17+13.04)=34.79`%

Discussion

51972.	0.465 g of an organic substance gave on combustion 1.32 g of CO_(2) and 0.315g of H_(2)O. Calculate the percentage of carbon and hydrogen in the compound.
Answer» ANSWER :C = 77.42%, H = 7.53 %

Discussion

51973.	0.45g organic compound by combustion, give 0.792g CO_(2) and 0.324g water. For this compound ammonia obtain from 0.24g compound in Kjeldahl's method absorb in 50 ml, 0.25N, H_(2)SO_(4). In neutralisation 77.0 mL, 0.25N is used. Determine the empirical formula of compound NaOH
Answer» SOLUTION :`C_(6)H_(5)NO`

Discussion

51974.	0.456 gof a metal gives 0.606gof its chloride. Calculate the equivalent mass of the metal.
Answer» Solution :Mass of the metal `=W _(1)=0.606g` `THEREFORE` Mass of CHLORINE `=W_(2)=0.606-0.456=0.15g` `0.15g ` of chlorine combine with `0.456 g` of metal `therefore 35.46 g` of chlorine will combine with `(0.456)/(0.15)xx35.46=107.76g "eq"^(-1)` Mass of chloride `=0.606-0.456=0.146g` `0.146 g` of chlorine combines with `0.456 g` of metal. `therefore35 .5 g` of chlorine will combines with `=(35.5xx0.456)/(0.146)` `=110.8g` of metal `therefore` equivalent mass of metal `=110 .8 g "EQU"^(-1)` .

Discussion

51975.	0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
Answer» Solution :Mass of the metal = `W_i = 0.456 G` Mass of the metal chloride =` W_2 =0.606 g` `therefore ` Mass of chloride `= W_2 - W_1` `=0.606 - 0.456 = 0.15 g` 0.15 g of chloride COMBINE with 0.456g of metal. `therefore ` 35.46 g of chloride will combine with `(0.456)/(0.15) xx 35.46= 107.79g eq^(-1)`.

Discussion

51976.	0.45 N & 0.6 N NaOH solution are mixed in 2:1 by volume. The amount of solute present in 1 lit of this solution is
Answer» 0.5 gm 25 gm 20 gm 5 g Solution :`(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))=N` `N=(0.45xx2+0.6xx1)/(3)=0.5N` implies 0.5 MOLES in 1 lit. = 20G

Discussion

51977.	0.45 g of an organic compound when analysed by combustion gave 1.10 g carbon dioxide and 0.30 g water. Calculate the persentage of carbon and hydrogen in it.
Answer» SOLUTION :Mass of the organic compound =0.45 g Mass of `CO_(2)` formed = 1.10 g Mass of `H_(2)O` formed = 0.30 g Percentage of `C=12/44 xx("Mass of carbon dioxide formed")/("Mass of compound")xx100` `=12/44xx((1.10 g))/((0.45 g))xx100=66.67 %` Percentage of `H=2/18 xx ("Mass of WATER formed")/("Mass of compound")xx100` `=2/18xx((0.30 g))/((0.45 g))xx100=7.41 %`.

Discussion

51978.	0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calculate the equivalent weight of the metal.
Answer» SOLUTION :Mass of the metal nitrate taken = 0.452 g Mass of the metal sulphate obtained = 0.4378 g SUPPOSE, equivalent mass of metal = E Using the relation, `("Mass of metal nitrate")/("Mass of metal sulphate") = ("Equivalent WEIGHT of metal nitrate")/("Equivalent weight of metal sulphate")` `0.452/0.4378 = (E+ 62)/(E + 48)` [Eq. weight of `NO_(3)^(-) = (14 + (16 xx 3))/1 = 62` [Eq. weight of `SO_(4)^(2-) = (32 + (16 xx 4))/2 = 48` or `0.452-0.4378 xx E = (0.4378 xx 62) -(0.452 xx 48) = 5.4476` or `E=5.4476/(0.452-0.4378) = 5.4476/0.0142 = 383.6`

