52671 + Interview Questions in Chemistry in Class 11 Page 1041 InterviewSolution

52001.	0.37 gm of alcohol when treated with CH_3MgI liberates 112 ml of CH_4 at STP. The molecularformula of alcohol is
Answer» `C_4H_9OH` `C_5H_(11)OH` `C_3H_7OH` `C_2H_5OH` SOLUTION :Molecilar FORMULA

Discussion

52002.	0.376g of aluminium reacted with an acid to displace 0·468 litre of hydrogen at NTP. Find the equivalent volume of hydrogen if the equivalent weight of Al is 9.
Answer» Solution :EQUIVALENT of Al = equivalent of HYDROGEN. `therefore ("weight of Al")/("EQ. weight of Al")=("volume of hydrogen at NTP")/("equivalent volume of hydrogen")` `(0.376)/(9) = (0.468)/("V(litre)")` `therefore"" V=11.2` LITRES.

Discussion

52003.	0.36g of an organic compound on combustion gave 1.1g of CO_(2) and 0.54g of H_(2)O. The percentages of carbon and Hydrogen in the compound are
Answer» 75,25 60,40 83.33,16.67 77.8,22.2 Answer :C

Discussion

52004.	0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be
Answer» 1 5 8 13 Solution :`0.365 g HCL =(0.365)/(36.5) "mole" = 0.01 "mole"` `100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100` `=0.02 ` mole NaOH left unneutralized = 0.01 mole Volume of solution = 100 ml `:.` Molarity of NaOH in the solution `=(0.01)/(100)xx1000=0.1M = 10^(-1)M` `:. [H^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13`

Discussion

52005.	0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into 100 ml of N/5 H_(2)SO_(4). The excess acid required 154 ml of N/10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
Answer» Solution :Volume of `N//5 H_(2)SO_(4)` solution taken = 100 mL The volume of `N//5 H_(2)SO_(4)` neutralised by N/10 NAOH can be obtained as follows : `N_("acid")xxV_("acid")=N_("ALKALI")xxV_("alkali")` `(N)/(5)xx V_("acid")=(N)/(10)xx154mL` `V_("acid")=(154)/(10)xxmL` = 77 NL. Therefore, Volume of `N//% H_(2)SO_(4)` used for neutralisting ammonia `=(100-77)mL` `=23mL` Then, `"Percentage of nitrogen in the sample "="1.4"xx"Normality volume of acid"` `("used for neutralising "NH_(3))/("Mass of the compound taken")` Percentage of nitrogen in the sample `=(1.4xx1//5xx23)/(0.35)=18.4`

Discussion

52006.	0.35 g of an organic acid was kjeldahlised and the ammonia obtained was passed into 100 cm^(3) of N//5H_(2)SO_(4). The excess of the acid required 154 cm^(3) of N//10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
Answer» Solution :Volume of unused ACID. `ubrace(N_(1)V_(1))_("Acid") EQUIV ubrace(N_(2)V_(2))_("Base")` `1/5xxV=1/10xx154` or `V=77 cm^(3)`. Step II. PERCENTAGE of NITROGEN. Volume of acid used `=100-77=23 cm^(3)` Percentage of `N=(1.4xx"Normality of acid used"xx"Volume of acid used")/("Mass of compound")=(1.4xx0.2xx23)/035=18.4 %`

Discussion

52007.	0.35 g of a metal was dissolved in 50 mL N-acid. The whole solution then required 20.85 mL of normal alkaline solution to neutralise the excess of the acid. Calculate the equivalent mass of the metal.
Answer» SOLUTION :N//A

Discussion

52008.	0.33g of an organic compound containing phosphorous gave 0.397 g of Mg_(2)P_(2)0_(7) by the analysis. Calculate the percentage of P in the compound.
Answer» SOLUTION :WEIGHT of organic compound = 0.33g , Weight of `Mg_(2)P_(2)0_(7)`= 0.397g 222 g of `Mg_(2)P_(2)0_(7)` contains 62 g of phosphorous. `:. 0.397 "g of" Mg_(2)P_(2)0_(7) "will contain" (62)/(222) xx 0.397 "g of P"`. `0.33 "g of organic compound contains" (62)/(222) xx 0.397 "g of P"` `:. 100 "g of organic compound will contain" (62)/(222) xx (0.397)/(0.33) xx 100=(2,461.4)/(73.26)=33.59%` Percentage of phosphorous=33.59%

