52671 + Interview Questions in Chemistry in Class 11 Page 1042 InterviewSolution

52051.	0.25 g of an organic compound gave 30 cm^(3) of moist dinitrogen at 288 K and 745 mm pressure. Calculate the percentage of nitrogen . (Aq tension of 288 K =12.7 mm)
Answer» Solution :Mass of substance =0.25 g Volume of moist dinitrogen =`30cm^(3)` Temperature =288 K Pressure= `745-12.7 =732.3 mm` STEP I . To reduce the volume of `N_(2)` at S.T.P We KNOW, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `P_(2),T_(2),V_(2)` refer to S.T.P conditions) `V_(2)=(P_(1)V_(1))/(T_(1))xx (T_(2))/(P_(2))==(732.3xx30xx273)/(288xx760) =27.4 cm^(3)` Step II: Calculation of percentage of NITROGEN `22400 cm^(3)` of dinitrogenat S.T.P weight =28 g `27.4 cm^(3) ` of dinitronge at S.T.P weight `(28xx274)/22400 =0.034 g ` Percentage of nitrogen in organic compound `=0.034/0.25 xx100=13.6`

Discussion

52052.	0.2475 g of an organic substance gave on combustion 0.495 g of CO_(2) and 0.2025g of H_(2)O. Calculate the percentage of carbon and hydrogen in it.
Answer» ANSWER :C = 54.54%, H = 9.09%

Discussion

52053.	0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
Answer» SOLUTION :`P _("solute") = K_(H)x _("solute in solution")` At pressure `1.5` atm, `P_(1) = K_(H) x _(1)""…(1)` At pressure `6.0` atm,` P _(2) = K _(H) x _(2)""…(2)` Dividing equation (1) by (2) We get `(p _(1))/( p _(2)) = (x _(1))/( x _(2))` `(1.5)/(6.0) = (0.24)/(x _(2))` THEREFORE `x _(2) = (0.24 xx 6. 0)/(1.5) =0.96 g //L`

Discussion

52054.	0.2475 g of an organic compound gave on combustion 0.4950 g of carbon dioxide and 0.2025 g of water. Calculate the percentage of carbon and hydrogen in it.
Answer» Solution :WT. of organic COMPOUND `= 0.2475` g Wt. of `CO_2` produced `= 0.4950` g Wt. Of `H_2 O` produced `= 0.20 25` g Percentage of carbon `= (12)/(44) xx ("Wt. Of" CO_2)/("Wt. OFCOMPOUND") xx 100` `= (12)/(44) xx (0.34950)/(0.2475) xx 100 = 54.54` Percentage of hydrogen `= (2)/(18) xx ("Wt. of" H_2 O )/("Wt. of compound") xx 100` `= (2)/(18) xx (0.2025)/(0.2475) xx 100 = 9.09`

Discussion

52055.	0.246 g of the organic compound gave 22.4 cm^(3) of nitrogen gas at STP as determined by Dumas method. The percent of nitrogen in the compound is
Answer» 11.38 17.07 22.76 34.14 Answer :A

Discussion

52056.	0.246 g of an organic substance when heated with excess of fuming nitric acid and silver nitrate gave 0.2584g of silver bromide residue. Calculate the weight percentage of bromine in the organic compound
Answer» ANSWER :0.4467

Discussion

52057.	0.24 gm of an organic substance gave 45.6 c.c. of nitrogen at N.T.P. The percentage of nitrogen is
Answer» `63.6%` `23.8%` `53.6% ` `53.6%` Solution :PERCENTAGE of `N=(45.6 XX 28)/(22400)xx (100)/(0.24) = 23.8 %`

Discussion

52058.	0.24 g of organic compound containing phosphorous gave 0.66 g of Mg_(2)P_(2)0_(7) by the usual analysis. Calculate the percentage of phosphorous in the compound
Answer» Solution :Weight of an ORGANIC compound = 0.24 G Weight of `Mg_(2)P_(2)0_(7)`= 0.66 g 222 g of `Mg_(2)P_(2)0_(7)` contains= 62 g of P `"0.66 g contains"=(62)/(222) xx 0.66 "g of P"` `"PERCENTAGE of P"=(62)/(222) xx (0.66)/(0.24) xx 100=76.80%`