Discussion

51979.	0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula.
Answer» Solution :Step I. Percentage of elements Percentage of carbon `= (12)/(44) XX ("Mass of "CO_(2))/("Mass of compound")xx100` `=(12)/(44)xx(0.44)/(0.45)xx100=26.67` Percentage of HYDROGEN `=(2)/(18)xx("Mass of" H_(2)O)/("Mass of compound")xx100` `=(2)/(18)xx(0.09)/(0.45)xx100=2.22` Percentage of oxygeb `= 100 - (26.67 + 2.22) = 100-28.89 = 71.11` Step II. Empirical formula of the compound `{:("ELEMENT","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",26.67,12,(26.67)/(12)=2.22,(2.22)/(2.22)=1.0,1),("H",2.22,1,(2.22)/(1)=2.22,(2.22)/(2.22)=1.0,1),("O",71.11,16,(71.11)/(16)=4.44,(4.44)/(2.22)=2.0,2):}` Empirical formula of the compound `= CHO_(2)` Step III. Molecular formula of the compound Empirical formula mass `= 12+1 +2 xx 16 = 45 u` Molecular mass = 90 AMU (Given) `n=("Molecular mass")/("Empirical formula mass")=((90u))/((45u))=2` Molecular formula `= n xx` Empirical formula `= 2 xx CHO_(2)=C_(2)H_(2)O_(4)`.

Discussion

51980.	0.4422g of an organic compound was Kjeldahlised and ammonia evolved was absorbed in 50 mL of semi-molar (0.5 M) H_(2)SO_(4). The residual acid required 131 mL of 0.25 M NaOH. Determine the percentange of nitrogen in the compound.
Answer» ANSWER :`54.61%`

Discussion

51981.	0.4037 g of an organic substance containing sulphur was heated with conc. nitric acid in a carius tube. On precipitation with BaCl_(2), 0.1963 g of BaSO_(4) was produced. Determine the percentage of sulphur in the compound.
Answer» SOLUTION :`S = 7.66%`

Discussion

51982.	0.40 g of an iodo-substituted organic compound gave 0.235 g of AgI by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).
Answer» SOLUTION :Weight of ORGANIC substance (w) = 0.40 g Weight of silver iodine (x) = 0.235 g `"Percentage of iodine"=(127)/(235) xx (x)/(w) xx 100=(127)/(235) xx (0.235)/(0.40) xx 100=31.75%`

Discussion

51983.	0.40 g of an organic compound containing phosphorus gave 0.555 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentange of phosphorusin the organiccompound.
Answer» ANSWER :`38.75%`

Discussion

51984.	0.4 mole of orthophosphoric acid and 1.0 mole of calcium hydroxide were allowed to react. Calculate the maximum number of moles of calcium phosphate formed.
Answer» Solution :The stoichiometric equation of `3Ca(OH)_(2)+2H_(3)PO_(4) to Ca_(3)(PO_(4))_(2)+3H_(2)O` 3 MOLES of `CA(OH)_(2)="2 moles of "H_(3)PO_(4)` 1 mole of `Ca(OH)_(2)="0.67 mole of "H_(3)PO_(4)` 0.6 mole of `Ca(OH)_(2)="0.4 mole of "H_(3)PO_(4)` `H_(3)PO_(4)` is the limiting regent. Hence, salt formed is dependent on the availability of acid only. 2 moles of `H_(3)PO_(4)="1 mole of "Ca_(3)(PO_(4))_(2)` 0.4 moles of `H_(3)PO_(4)=?` Number of moles of calcium phosphate formed `=0.4 xx 1/2=0.2"mole"`

Discussion

51985.	0.4 gm of polybasic acid H_(n)A (M.wt = 96) requires 0.5 gm NaOH for complete neutralisation. The number of replacable hydorgen atoms are (all the hydrogens are acidic)
Answer» SOLUTION :m EQ ACID = M eq BASE `(0.4)/(Nw//n)xx1000=(0.5)/(40//1)xx1000impliesn=3`

Discussion

51986.	0.4 g of an organic compound in Dumas method, gave 50 ml of nitrogen collected at 300 K and 745 mn pressure Calculate the percentage composition of nitrogen in the compound. Aqueous tension at 300 K is 15 mm.
Answer» 0.4332 0.6439 0.2242 0.1367 Solution :`V_0 =(739xx50 xx273)/(300 XX 760)=43.7 " ML"` `" % of " N_2=(28xx43.7 XX100)/(22,400 xx 0.4) =13.67`