Discussion

52009.	0.32g of metal gave on treatment with an acid 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal
Answer» 58 32 11.2 24 Solution :EQUIVALENT WEIGHT `=("Mass of METAL" xx 11200)/("Volume in mL of HYDROGEN")` Given, mass of metal `= 0.32g` volume of hydrogen at NTP `= 112` mL Equivalent weight`=(0.32xx11200)/(112)=32`

Discussion

52010.	0.33 g of an organic compound was analysed for nitrogen by Kjeldhal's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4).The excess acid required 25 ml of 0.1 M NaOH for neutralization. The Percentage of nitrogen in the given compound is
Answer» 11.5 23 12. 5 14.5 Solution :VOLUME of `H_(2)SO_(4)` used = 25 ml NORMALITY of `H_(2)SO_(4) = 0.1` N

Discussion

52011.	0.33 mol cholesterol gives 9 mole CO_(2) on combustion. It was observed that cholesterol contains 83.85%C, 12%H and 4.15%O. Find its molecular formula and molecular mass.
Answer» Answer :`C_(27)H_(46)O; 386`

Discussion

52012.	0.32g of an organic compound containing sulphur produces 0.233g of BaSO_(4). Percentage of sulphur in the compound is
Answer» 20 50 10 80 Answer :C

Discussion

52013.	0.324 g of copper was dissolved in nitric acid and the copper nitrate so produced was burnt till all copper nitrate converted to 0.406 g of copper oxide. Calculate the equivalent weight of copper.
Answer» Solution :`UNDERSET(0.324 "g")(CU+ HNO_3) to "copper nitrate" tounderset(0.406 "g") "copper oxide"` Suppose the EQUIVALENT weight of copper `(Cu)` is `E_(Cu)`. Equivalent of Cu `=` equivalent of copper oxide `(0.324)/(E_(Cu) )= (0.406)/("eq. wt. of copper oxide")` `(0.324)/( E_(Cu) ) = ( 0.406)/("eq. wt. of Cu + eq. wt. of O")= ( 0.406)/( E_(Cu ) +8)` Hence the equivalent weight of Cu is `31.60`.

Discussion

52014.	0.32 g of metal on treatment with acid gave 112mL of hydrogen at NTP. The equivalent weight of metal is
Answer» 58 32 11.2 24 Solution :0.32 gm METAL `RARR` 112 ml of hydrogen ? `"" rarr` 11,200 ml of hydrogen = 32 gm

Discussion

52015.	0.3160 g of an organic substance was heated with fuming nitric acid in a carius tube to convert phosphorus present into phosphoric acid. Addition of magnesia mixture formed MgNH_(4)PO_(4) which upon heating gave 0.1697 g Mg_(2)P_(2)O_(7). Calculate the percentage of phosphorus in the compound.
Answer» <P> Solution :Percentage of PHOSPHORUS `=62/222xx("Mass of "Mg_(2)P_(2)O_(7))/("Mass of compound")xx100=62/222xx0.1697/0.3160xx100=15 %`

Discussion

52016.	0.316 g of an organic compound gives 0.466 g of barium sulphate by carius method. Calculate the percentage of sulphur?
Answer» SOLUTION :`S = 20.25%`

Discussion

52017.	0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).
Answer» Solution :MASS of the substance taken = 0.316 g Mass of `BaSO_(4)` FORMED = 0.466 g Molecular mass of `BaSO_(4) = 137 + 32 + 64 = 233` Then, mass of S in 0.466 g of `BaSO_(4)` `=(0.466xx32)/(233)g` Percentage of S in compound `=(0.466xx32xx100)/(233xx0.316)` = 20.25

Discussion

52018.	0.303 g of an organic compound was analysed for nitrogen by Kjeldahl's method. The ammonia evovled was absorbed in 50 ml. of 0.1N H_(2)SO_(4). The excess acid required 25ml of 0.1N NaOH for neutralisation. Calculate the percentage of nitrogen present in the compound
Answer» ANSWER :0.1155

Discussion

52019.	0.303 g of an organic compound was analysed for nitrogen by Kjedahl's method. The ammonia gave 0.2584 g of silver bromide Calculate the weight percentage of bromine in the organic compound.
Answer» ANSWER :0.1155