Discussion

52059.	0.24 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively
Answer» 4.58, 21.95 21.95,4.58 45.8,2.195 2.195,45.8 Solution :% of `C = ("wtof" CO_(2))/(44) xx (100)/(w) xx 12` % of `H = ("wtof" H_(2)O)//(18) xx (100)/(w) xx2`

Discussion

52060.	0.24 g of an organic compound gave 0.287 g of AgCl in the carius method. Calculate the percentage of chlorine in the compound.
Answer» SOLUTION :`CL = 25%`

Discussion

52061.	0.234g of an organic compound on heating with conc nitric acid and silver nitrate in carius furnace gace 0.84g or siver chloride. Calculate the weight percentageof chlorine in the compound.
Answer» ANSWER :80.8

Discussion

52062.	0.2346 g of an organic compound containing C, H & 0, on combustion gives 0.2754 g of H_(2)0 and 0.4488 g CO_(2) . Calculate the % composition of C, H & 0 in the organic compound.
Answer» Solution :WEIGHT of organic substance (W) = 0.2346 g Weight of water (x) = 0.2754 g Weight of `CO_(2)` (y) = 0.4488 g `"Percentage of CARBON" = (12)/(44) xx (y)/(w) xx 100 = (12)/(44) xx (0.4488)/(0.2346) xx 100 = 52.17%` `"Percentage of hydrogen" =(2)/(18) xx (x)/(w) xx 100=(2)/(18) xx (0.2754)/(0.2346) xx 100=13.04%` Percentage of oxygen =[100-(52.17+13.04)]=100-65.21=34.79%

Discussion

52063.	0.2325 g oforganiccompound wasanalysedfornitrogenby Dumas method. 31.7mLofmoist nitrogenwascolletedat 25^(@) Cand 755.8 mm of Hgpressure, Calculatethepercentageof nitrogenin thesample(Aqueoustensionof waterat 25^@Cis 23.8mm of Hg)
Answer» `18%` `19.7%` `15.1%` `20.2%` ANSWER :C

Discussion

52064.	0.2325 g of an organic compound was analysed for nitrogen by Duma's method. 31.7 mL of moist nitrogen gas was collected at 25^(@)C and 755.8 mm Hg pressure. Determine the percentage of nitrogen in the compound. The aqueous tension of water is 23.8 mm Hg at 25^(@)C.
Answer» Solution :Step I. Volume of `N_(2)` at N.T.P. `{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=31.7 mL,V_(2)=?),(P_(1)=755.8-23.8=732 mm,P_(2)=760 mm),(T_(1)=25+273=298 K,T_(2)=273 K):}` `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((732mm)XX(31.7 mL)xx(273 K))/((298 K)xx(760 mm))=27.97 mL` Step II. Percentage of NITROGEN `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("MASS of COMPOUND")xx100=28/22400xx27.97/0.2325xx100=15.1 %`

Discussion

52065.	0.22g of a metal chloride required 0.51g of AgNO_(3) to precipitate chloride completely. If the specific heat of metal is 0.057 ca g^(-1), find the molecular formula of metal chloride?
Answer» ANSWER :`MCl_(3)`

Discussion

52066.	0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to Fe^(2+) with SnCl_(2). The reduced solution required 20 mL of 0.1 N KMnO_(4) solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :
Answer» `55.5 %` `45.0%` `50.0 %` `40.0 %` Solution :Number of EQUIVALENT of IRON =Number of equivalents of `KMnO_(4)` `(NV)/(1000)=(0.1xx20)/(1000)=0.002` Mass of iron (pure)`=0.002xx55.5=0.111 G` % purity `=("Mass of pure iron")/("Mass of iron ore")xx100=(0.111)/(0.222)xx100=50%`

Discussion

52067.	0.22g of an organic compound on combustion in an atomosphere of carbon -dioxide gave 34 cm^(3) of moist nitrogen at 17^(0)C and 733.4mm pressure. It the aqueous tension at 17^(0)C is 13.4 mm, calculate the percentage of nitrogen in the compound
Answer» ANSWER :0.1727

Discussion

52068.	0.2174 g of the substance gave 0.5825 g of BaSO_(4) by Carius method. Calculate the percentage of sulphur.
Answer» ANSWER :`36.78%`