Discussion

51987.	0.4g of a compound on complete combustion gave 56ml of CO_(2) at 760mm and 0^(@)C. The percentage of carbon in the compound is
Answer» 50 60 27.5 7.5 Answer :D

Discussion

51988.	0.4 g of an organic compound gave 0.188g of silver bromide by a halogen estimation method. The percentage of bromine in the compound is (at . Wts of Ag= 108, Br = 80)
Answer» 20 40 46 60 Answer :B

Discussion

51989.	0.4g ofan organiccompoundgave0.188g ofsilverbromidebyhalogenestimationmaethodThepercentageof brominein thecompoundis ___.( At .Massof Ag =108 , Br=80)
Answer» 0.398 0.46 0.2 0.4 Solution :`0.188 " G of SILVER" -= (0.188xx80)/(188) =0.08g " of "Bra_2` `therefore% of B r_2 =(0.08)/(0.4) xx 100=20`

Discussion

51990.	0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound
Answer» Solution :Weight of the compound =W = 0.3g Weight of `CO_(2)= W_(1) = 0.44g` Weight of `H_(2)O = W_(2) = 0.18g` % of carbon `=(W_(1) xx 12 xx 100)/(W xx 44) = (0.44 xx 12 xx 100)/(0.3 xx 44) = 40.0%` % of hydrogen `= (W_(2) xx 2xx 100)/(W xx 18) = (0.18 xx 2 xx 100)/(0.3 xx 18) = 6.67%` After calculating the percentage COMPOSITION of the given elements in the compound, if their sum is not 100, then the difference is assumed as oxygen `:.` % of oxygen `=[100-(40+6.67)]= 53.33%`

Discussion

51991.	0.395 g of an organic compound by carius method for the estimation of sulphur gave 0.582 g of BaSO_(4). Calculate the percentage of sulphur in the compound.
Answer» SOLUTION :Percentage of sulphur `=32/233xx("MASS of "BaSO_(4))/("Mass of COMPOUND")xx100=32/233xx0.582/0.395xx100=20.24 %`

Discussion

51992.	0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. Structure of B is
Answer» `CH_(3)-CH_(2)-CH=CH-CH_(3)` `CH_(3)-overset(CH_(3))overset(\|)(CH)-CH_(2)OH` `CH_(3)-CH_(2)-CH=CH_(2)` Answer :B

Discussion

51993.	0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. Structure of A is
Answer» `CH_(3)-CH_(2)-CH-CH_(2)-OH` `CH_(3)-overset(CH_(3))overset(\|)(C)H-CH_(2)OH` `CH_(3)-CH_(2)-UNDERSET(OH)underset(\|)(CH)-CH_(3)` `CH_(3)-underset(OH)underset(\|)overset(CH_(3))overset(\|)(C)-CH_(3)` ANSWER :B

Discussion

51994.	0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. From given data molecular formula of (A) is
Answer» `C_(5)H_(11)OH` `C_(4)H_(9)OH` `C_(6)H_(13)OH` `C_(4)H_(7)OH` Answer :B

Discussion

51995.	0.3780g of an organic chloro compound gave 0.5740g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound (Ag= 108, Cl= 35.5)
Answer» SOLUTION :`underset(35.5g)(Cl) overset(+)rarr underset(143.5g)(AgCl)` molecular mass of AgCl= 108 + 35.5 = 143.5g 143.5g AgCl contain 35.5g CHLORINE. `therefore` 0.5740g AgCl contain, Chlorine `=(35.5 xx 0.5740)/(143.5)gm` % `Cl = (35.5)/(143.5) xx (0.5740)/(0.3780) xx 100` =37.5661= 37.57

Discussion

51996.	0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer» SOLUTION :Mass of the COMPOUND `= 0.3780 g` Mass of SILVER chloride `=0.5740 g` PERCENTAGE of CHLORINE `=35.5/143.5xx("Mass of silver chloride")/("Mass of compound")xx100` `=35.5/143.5xx((0.5740 g))/((0.3780 g))xx100=37.57 %`