Discussion

52020.	0.303 g fo sample was analysed for nitrogen by Kjeldahl's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4). The excess acid required 25 ml of 0.1 M NaOH for neutralisation.
Answer» `11.5` 23 `12.5` `14.5` Solution :No. of milli eq of `H_(2)SO_(4)` LEFT after NEUTRALISATION. = No. of m. eq of NaOH `= 25xx0.1xx1=2.5` m eq. No. of m eq of `H_(2)SO_(4)` taken `= 50xx0.05xx2=5m. eq.` No. of eq of `H_(2)SO_(4)` ultilised for neutralisations `= 5-2.5=2.5` m.eq. `THEREFORE` No. of m. eq. of `NH_(3)` LIBERATED = 2.5 m.q. % of `N=("No. of meq. of "NH_(3)xx1.4)/("wt.of org. COMPOUND")` `= (2.5xx1.4)/(0.303)=11.55%`.

Discussion

52021.	0.302 g of organic compound gave 0.268g of silver bromide. The % of bromine in the sample is
Answer» 20 50 37.75 75 Answer :C

Discussion

52022.	0.301 g of an organic compound gave 0.282 g of silver bromide by a halogen estimation method. Find the percentage of bromine in the compound.
Answer» ANSWER :`39.87%`

Discussion

52023.	0.301 g of an organic compound gave 0.282 g of silver bromide by carius method. Find the percentage of bromine.
Answer» SOLUTION :`BR = 39.83%`

Discussion

52024.	0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO_(2) and 0.18 g H_(2)O. If one mole of compound weighs 60, then molecular formulaof the compound is
Answer» `C_(3)H_(8)O` `C_(2)H_(4)O_(2)` `CH_(2)O` `C_(4)H_(6)O` Solution :`%C = (12)/(44) XX (0.44)/(0.30) xx 100 = 40` `%H = (2)/(18) xx (0.8)/(0.30) xx 100 = 6.66` `%O = 100 - (40.0 + 6.66) = 53.34` Now `C:H:O = (40)/(12) : (6.66)/(1.0) : (53.34)/(16)` = 3.33: 6.66 : 3.33 `:. E.F. = CH_(2)O` E.F. wt. = 12+2+16 = 30 But Mol. wt. = 60 `:. M.F. = CH_(2)O xx (60)/(30) = C_(2)H_(4)O_(2)`

Discussion

52025.	0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water. Calculate the percentage of carbon and hydrogen in it
Answer» Solution :(i) Weight of organic compound = 0.30 G Weight of carbon-dioxide= 0.88 g Weight of water = 0.54 g Percentage of HYDROGEN : 18 g water contains 2 g of hydrogen `:."0.54" g"` of water contain `=2/(18)xx0.54g` of hydrogen `:.""%` of hydrogen `=2/(18)xx(0.54)/(0.30)xx100=(2)/(18)xx(54)/(0.3)` `""%` of H = `0.111xx180=19.888~~20%` Percentage of carbon : 44 g of `CO_(2)` contains 12 g of carbon 0.88 g of `CO_(2)` contains = `(12)/(44)xx0.88g` of carbon `:.""%` of carbon `=(12)/(44)xx(88)/(0.3)=(24)/(0.3)` `""%` of carbon `=80%` (ii)

Discussion

52026.	0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g water. Calculate the percentage of carbon and hydrogen in it.
Answer» SOLUTION :WEIGHT of organic compound = 0.30 g Weight of carbon dioxide = 0.88 g Weight of water = 0.54 g Percentage of carbon : 44 g if carbondioxide contains, carbon = 12 g 0.88 g of carbon dioxide contains, carbon `= (12xx0.88)/(44)g` 0.30 g substance contains, carbon `= (12xx0.88)/(44)g` 100 g substance contains `= (12xx0.88)/(44)XX(100)/(0.30)=80 g` of carbon Percentage of carbon = 80 Percentage of HYDROGEN : 18 g of water contains, hydrogen = 2g 0.54 g of water contains, hydrogen `= (2xx0.54)/(18)g` 0.30 g of substance contains hydrogen `= (2xx0.54)/(18xx0.30)g` 100 g of substance contains `=(2xx0.54)/(18xx0.30)xx100 g` = 20 g of hydrogen Percentage of hydrogen = 20