Discussion

52069.	0.2056 g of an organic compound gave on combustion 0.114 g of H_(2)O and 0.880 g of CO_(2_. Find the percentage of hydrogen and carbon in the organic compound.
Answer» SOLUTION :`C = 93.76%, H = 6.128`

Discussion

52070.	0.2046 g of an organic compound gave 30.4 mL of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the substance (Aqueous tension at 288 K is 12.7 mm).
Answer» ANSWER :`16.68%`

Discussion

52071.	0.2046 g of an organic compound gave 30.4 cm^(3) of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the compound (Aqueous tension at 288 K is 12.7 mm)
Answer» Solution :Step I. Volume of `N_(2)` at N.T.P. `{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=30.4 cm^(3),V_(2)=?),(P_(1)=732.7-12.7=720 mm,P_(2)=760 mm),(T_(1)=288 K,T_(2)=273 K):}` `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((720mm)XX(30.4 cm^(3))xx(273 K))/((288 K)xx(760 mm))=27.3 cm^(3)` Step II. Percentage of nitrogen `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of COMPOUND")xx100=28/22400xx27.3/0.2046xx100=16.68 %`

Discussion

52072.	0.2033 g of an organic compound in Dumas method gave 31.7 mL of moist N_(2) at 14^(@)C and 758 mm pressure. Percentage of N_(2) in the compound is (Aq. Tension at 14^(@)C = 14mm)
Answer» 0.1844 0.1689 0.156 0.16 Solution :`p_(N_(2)) = 758 - 14` `("at" 14^(@)C) (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` (at STP) % N ` = ("Volume of" N_(2) "at STP")/(22,400) xx (100)/(W) xx 28`

Discussion

52073.	0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?
Answer» `1.3g` `0.130g` `0.013g` `13g` Solution :After an equilibrium between `0.200g` of iodine and `100mL` of water is reached, the ADDITION of `150ML` of water will result in the further dissolution of `I_2`. VOLUME of solution after the addition of `150mL` of water `=100mL+150mL` `=250ML` According, `100mL` of solution contains `0.28g I_2`. `1mL` solution contains `(0.28g)/(1000ML)` Therefore, `250mL` of solution will contain `(0.28g)/(1000ml)xx250mL=0.070g` of `I_2` Undissolved iodine `=(0.200g)-(0.070g)` `=0.130g`

Discussion

52074.	0.20 g of an organic compound forms 0.5764 g of CO_(2) and 0.1512 g of water on combustion. Calculate the percentage of carbon and hydrogen in the compound.
Answer» Solution :Mass of the organic COMPOUND =0.20 g Mass of `CO_(2)` FORMED = 0.5764 g Mass of `H_(2)O` formed = 0.1512 g Percentage of `C=12/44 xx("Mass of CARBON DIOXIDE formed")/("Mass of compound")xx100` `=12/44xx((0.5764 g))/((0.20 g))xx100=78.6 %` Percentage of `H=2/18xx("Mass of water formed")/("Mass of compound")xx100` `=2/18xx((0.1512 g))/((0.20 g))xx100=8.4 %`

Discussion

52075.	0.2004 g of glucose gave on combustion 0.2940 g of CO_(2) and 0.1202 g of H_(2)O. Find the percentage composition.
Answer» Solution :Weight of ORGANIC compound = 0.2004 g Weight of carbon dioxide = 0.2940 g Weight of water = 0.1202 g Percentage of carbon 44 g of CO2 contains, carbon = 12 g `"0.2940 g of "CO_(2)" contains, carbon"=(12xx0.2940)/(44)xx(100)/(0.2004)=40.01` Percentage of HYDROGEN 18 g of `H_(2)O` contains, hydrogen = 2 g 0.1202 of `H_(2)O` contains, hydrogen `=(2xx0.1202)/(18)` Percentage of hydrogen `=(2xx0.1202)/(18)xx(100)/(0.2004)=6.66` Percentage of OXYGEN is therefore by difference, `=[100-(40.01+6.66)]=53.33`

Discussion

52076.	0.2 of an organic compound containing phosphorusgave 1.877 g of ammonium phosphomolybdate by usual analysis. Calculate the percentage of phosphorus in the organic compound.
Answer» <P> SOLUTION :`UNDERSET(1877g)((NH_(3))_(3)PO_(4). 12MoO_(3)) -= underset(31g)(P)` `:. % P = (31)/(1877) xx (1.877)/(0.2) = 15.5`