Discussion

51997.	0.378 g of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapour . Calculate the percentage of C and H.
Answer» Solution :MASS of ORGANIC compound =0.378 g Mass of `CO_(2)` formedc =0.264 g Mass of `H_(2)O` formed=0.162 g (i) PERCENTAGE of carbon 44 g of `CO_(2)` contains carbon =12 g 0.264 g of `CO_(2)` contains carbon =`12/44 xx 0.264 =0.072 g ` Percentage of hydrogen 18 g of `H_(2)O` contains hydrogen = 2g 0.16.2 of `H_(2)O` contains hydrogen =`2/18xx0.162 =0.018 g ` Percentage of hydrogen =`0.018/0.378 xx100 =4.76 % `

Discussion

51998.	0.3780 g of an organic chloro compound gave 0.5740 g silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Answer» Solution :Here, the MASS of the substance taken = 0.3780 g Mass of AGCL FORMED = 0.5740 g Now 1 MOLE of `AgCl -= 1` g atom of Cl or (108 + 35.5) = 143.5 g of `AgCl -= 35.5 g` of Cl Applying the relation, Percentage of chlorine `= (35.5)/(143.5) xx ("Mass of AgCl formed")/("Mass of substance taken") xx 100` `= (35.5)/(143.5) xx (0.5740)/(0.3780) xx 100 = 37.566%`.

Discussion

51999.	0.3780 g an organic compound gave 0.5740 g AgCl in Carius method. Calculate the percentage of chlorine in it.
Answer» ANSWER :`(35.5)/143.5 XX (0.5740)/0.3780 xx 100 = 37.57% CHLORINE'

Discussion

52000.	0.378 g od an organic compound containing carbon and hydrogen was subjected to cobustion by Liebig's method , the CO_(2) and H_(2)Oformed were passed through potash bulbs and CaCl_(2) (anhydrous) tube. At end of the experiment, the increase in the respective weights were 0.264 g and 0.162 g. Calculate the percentage of carbon and hydrogen.
Answer» Solution :Increase in weight of POTASH BULBS (mass pf `CO_(2)` evolved) `=0.264 G` Increase in weight of `CaCl_(2)` tube (mass of `H_(2)O` forward) `=0.162 g` `%` of `C=12/44xx("Mass of "CO_(2))/("Mass of compound")xx100=12/44xx0.264/0.378xx100= 19.05%` `%` of `H=12/18xx("Mass of "H_(2)O)/("Mass of compound")xx100=2/18xx0.162/0.378xx100= 4.76 %`

Discussion

Explore topic-wise InterviewSolutions in .

0.5 mole each of two ideal gases A(C_(v,m)=(5)/(2)R) and B(C_(v,m)=3R) are taken in a container and expanded reversibely and adibatically , during this process, temperature of gaseous mixture decreased from 350 K and 250 K. Find DeltaH (in "cal"//"mol") for the process :

0.5 mole each of H_2SO_4and CH_4 are kept in a container. A hole was made in the container. After 3 hours the order of partial pressures in the container will be:

0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 400 K reaction CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) equilibrium constant of K_c = 0.9 "mol L"^(-1). Calculate mol of CO_2 at equilibrium how much percentage of reaction completed ?

0.5 litres of an ideal gas is present at a pressure of 700 mm of Hg and 30^@C. Find its volume at standard temperature and pressure.

0.5 litres of an ideal gas is present at a pressure.

0.5 g of organic compound in Kjeldhal's method liberated ammonia, which nutralised 60 ml of 0.1 N H_(2)SO_(4) solution. The percentage of nitrogen in the compound is

0.5 g of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in 150 mL of N//5 H_(2)SO_(4) solution. Excess sulphuric acid required 20 mL of 1N NaOH for complete neutralization. The percentage of NH_(3) in the ammonium chloride is :

0.5 g of bleaching powder was suspended in water an excess of KI added. On acidifying with dilute H_(2)SO_(4) iodine was liberated which required 50 mL of N//10 hypo solution. Calculate the percentage of available chlorine in bleaching powder.

0.5 g of an organic compound cantaining N on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralization of NH3 . The percentage of N in the compound is

0.5 g of an organic compound containing nitrogen on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralisation of ammonia. The percentage of nitrogen in the compound is

0.5 g of a sample of bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4). I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O). Calculate the percentage of available Cl_(2) in bleaching powder.

