94420 + Interview Questions in Chemistry in Class 12 Page 1871 InterviewSolution

93501.	0.2 gm of fine animal charcoal is mixed with ahlf litre of acetic acid (-SM) solutioin and shaken for 30 minutes
Answer» The CONCENTRATION of the solution decreases Concentration increases Concentration remains None of these Solution :ACTIVATED CHARCOAL adsorbed the IMPURITY of acetic ACID by which the concentration of acetic acid solution decrease.

Discussion

93502.	0.2 gm of a mixture of NaOH and Na_(2)CO_(3) and inert impurities was first titrated with phenolphthalein and n/10HCl, 17.5 ml of HCl was required at the end point. After this methyl organge was added and 2.5 ml of same HCl was again required for next end point. Find out percentage of NaOH and Na_(2)CO_(3) in the mixture.
Answer» Solution :Let , `W_(1)` gm NaOH and `W_(2) gmNa_(2)CO_(3)` was present in MIXTURE . At phenophthaiein end point , `(W_(1)/40 + 1/2xx W_(2)/53 ) = 1/10 xx 17.5 xx 10^(-3)` At SECOND end point following reaction takes place Eq. of `NaHCO_(3) = Eq`. of HCL used (in second titration ) ` = 1/2 " Eq. of " Na_(2)CO_(3)` ` 1/2 xx W_(2)/53 =25 xx 1/10 xx 10^(-3) ` `W_(2) = 0.0265` gm Putting the value of `W_(2) ` in Eq. (1) , we get `W_(1) = 0.06` gm Percentage of NaOH = `(0.06)/(0.2) xx 100 = 30 %` Percentage of `NaCO_(3) = (0.0265)/(0.2) xx 100 = 13.25 %`

Discussion

93503.	0.2 g of a sample of H_(2)O_(2) required 10 mL of N KMnO_(4) in a titration in the presence of H_(2)SO_(4) .Purity of H_(2)O_(2) is
Answer» 0.25 0.85 0.65 0.95 Answer :B

Discussion

93504.	0.2g of an organic compoundcontaningC, Hand O on combustion gave 0.147 g of CO_(2)and0.12g of water . Thepercentage content ofoxygen in the compoundis :
Answer» `73.29` `86.36` `83. 46` `74.92` ANSWER :A

Discussion

93505.	0.2 g of an organic compound containing C, Hand O ,on combustion yielded 0.147 g Co_2 and 0.12 g water .The percentage of oxygen in it is :
Answer» 73,29% 0.7845 0.8323 0.895 Answer :A

Discussion

93506.	(0.1M "urea")/((A)),(0.1M NaCl)/((B)),(0.1MBaCl_(2))/((C))
Answer» Solution :`{:("ORDER of" PI,CgtBgtA),("order of" R.L.V.P,C GT B gt A),("order of" V.P,Agt B gt C),("order of" DeltaT_(B),C gt B gt A),("order of" T_(B) "of solution", C gt B gt A),("order of" DeltaT_(F),C gt B gt A),("order of" T_(F) " of solution",A gt B gt C):}`

Discussion

93507.	0.1M solution of which of the following compounds will have the lowest freezing point ?
Answer» `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O` `[Co(Cl)(NH_(3))_(5)]Cl_(2)` `K_(3)[FE(CN)_(6)]` `[CrCl_(2)(NH_(3))]Cl` Solution :Freezing POINT `(1)/("No. of Pparticles")`

Discussion

93508.	0.1m solution of KCI and BaCI_(2) are preparred. The freezing point of KCI solution is found to be- 4.0^(@)C.What will be the freezing point of BaCI_(2) solutionassuming that both KCI and BaCI_(2) solution are sompletely ionised in solution ?
Answer» `-3^(@)C` `-4^(@)C` `-5^(@)C` `-6^(@)C` SOLUTION :`"For" KCI,DeltaT_(f)=iK_(f)xxm` `DeltaT_(f)=(0.0.4^(@)C)=4^(@)C` `4K=2xxK_(f)xxm` `"For " BaCI_(2),DeltaT_(f)=iK_(f)xxm` `=3xxK_(f)xxm` Both `K_(f)` and m are same for the SOLUTE, `DeltaT_(f)=(BaCI_(2))=3` FREEZING POINT of `BaCI_(2)=0-3=-3^(@)C`