Discussion

52077.	0.2 mole of an alkane on complete combustion gave 26.4g of CO_(2). The molecular weight of alkane is
Answer» 16 30 44 58 Answer :C

Discussion

52078.	0.2 gram -mole of an unsaturated hydrocarbon on complete combustion produces 26.4 g of CO_(2) . The molecular weight of the hydro-carbon is
Answer» 42 88 46 30 Answer :A

Discussion

52079.	0.2 g off an organic compound contains C, H and O. On combustion, it yields 0.15 g CO_(2) and 0.12" g "H_(2)O. The percentage of C, H and O respectively is
Answer» `C=15%,H=20%,O=65%` `C=10%,H=8.2%,O=81.8%` `C=12.2%,H=8.8%,O=79%` `C=20%,H=6.66%,O=73.34%` SOLUTION :% of `C=(12)/(44)xx(0.15)/(0.2)xx100=20%` % of `H=(2)/(18)xx(0.12)/(0.2)xx100=6.66%` % of `O=100-(20+6.66)=73.34%`

Discussion

52080.	0.2g of an organic compound on analysis give 0.147g of carbondioxide, 0.12 g of water and 74.6 c,c of nitrogen at S.T.P. Calculate the weight percentages of constituents.
Answer» Solution :Weight of COMPOUND = w = 0.2 h , volume of `N_(2)` at STP = `v_(2)` = 74.6 cc `%C = (w_(1)xx12xx100)/(wxx44) = (0.147xx12xx100)/(0.2xx44) = 20.04%` `%H = (w_(2)xx2xx100)/(wxx18) = (0.12xx2xx100)/(0.2xx18) = 6.66%` `%N = (v_(2)xx25 xx100)/(wxx22400) = (0.74.6)/(8xx0.2) = 46.63%` Weight percentage of oxygen is obtained INDIRECTLY `%O = 100 = (%C+%H+%N) = 100-73.33 = 26.67%`

Discussion

52081.	0.2 g of an organic compound on complete combustion produces 0.18 g of water. The precentage of hydrogen in it is:
Answer» 5 10 15 20 Solution :Percentage of `H` `=2/18xx("MASS of "H_(2)O)/("Mass of COMPOUND")xx100` `=2/18xx0.18/0.20xx100=10`

Discussion

52082.	0.2 g ofa gas 'x' occupy a volume of 440 ml . If 0.1 g of CO_2 gas occupy a volume of 320 ml at the same temperature and pressure , then the gas 'x' could be
Answer» `C_(4) H_(10)` `NO_(2)` `O_(2)` `SO_(2)` ANSWER :D

Discussion

52083.	0.1M CH_(3)COOH solution is titrated against 0.05M NaOH solution. Calculate pH at 1//4th and 3//4th stages of neutralization of acid. The pH for 0.1M CH_(3)COOH is 3.
Answer» ANSWER :`4.5228, 5.4771;`

Discussion

52084.	In an estimation of sulphur by Carius method, 0.2175 g of the substance gave 0.5825 g of BaSO_(4) , calculate the percentage composition of S in the compound.
Answer» SOLUTION :`C = 30.6%, 4 = 3.8%, CL = 45.2%, O = 20.4%`

Discussion

52085.	0.196 g of an organic compound gave 0.22 g of CO_(2) and 0.0675 g of H_(2)O. In carius determination, 0.3925 g of the substance gave 0.717 g of dry AgCl. Find the percentage composition of the substance.
Answer» SOLUTION :`C = 30.6%, 4 = 3.8%, CL = 45.2%, O = 20.4%`

Discussion

52086.	0.189g of an organic substance gave in a Carius determination 0.287 g of silver chloride. What is the percentage of chlorine in the given compound ?
Answer» ANSWER :`37.57%`

Discussion

52087.	0.1890 g of organic compound containing chlorine gave in carius method 0.2870 g of silver chloride. Find the percentage of chlorine in the compound.
Answer» Solution :PERCENTAGE of bromine `=35.5/143.5xx ("MASS of AGCL")/("Mass of compound")xx100=35.5/143.5xx0.2870/0.1890xx100=37.56%`