0.5 g o fan impure sample of oxalate was dissolved in water and the solution made 100 mL on titration 10 mLof this solution requared 15 mL of N//20 KMnO_(4) solution . Calculate the percentage of pure oxalate in the sample

0.5 g mixture of oxalic acid (H_(2)C_(2)O_4)and some sodium oxalate (Na_(2)C_(2)O_4)with some impurities requires 40 ml of 0.1M NaOH for complete neutralization and 6ml of 0.2 M KMnO_4for complete oxidation. Calculate the % of Na_2C_(2)O_4in the mixture

0.482 mol N_2 and 0.933 mol O_2 are placed in a 10 L vessel and allowed to form N_2O at constant temperature. Numerical valueof K_c for the reaction 2N_2(g)+O_2(g) leftrightarrow 2N_2O(g) is 2 times 10^-37 K mol^-1. Calculate the equilibrium concentration of nirous oxide.

0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 g of oxygen. What is the percentage composition of the compound ?

0.46g of an organic compound was analysed. The increase in mass of CaCl_(2) U-tube was 0.54g and potash bulb was 0.88g. The percentage composition of the compound is

0.465 g of an organic substance gave on combustion 1.32 g of CO_(2) and 0.315g of H_(2)O. Calculate the percentage of carbon and hydrogen in the compound.

0.45g organic compound by combustion, give 0.792g CO_(2) and 0.324g water. For this compound ammonia obtain from 0.24g compound in Kjeldahl's method absorb in 50 ml, 0.25N, H_(2)SO_(4). In neutralisation 77.0 mL, 0.25N is used. Determine the empirical formula of compound NaOH

0.456 gof a metal gives 0.606gof its chloride. Calculate the equivalent mass of the metal.

0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.

0.45 N & 0.6 N NaOH solution are mixed in 2:1 by volume. The amount of solute present in 1 lit of this solution is

0.45 g of an organic compound when analysed by combustion gave 1.10 g carbon dioxide and 0.30 g water. Calculate the persentage of carbon and hydrogen in it.

0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calculate the equivalent weight of the metal.

0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula.

0.4422g of an organic compound was Kjeldahlised and ammonia evolved was absorbed in 50 mL of semi-molar (0.5 M) H_(2)SO_(4). The residual acid required 131 mL of 0.25 M NaOH. Determine the percentange of nitrogen in the compound.

0.4037 g of an organic substance containing sulphur was heated with conc. nitric acid in a carius tube. On precipitation with BaCl_(2), 0.1963 g of BaSO_(4) was produced. Determine the percentage of sulphur in the compound.

0.40 g of an iodo-substituted organic compound gave 0.235 g of AgI by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).

0.40 g of an organic compound containing phosphorus gave 0.555 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentange of phosphorusin the organiccompound.

0.4 mole of orthophosphoric acid and 1.0 mole of calcium hydroxide were allowed to react. Calculate the maximum number of moles of calcium phosphate formed.

0.4 gm of polybasic acid H_(n)A (M.wt = 96) requires 0.5 gm NaOH for complete neutralisation. The number of replacable hydorgen atoms are (all the hydrogens are acidic)

0.4 g of an organic compound in Dumas method, gave 50 ml of nitrogen collected at 300 K and 745 mn pressure Calculate the percentage composition of nitrogen in the compound. Aqueous tension at 300 K is 15 mm.

0.4g of a compound on complete combustion gave 56ml of CO_(2) at 760mm and 0^(@)C. The percentage of carbon in the compound is

0.4 g of an organic compound gave 0.188g of silver bromide by a halogen estimation method. The percentage of bromine in the compound is (at . Wts of Ag= 108, Br = 80)

0.4g ofan organiccompoundgave0.188g ofsilverbromidebyhalogenestimationmaethodThepercentageof brominein thecompoundis ___.( At .Massof Ag =108 , Br=80)

0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound

0.395 g of an organic compound by carius method for the estimation of sulphur gave 0.582 g of BaSO_(4). Calculate the percentage of sulphur in the compound.

0.3780g of an organic chloro compound gave 0.5740g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound (Ag= 108, Cl= 35.5)

0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.

0.378 g of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapour . Calculate the percentage of C and H.

0.3780 g of an organic chloro compound gave 0.5740 g silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

0.3780 g an organic compound gave 0.5740 g AgCl in Carius method. Calculate the percentage of chlorine in it.