Discussion

52027.	0.3 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise 30 mL of 0.IN H_(2)SO_(4) . Calculate the percentage of nitrogen in the compound.
Answer» Solution :Weight of organic compound (w) = 0.3 g Strength of SULPHURIC acid used (N) = 0.1 N Volume of sulphuric acid used (V) = 30 ML 30 ml of 0.1 N sulphuric acid = 30 ml of 0.1 N ammonia `:. "Percentage of nitrogen" =((14 xx NV)/(1000 xx w)) xx 100=((14 xx 0.1 xx 30)/(1000 xx 0.3)) xx 100=14%`

Discussion

52028.	0.3 g of KI is dissolved in 25 " mL of " water. After adding to this solution double its volume of concentration HCl, a solution of KIO_(3) is graduaaly added with stirring. Iodine is liberated as first but redissolved. It is observed that 24.1 " mL of " iodate solution is just sufficient to dissolve the iodine. If the iodate solution contains 0.8 g per 100 mL formulate the reaction that has taken place.
Answer» Solution :0.3 of KI CONTAINS `(0.3)/(166)=1.807xx10^(-3)` " mol of "iodide ion, 24.1 " mL of " iodate SALUTION contains `(24.1)/(100)xx(0.8)/(214)=9.01xx10^(-4)` mol. It is clear from these figures that 2 moles of `I^(ɵ)` ions react with 1 MOLE of iodate ion. In thus PROCESS `I^(ɵ)` is oxidised and `IO_(3)^(ɵ)` is reduced to form some common species in which the oxidation number of iondine is x (say). Thus. `2(x-(-1)=(5-x)` i.e., `2x+2=5-x` `therefore3x=3impliesx=+1` In presence of excess HCl, this would correspond to the formation of ICL. `therefore` the reaction is formulated as follows, `2I^(ɵ)+IO_(3)^(ɵ)+3Cl^(ɵ)+6H^(o+)to3ICl+3H_(2)O`

Discussion

52029.	0.3 g of a gas has a volume of 112 ml at 0^@C and 2atm pressure. Its Molecular weight is
Answer» 60 30 44 28 Answer :C

Discussion

52030.	0.2m aqueous solution of KCl freezes at -0.68^(@)C calculate van't Hoff factor. K _(f) for water is 1.86 K kg mol ^(-1).
Answer» Solution :`I = ("Observed propoerty")/("Theoretical PROPERTY (CALCULATED)")` Given `Delta T_(f) = 0.680 Km ,m=0.2 m, Delta T_(f) ` (observed) `=0.680K,` `Delta T _(f)` (colculated) `=K _(f). M = 1. 86 K kg mol ^(-1) XX 0.2 mol kg ^(-1) = 0.372K` `i = ((Delta T _(f)) "observed")/((Delta T _(f))"calculated") = (0.680K )/(0.372K) =1.82`

Discussion

52031.	0.2g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 20 mL of 0.1 N H_(2)SO_(4). Calculate the percentage of nitrogen in the compound.
Answer» ANSWER :0.14

Discussion

52032.	0.2g of a sample of H_2O_2required 10mL of N KMnO_4in a titration in the presence of H_2SO_4Purity of H_2O_2 is
Answer» 0.25 0.85 0.65 0.95 Solution :`W/17xx1000=10xx1` `W = (170)/(1000) = 0.17 ` gm 0.2 gm `rarr` 0.17 gm PURE 100 gm `rarr` ? `=(100xx0.17)/(0.2) =85%`

Discussion

52033.	0.29 g of an organic compound were analysed by Liebig's method. The increase in the mass of U-tube and the potash bulbs at the end of the experiment were found to be 0.27g and 0.66g respectively. Calculate the percentage of carbon and hydrogen in it
Answer» SOLUTION :%C = 62.1 and %H = 11.1

Discussion

52034.	0.28g of an organic compound in Dumas method liberated 22.4 ml of nitrogen at STP. The percentage of nitrogen in the compound is
Answer» 20 10 80 50 Answer :A

Discussion

52035.	0.2828g of iron wire was dissolved in excess dilute H_(2)SO_(4) and the solution was made up to 100 mL. 20mL of this solution required 30mL of N//30 K_(2)Cr_(2)O_(7) doluyion for exact oxidation. Calculate % purity of Fe in wire.
Answer» ANSWER :`99%`