Discussion

52088.	0.185 g of an organic compound when treated with Conc. HNO_(3) and silver nitrate gave 0.320g of silver bromide. Calculate the % of bromine in the compound. (Ag=108, Br=80).
Answer» SOLUTION :Weight of organic substance (w) = 0.185 g , Weight of silver bromide (X) = 0.320 g `"PERCENTAGE of bromine"=(80)/(188) xx (x)/(w) xx 100=(80)/(188) xx (0.32)/(0.185) xx 100=73.6%`

Discussion

52089.	0.16g of methane was subjected to combustion at 27^(@)C in a bomb calorimeter system. The temperature of the calorimeter system ( including water ) was found to rise by 0.5^(@)C. Calculate the heat of combustion of methane at (i) constant volume, and (ii) constant pressure . The thermal capacity of the calorimeter system is17.7 k J K^(-1) ( R = 8.314 kJ K^(-1) mol^(-1))
Answer» <P> Answer :`q_(v) = - 885 kJ mol^(-1) , q_(p) = -890 kJ mol^(_1)`

Discussion

52090.	0.16g of an organic substance was heated in Carius tube and the sulphuric acid formed was precipitated as BaSO_(4) with BaCl_(2). The weight of the dry BaSO_(4) was 0.35 g. Find the percentage of sulphur.
Answer» ANSWER :`30.04%`

Discussion

52091.	0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14^(@)C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14^(@)C =12 mm of Hg).
Answer» SOLUTION :Weight of Organic compound =0.168 g `"Volume of moist nitrogen" (V_(1))=31.7 ml=31.7 xx 10^(-3) L` `"Temperature" (T_(1)) = 14^(@)C = 14 + 273 = 287 K` Pressure of Moist nitrogen (P) =758 mm Hg Aqueous tension at `14^(@)C` = 12 mm of Hg `:.` Pressure of DRY nitrogen=(`P-P^(1)`)=758-12=746 mm of Hg `(P_(1)V_(1))/(T_(1))=(P_(0)V_(0))/(T_(0))` `V_(0)=(746 xx 31.7 xx 10^(-3))/(287) xx (273)/(760)` `V_(0)=29.8 xx 10^(-3) L` `"Percentage of nitrogen"=((28)/(22.4) xx (V_(0))/(W)) xx 100` `=(28)/(22.4) xx (29.58 xx 10^(-3))/(0.1688) xx 100=21.90%`

Discussion

52092.	0.16 g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 ml. Calculate the percentageof N_(2)H_(4) that has reacted with water in this solution . The K_(b) for N_(2)H_(4) is 4.0xx10^(-6)M.
Answer» Solution :`N_(2)H_(4)+H_(2)O hArr N_(2)H_(5)^(+) + OH^(-)` Conc. Of `N_(2)H_(4) = 0.16 g ` in500 ML = 0.32 g `L^(-1) = (0.32)/(32) ` mol `L^(-1) = 0.01 M = 10^(-2) M` Suppose x mol `L^(-1)` is the amount of hydrazine reacted. Then `{:(,N_(2)H_(4) ,+ ,H_(2)O,hArr,N_(2)H_(5)^(+),+,OH^(-),,),("Initial conc.",10^(-2)M,,,,,,,,),("Conc. after reaction",10^(-2) - x ,,x,,x,,,,):}` `K_(b) = ([N_(2)H_(5)^(+)][OH^(-)])/([N_(2)H_(4)])` `4.0xx10^(-6) = (x^(2))/(10^(2)-x) ~= (x^(2))/(10^(2)) or x^(2)=4.0xx10^(-8) or x=2.0xx10^(-4)` mol `L^(-1)` `:.` % of hydrazine reacted with water `=(2.0xx10^(-4))/(10^(-2))xx100=2%`

Discussion

52093.	0.16 g of an organic compound was heated in a carius tube and H_(2)SO_(4) acid formed was precipitated with BaCl_(2). The mass of BaSO_(4) was 0.35 g. Find the percentage of sulphur.
Answer» Solution :Weight of organic SUBSTANCE (w) = 0.16 G Weight of BARIUM SULPHATE (x) = 0.35 g `"Percentage of Sulphur"=(32)/(233) xx (x)/(w) xx 100=(32)/(233) xx (0.35)/(0.16) xx 100=30.04%`

Discussion

52094.	0.16 g a dibasic acid required 25 mL of decinormal NaOH solution for complete neutralisation. The molecular mass of the acid is :
Answer» 32 64 128 256 Solution :N//A