Discussion

52036.	0.28 g of nitrogenous compound was subjected to Kjeldahl's process to produce 0.17 g of NH_(3) . The percentage of nitrogen in the organic compound is
Answer» 5 2 50 80 Solution :0.17 G `NH_(3)` will CONTAINS `(14/17 xx0.17)` g of nitrogen , i.e., 0.14 g of nitrogen % Nitrogen =`("MASS of nitrogen ")/("Mass of compound ")xx100=0.14/0.28xx100=50%`

Discussion

52037.	0.27g of an organic compound gave on combustion 0.396g of CO_(2) and 0.216g of H_(2)O. 0.36g of the same substance gave 48.88 mL of N_(2) at 290 K and 740 mm pressure. Calculate the percentage composition of the compound.
Answer» ANSWER :C = 40%, H = 8.89%, N = 15.56% and O = 35.55%

Discussion

52038.	0.28 g of a commercial sample of K_(2)Cr_(2)O_(7) was dissolved in water. Excess of KI was added to it along with dilute H_(2)SO_(4). Iodine liberated was then titrated against sodium thiosulphate solution containing 24.82 g of Na_(2)S_(2)O_(3). 5H_(2)O per litre. The thiosulphate solution required was 50 mL. Find the percentage purity of the sample of K_(2)Cr_(2)O_(7).
Answer» SOLUTION :N//A

Discussion

52039.	0.262 g of a substance gave, on combustion 0.361 g of CO_(2) and 0.147 g of H_(2)O. What is the empirical formula of the substance
Answer» `CH_(2) O` `C_(3) H_(6) O` `C_(3) H_(6) O_(2)` `C_2 H_(6) O_(2)` Answer :A

Discussion

52040.	0.26 g of an organic compundgave 0.039 g of water and 0.245 g of CO_2 on combustion. Calculate the percentage of carbon and hydrogen.
Answer» Solution :WEIGHT of organic compound =0.26 G Weight of water =0.039 g Weight of `CO_2`= 0.245 g Percentage of hydrogen 18 g of watercontains 2 g of hydrogen 0.039 mg of water contains`-(2)/(18)xx(0039)/(0.26)`of H % of hydrogen `-(0.39)/(0.26)xx(2)/(18)xx100=1.66%` Percentage of csarbon 44 g of `CO_2` contains 12 g of C 0.245 g `CO_2` contains=`(12)/(44)xx(0.245)/(0.26)g` of C %of CARBON =`(12)/(44)xx(0.245)/(0.26)xx100=25.69%`

Discussion

52041.	0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer» Solution :Weight of organic compound = 0.26 g Weight of WATER = 0.039 g Weight of `CO_(2)` =0.245 g Percentage of HYDROGEN 18 g of water contains 2 g of hydrogen `"0.039 g of water contains" =(2)/(18) xx (0.039)/(0.26) "of H"` `:. % "of hydrogen"=(0.039)/(0.26) xx (2)/(18) xx 100=1.66%` Percentage of CARBON 44g of `CO_(2)` contains 12 g of C `0.245 "g of" CO_(2) "contains"=(12)/(44) xx (0.245)/(0.26) "g of C"` `% "of Carbon"=(12)/(44) xx (0.245)/(0.26) xx 100=25.69%`

Discussion

52042.	0.25g of an organic compound an analysis by Duma,s method gave 32cc nitrogen at STP. What is the weight percentage of nitrogen in the given compound?
Answer» ANSWER :16

Discussion

52043.	0.25g of an organic compound gave 30cm^(3) of moist dinitrogen at 288K and 745mm pressure. Calculate the percentage of nitrogen. (Aqueous tension at 288K = 12.7mm)
Answer» <P> SOLUTION :Mass of the substance = 0.25g Volume of moist dinitrogen = `30cm^(3)` Temperature = 288K Pressure `= 745- 12.7 = 732.3mm` Volume of dinitrogen at S.T.P `V_(2) = (P_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2)) = (732.3 xx 30 xx 273)/(288 xx 760) = 27.4 CM^(3)` `22400cm^(3)` of dinitrogen at STP weight = 28g `27.4 cm^(3)` of dinitrogen at STP weight `= (28 xx 27.4)/(22400)` = 0.034g Percentage of NITROGEN in organic compound `=(0.034)/(0.25) xx 100 = 13.6`

Discussion

52044.	0.25g compound give 0.350g, BaSO_(4). What is the percentage of sulphur?
Answer» ANSWER :0.1856

Discussion

52045.	0.2595 g of organic compound when treated by carius method gave 0.35 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=32/233xx0.35/0.2595xx100=18.52%`