Discussion

52095.	0.156g of an organic compound in carius tube method gave 0.235g of silver iodide. Calculate the percent weight of halogen in the compound .
Answer» ANSWER :81.4

Discussion

52096.	0.157g of a certain gas collected over water occupies a volume of 135ml at 27°C and 750mm of Hg. Assuming ideal behaviour, the molecular weight of the gas is (aqueous tension at 27°C is 26.7 mm of Hg)
Answer» 30 32 28 16 Answer :C

Discussion

52097.	0.1570 g of an organic compound was heated with fuming nitric acid in carius tube. Addition of excess of barium chloride yielded 0.4813 g of barium sulphate as a white precipitate. Determine the percentage of sulphur in the compound.
Answer» Solution :PERCENTAGE of SULPHUR `=32/233xx("MASS of "BaSO_(4))/("Mass of COMPOUND")xx100=32/233xx0.4813/0.1570xx100=42.1 %`

Discussion

52098.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)
Answer» SOLUTION :Pyridiniumchloride = 0.15 mole Pyridine 500 ` cm ^(3)"of "0.2 M = 0.1 ` mole ` pOH = pK_b +log "" ([s]) /(["Base" ]) ` ` =- log (1.5 xx 10 ^(-9) ) +log"" (0.15)/(0.1) ` ` =9- log 1. 5+ log 1.5 =9` ` PH =14- 9 = 5`

Discussion

52099.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution . Calculate pH and hydroxyl ion concentration in the resulting solution assuming no change in the volume (k_(b) forpyridine = 1.5xx10^(-9)).
Answer» Solution :Pyridine is a weak base. Thus, pyridine + pyridine chloride solution is a basic BUFFER. Hence, `pOH = pK_(b) + log .(["Salt"])/(["Base"])` `pK_(b) = - log K_(b) = - log (1.5xx10^(-9))=9-0.1761=8.8239` [Pyridine ] = 0.23 M (Given ), [ Pyridinium chloride ] = `(0.15)/(500) xx 1000 = 0.30 M` `:. pOH = 8.82+ log. (0.30)/(0.20) = 8.82 + 0.1761 = 8.896` i.e., `- log [OH^(-)]= 8.896or log [OH^(-)] = - 8.896 = BAR(9) . 104 or [OH^(-) ] = 1.271xx10^(-9)` `[OH^(-)] ` from `H_(2)O = 10^(-7)` M cannot be neglected. Hence, total `[OH^(-)]=1.27xx10^(-9) + 10^(-7)= 10^(-9) (1.27+100) = 111.27 xx 10^(-9) M = 1.1127 xx 10^(-7) M` `[H^(+)] = (K_(w))/([OH^(-)])=(10^(-14))/(1.1127xx10^(-7))=8.987 xx 10^(-8)M` `PH = - log [H^(+)] = - log (8.987 xx 10^(-8))=8-0.9536= 7.0464`.

Discussion

52100.	015mole of CO taken in 25 l flask is maintained at 750 K along with catalyst so that the following reaction can take place : CO (g) + 2 H_(2) (g) hArr CH_(3)OH (g) Hydrogen is introduced until the total pressure of the system is 85 atmosphere at equilibrium and 008 mole of methanol is formed . Calculate (i) K_(p) and K_(c) and (ii) the final pressure if the same amount of CO and H_(2) as before are used but with no catalyst so that the reaction does not take place.
Answer» Solution :`{:((i),CO,+,2H_(2),hArr,CH_(3)OH) ,("Intialy :",015 "mole",,,,) ,("At eqm:",015 - 008 "mole",,,,), (,=0017 "mole",,,,):}` Total volume, `V= 25 L` Total PRESSURE` P+ 85 "atm", T= 750 "K" ` `" APPLYING "PV= nRT,` we get` 85 xx25 = n xx0* 0821 * 750or n = 0* 345 "mole" ` ` :."No. of moles of " H_(2) " at equilibrium "= 0* 345 - ( 0* 017 + 008 ) = 0 248 " mol "` ` P_(CO)= (0* 017 )/(0345 ) xx 85 "atm"= 0* 42 "atm" ` ` p_(H_(2)) = ( 0248 )/(0 345) xx85 "atm " = 6 11 "atm "` ` P _(CH_(3)OH) = (08)/(0345) xx 85 "atm"= 1 97 "atm" ` ` K_(p) = (P_(CH_(3)OH))/(P_(CO) xx P_(H_(2)))= (197 )/(042 xx (6* 11)^(2) )= 0* 1256 ` ` K_(c)= ([CH_(3) OH])/([CO][H_(2)]^(2))=(0* 08 //25)/((0 017 //25 )(0248 //25)^(2)) = 478 2` n(ii) No. of moles of `H_(2) " taken intially " = 0* 248 + 2 xx 0* 08 = 0* 308 ` No. of moles of CO taken intially `= 015` ApplyingPV = nRt , ` P xx 25 = 0* 458 xx 0* 0821 xx 750or P = 11* 28 "atm " `.

Discussion

Explore topic-wise InterviewSolutions in .

0.25 g of an organic compound gave 30 cm^(3) of moist dinitrogen at 288 K and 745 mm pressure. Calculate the percentage of nitrogen . (Aq tension of 288 K =12.7 mm)

0.2475 g of an organic substance gave on combustion 0.495 g of CO_(2) and 0.2025g of H_(2)O. Calculate the percentage of carbon and hydrogen in it.

0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.

0.2475 g of an organic compound gave on combustion 0.4950 g of carbon dioxide and 0.2025 g of water. Calculate the percentage of carbon and hydrogen in it.

0.246 g of the organic compound gave 22.4 cm^(3) of nitrogen gas at STP as determined by Dumas method. The percent of nitrogen in the compound is

0.246 g of an organic substance when heated with excess of fuming nitric acid and silver nitrate gave 0.2584g of silver bromide residue. Calculate the weight percentage of bromine in the organic compound

0.24 gm of an organic substance gave 45.6 c.c. of nitrogen at N.T.P. The percentage of nitrogen is

0.24 g of organic compound containing phosphorous gave 0.66 g of Mg_(2)P_(2)0_(7) by the usual analysis. Calculate the percentage of phosphorous in the compound

0.24 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively

0.24 g of an organic compound gave 0.287 g of AgCl in the carius method. Calculate the percentage of chlorine in the compound.

0.234g of an organic compound on heating with conc nitric acid and silver nitrate in carius furnace gace 0.84g or siver chloride. Calculate the weight percentageof chlorine in the compound.

0.2346 g of an organic compound containing C, H & 0, on combustion gives 0.2754 g of H_(2)0 and 0.4488 g CO_(2) . Calculate the % composition of C, H & 0 in the organic compound.

0.2325 g oforganiccompound wasanalysedfornitrogenby Dumas method. 31.7mLofmoist nitrogenwascolletedat 25^(@) Cand 755.8 mm of Hgpressure, Calculatethepercentageof nitrogenin thesample(Aqueoustensionof waterat 25^@Cis 23.8mm of Hg)

0.2325 g of an organic compound was analysed for nitrogen by Duma's method. 31.7 mL of moist nitrogen gas was collected at 25^(@)C and 755.8 mm Hg pressure. Determine the percentage of nitrogen in the compound. The aqueous tension of water is 23.8 mm Hg at 25^(@)C.

0.22g of a metal chloride required 0.51g of AgNO_(3) to precipitate chloride completely. If the specific heat of metal is 0.057 ca g^(-1), find the molecular formula of metal chloride?

0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to Fe^(2+) with SnCl_(2). The reduced solution required 20 mL of 0.1 N KMnO_(4) solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :

0.22g of an organic compound on combustion in an atomosphere of carbon -dioxide gave 34 cm^(3) of moist nitrogen at 17^(0)C and 733.4mm pressure. It the aqueous tension at 17^(0)C is 13.4 mm, calculate the percentage of nitrogen in the compound

0.2174 g of the substance gave 0.5825 g of BaSO_(4) by Carius method. Calculate the percentage of sulphur.

0.2056 g of an organic compound gave on combustion 0.114 g of H_(2)O and 0.880 g of CO_(2_. Find the percentage of hydrogen and carbon in the organic compound.

0.2046 g of an organic compound gave 30.4 mL of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the substance (Aqueous tension at 288 K is 12.7 mm).