Discussion

52046.	0.2595 g of an organic substance, when treated by Carius method, gave 0.25 g of BaSO_(4). Calculate the percentage of sulphur in the compound.
Answer» ANSWER :`13.23%`

Discussion

52047.	0.257 g of an organic substance was Kjeldahlised and ammonia evolved was absorbed in 50 mL of N/10 HCl which required 23.2 ml of N/10 NaOH for neutralization. Determine the percentage of nitrogen in the compound.
Answer» SOLUTION :`N = 14.6%`

Discussion

52048.	0.25 g of an organic compound 'X' containing carbon, hydrogen and oxygen was analysed by the combustion method. The increase in mass of calciumchloride tube and potash bulbs at the end of the experiment was found to be 0.15 g and 0.1837 g respectively. Calculate the percentage composition of the compound.
Answer» Solution :Wt. of `CO_2` FORMED = Increase in potash BULBS `= 0.1837` g Wt. of organic COMPOUND = 0.25 `% C = *(12)/(44) xx (0.1837)/( 0.25) xx 100` `= 20.04%` Wt. of `H_2 O` produced = Increase in `CaCl_2` TUBE `= 0.15` g `%H= (2)/( 16) xx (0.15)/( 0.25) xx 100` `= 6.66%` % of `O - 100 - ( 20. 04 + 6.66)` `73.30 %`

Discussion

52049.	0.25 g of an oxalate salt was dissolved in 100 mL of water. 10 mL of this solution required 8 mL of N//20 KMnO_(4) for its oxidation. Calculate the percentage of oxalate in the salt.
Answer» SOLUTION :N//A

Discussion

52050.	0.25 g of an organic compound was found to produce 0.35 g of AgCl after heating with fuming HNO_(3) and AgNO_(3) in a sealed carius method. Determine the percentage of chlorine in the compound.
Answer» SOLUTION :`CL = 34.75%`

Discussion

Explore topic-wise InterviewSolutions in .

0.37 gm of alcohol when treated with CH_3MgI liberates 112 ml of CH_4 at STP. The molecularformula of alcohol is

0.376g of aluminium reacted with an acid to displace 0·468 litre of hydrogen at NTP. Find the equivalent volume of hydrogen if the equivalent weight of Al is 9.

0.36g of an organic compound on combustion gave 1.1g of CO_(2) and 0.54g of H_(2)O. The percentages of carbon and Hydrogen in the compound are

0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be

0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into 100 ml of N/5 H_(2)SO_(4). The excess acid required 154 ml of N/10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.

0.35 g of an organic acid was kjeldahlised and the ammonia obtained was passed into 100 cm^(3) of N//5H_(2)SO_(4). The excess of the acid required 154 cm^(3) of N//10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.

0.35 g of a metal was dissolved in 50 mL N-acid. The whole solution then required 20.85 mL of normal alkaline solution to neutralise the excess of the acid. Calculate the equivalent mass of the metal.

0.33g of an organic compound containing phosphorous gave 0.397 g of Mg_(2)P_(2)0_(7) by the analysis. Calculate the percentage of P in the compound.

0.32g of metal gave on treatment with an acid 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal

0.33 g of an organic compound was analysed for nitrogen by Kjeldhal's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4).The excess acid required 25 ml of 0.1 M NaOH for neutralization. The Percentage of nitrogen in the given compound is

0.33 mol cholesterol gives 9 mole CO_(2) on combustion. It was observed that cholesterol contains 83.85%C, 12%H and 4.15%O. Find its molecular formula and molecular mass.

0.32g of an organic compound containing sulphur produces 0.233g of BaSO_(4). Percentage of sulphur in the compound is

0.324 g of copper was dissolved in nitric acid and the copper nitrate so produced was burnt till all copper nitrate converted to 0.406 g of copper oxide. Calculate the equivalent weight of copper.

0.32 g of metal on treatment with acid gave 112mL of hydrogen at NTP. The equivalent weight of metal is

0.3160 g of an organic substance was heated with fuming nitric acid in a carius tube to convert phosphorus present into phosphoric acid. Addition of magnesia mixture formed MgNH_(4)PO_(4) which upon heating gave 0.1697 g Mg_(2)P_(2)O_(7). Calculate the percentage of phosphorus in the compound.

0.316 g of an organic compound gives 0.466 g of barium sulphate by carius method. Calculate the percentage of sulphur?