0.2046 g of an organic compound gave 30.4 cm^(3) of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the compound (Aqueous tension at 288 K is 12.7 mm)

0.2033 g of an organic compound in Dumas method gave 31.7 mL of moist N_(2) at 14^(@)C and 758 mm pressure. Percentage of N_(2) in the compound is (Aq. Tension at 14^(@)C = 14mm)

0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?

0.20 g of an organic compound forms 0.5764 g of CO_(2) and 0.1512 g of water on combustion. Calculate the percentage of carbon and hydrogen in the compound.

0.2004 g of glucose gave on combustion 0.2940 g of CO_(2) and 0.1202 g of H_(2)O. Find the percentage composition.

0.2 of an organic compound containing phosphorusgave 1.877 g of ammonium phosphomolybdate by usual analysis. Calculate the percentage of phosphorus in the organic compound.

0.2 mole of an alkane on complete combustion gave 26.4g of CO_(2). The molecular weight of alkane is

0.2 gram -mole of an unsaturated hydrocarbon on complete combustion produces 26.4 g of CO_(2) . The molecular weight of the hydro-carbon is

0.2 g off an organic compound contains C, H and O. On combustion, it yields 0.15 g CO_(2) and 0.12" g "H_(2)O. The percentage of C, H and O respectively is

0.2g of an organic compound on analysis give 0.147g of carbondioxide, 0.12 g of water and 74.6 c,c of nitrogen at S.T.P. Calculate the weight percentages of constituents.

0.2 g of an organic compound on complete combustion produces 0.18 g of water. The precentage of hydrogen in it is:

0.2 g ofa gas 'x' occupy a volume of 440 ml . If 0.1 g of CO_2 gas occupy a volume of 320 ml at the same temperature and pressure , then the gas 'x' could be

0.1M CH_(3)COOH solution is titrated against 0.05M NaOH solution. Calculate pH at 1//4th and 3//4th stages of neutralization of acid. The pH for 0.1M CH_(3)COOH is 3.

In an estimation of sulphur by Carius method, 0.2175 g of the substance gave 0.5825 g of BaSO_(4) , calculate the percentage composition of S in the compound.

0.196 g of an organic compound gave 0.22 g of CO_(2) and 0.0675 g of H_(2)O. In carius determination, 0.3925 g of the substance gave 0.717 g of dry AgCl. Find the percentage composition of the substance.

0.189g of an organic substance gave in a Carius determination 0.287 g of silver chloride. What is the percentage of chlorine in the given compound ?

0.1890 g of organic compound containing chlorine gave in carius method 0.2870 g of silver chloride. Find the percentage of chlorine in the compound.

0.185 g of an organic compound when treated with Conc. HNO_(3) and silver nitrate gave 0.320g of silver bromide. Calculate the % of bromine in the compound. (Ag=108, Br=80).

0.16g of an organic substance was heated in Carius tube and the sulphuric acid formed was precipitated as BaSO_(4) with BaCl_(2). The weight of the dry BaSO_(4) was 0.35 g. Find the percentage of sulphur.

0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14^(@)C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14^(@)C =12 mm of Hg).

0.16 g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 ml. Calculate the percentageof N_(2)H_(4) that has reacted with water in this solution . The K_(b) for N_(2)H_(4) is 4.0xx10^(-6)M.

0.16 g of an organic compound was heated in a carius tube and H_(2)SO_(4) acid formed was precipitated with BaCl_(2). The mass of BaSO_(4) was 0.35 g. Find the percentage of sulphur.

0.16 g a dibasic acid required 25 mL of decinormal NaOH solution for complete neutralisation. The molecular mass of the acid is :

0.156g of an organic compound in carius tube method gave 0.235g of silver iodide. Calculate the percent weight of halogen in the compound .

0.157g of a certain gas collected over water occupies a volume of 135ml at 27°C and 750mm of Hg. Assuming ideal behaviour, the molecular weight of the gas is (aqueous tension at 27°C is 26.7 mm of Hg)

0.1570 g of an organic compound was heated with fuming nitric acid in carius tube. Addition of excess of barium chloride yielded 0.4813 g of barium sulphate as a white precipitate. Determine the percentage of sulphur in the compound.

0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)

0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution . Calculate pH and hydroxyl ion concentration in the resulting solution assuming no change in the volume (k_(b) forpyridine = 1.5xx10^(-9)).