0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).

0.303 g of an organic compound was analysed for nitrogen by Kjeldahl's method. The ammonia evovled was absorbed in 50 ml. of 0.1N H_(2)SO_(4). The excess acid required 25ml of 0.1N NaOH for neutralisation. Calculate the percentage of nitrogen present in the compound

0.303 g of an organic compound was analysed for nitrogen by Kjedahl's method. The ammonia gave 0.2584 g of silver bromide Calculate the weight percentage of bromine in the organic compound.

0.303 g fo sample was analysed for nitrogen by Kjeldahl's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4). The excess acid required 25 ml of 0.1 M NaOH for neutralisation.

0.302 g of organic compound gave 0.268g of silver bromide. The % of bromine in the sample is

0.301 g of an organic compound gave 0.282 g of silver bromide by a halogen estimation method. Find the percentage of bromine in the compound.

0.301 g of an organic compound gave 0.282 g of silver bromide by carius method. Find the percentage of bromine.

0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO_(2) and 0.18 g H_(2)O. If one mole of compound weighs 60, then molecular formulaof the compound is

0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water. Calculate the percentage of carbon and hydrogen in it

0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g water. Calculate the percentage of carbon and hydrogen in it.

0.3 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise 30 mL of 0.IN H_(2)SO_(4) . Calculate the percentage of nitrogen in the compound.

0.3 g of a gas has a volume of 112 ml at 0^@C and 2atm pressure. Its Molecular weight is

0.2m aqueous solution of KCl freezes at -0.68^(@)C calculate van't Hoff factor. K _(f) for water is 1.86 K kg mol ^(-1).

0.2g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 20 mL of 0.1 N H_(2)SO_(4). Calculate the percentage of nitrogen in the compound.

0.2g of a sample of H_2O_2required 10mL of N KMnO_4in a titration in the presence of H_2SO_4Purity of H_2O_2 is

0.29 g of an organic compound were analysed by Liebig's method. The increase in the mass of U-tube and the potash bulbs at the end of the experiment were found to be 0.27g and 0.66g respectively. Calculate the percentage of carbon and hydrogen in it

0.28g of an organic compound in Dumas method liberated 22.4 ml of nitrogen at STP. The percentage of nitrogen in the compound is

0.2828g of iron wire was dissolved in excess dilute H_(2)SO_(4) and the solution was made up to 100 mL. 20mL of this solution required 30mL of N//30 K_(2)Cr_(2)O_(7) doluyion for exact oxidation. Calculate % purity of Fe in wire.

0.28 g of nitrogenous compound was subjected to Kjeldahl's process to produce 0.17 g of NH_(3) . The percentage of nitrogen in the organic compound is

0.27g of an organic compound gave on combustion 0.396g of CO_(2) and 0.216g of H_(2)O. 0.36g of the same substance gave 48.88 mL of N_(2) at 290 K and 740 mm pressure. Calculate the percentage composition of the compound.

0.262 g of a substance gave, on combustion 0.361 g of CO_(2) and 0.147 g of H_(2)O. What is the empirical formula of the substance

0.26 g of an organic compundgave 0.039 g of water and 0.245 g of CO_2 on combustion. Calculate the percentage of carbon and hydrogen.

0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.

0.25g of an organic compound an analysis by Duma,s method gave 32cc nitrogen at STP. What is the weight percentage of nitrogen in the given compound?

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

0.25g compound give 0.350g, BaSO_(4). What is the percentage of sulphur?

0.2595 g of organic compound when treated by carius method gave 0.35 g of BaSO_(4). Calculate the percentage of sulphur in the compound.

0.2595 g of an organic substance, when treated by Carius method, gave 0.25 g of BaSO_(4). Calculate the percentage of sulphur in the compound.

0.257 g of an organic substance was Kjeldahlised and ammonia evolved was absorbed in 50 mL of N/10 HCl which required 23.2 ml of N/10 NaOH for neutralization. Determine the percentage of nitrogen in the compound.

0.25 g of an oxalate salt was dissolved in 100 mL of water. 10 mL of this solution required 8 mL of N//20 KMnO_(4) for its oxidation. Calculate the percentage of oxalate in the salt.

0.25 g of an organic compound was found to produce 0.35 g of AgCl after heating with fuming HNO_(3) and AgNO_(3) in a sealed carius method. Determine the percentage of chlorine in the compound.