Discussion

93509.	0.1m K_(2)SO_(4) ionises to 10 x% to boil at same temperature as 0.2 glucose. What is x ?
Answer» ANSWER :5

Discussion

93510.	0.1M CH_(3)COOH(aq.) Is mixed with 0.1M NaOH(aq.). The pH of solution on mixing will be (K_(a) ofCH_(3)COOH is 1.8xx10^(-5)
Answer» `4.74` `8.70` `9.75` NONE of the above Solution :`N//A`

Discussion

93511.	0.1g of an unknown substance was dissolved in 5g of camphor and it was found that the melting point of camphor was depressed by 5.3^(@)C. If K_(f) is 39.7, find the weight of 1 mole of the solute.
Answer» ANSWER :`150`

Discussion

93512.	0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrichemical equivalent of copper ?
Answer» Solution :Here, t = 50 minutes `=50xx60` seconds , I = 0.2 AMPERE Quantity of electricity used is `Q=Ixxt=0.2xx50xx60=600" coulomb"` AMOUNT of COPPER deposited by 600 COULOMBS = 0.1978 g Amount of copper deposited by 1 coulomb `=(0.1978)/(600)g=0.0003296g` Electrochemical EQUIVALENT of copper `=0.0003296=3.296xx10^(-4)gc^(-1)` `=3.206xx10^(-7)kgc^(-1)`

Discussion

93513.	0.1914g of an organic acid is dissolved in approx. 20 ml of water. 25 ml of 0.12 N NaOH required for the complete neutralization of the acid solution. The equivalent weight of the acid is
Answer» 65 64 63.8 62.5 Solution :VOLUME`=25ml=(25)/(1000)`LITRE NORMALITY`=(wt)/(eq.wtxxvolume)implies0.12=(0.1914xx1000)/(Exx25)` eq.wt.`=(0.1914xx1000)/(0.12xx25)=63.8`.

Discussion

93514.	0.1890g of an organic compound gave 0.2870g of silver chloride by Carius method. Find the percentage of chlorine in the compound
Answer» ANSWER :0.3757

Discussion

93515.	0.189 g of an organic substance gave in a carius determination 0.287 g of silver chloride. What is the % of Cl in the given compound?
Answer» SOLUTION :W= 0.189g WT. AgCl= 0.287 G % of Cl `+(35.5xx0.287xx100)/(143.5xx0.189) =37.57%`

Discussion

93516.	0.188gm of silver bromide is obtained from x gm of an organic compound and percentage of bromine in the compound is 4. Then the x value is
Answer» <BR> Solution :`% Br = (80 xx "weight of AGBR" xz 100)/(188 xx "weight of compound")` , valancy of x is 2

Discussion

93517.	0.188 g of silver bromide is obtained from 4 g of an organic compound. Calculate the percentate of bromine in the compound
Answer» <BR> Solution :`%Br=(80 xx "weight of AgBr" xx 100)/(188 xx "weight of a organic COMPOUND")`

Discussion

93518.	0.170 g of a volatile liquid displaced 69.1 mL air in Victor Meyor's method. The atmospheric temperature was 15^@ C and pressure was 755 mm. (Aqueous tension at 15^@C = 12.7 mm ). Find the molecular weight of the liquid.
Answer» ANSWER :59.52

Discussion

93519.	0.16g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of N_(2)H_(4) that has reacted with water in this solution. The K_(b) for N_(2)H_(4) is reacted with water in this solution. The K_(b) for N_(2)H_(4) is 4.0 xx 10^(-6) M
Answer» SOLUTION :`{:(,N_(2)H_(4)+,H_(2)O hArr,N_(2)H_(5)^(+)+,OH^(-)),("At t = 0",1,,0,0),("At eqm.",1-alpha,,alpha,alpha),(,,,,):}` `K_(B) = (C alpha^(2))/((1-alpha)) RARR K_(b) = C alpha^(2)` (`because alpha` is very small, so `1-alpha ~~ 1`) `[N_(2)H_(4)] = C = (0.16 xx 1000)/(32 xx 500) = 0.01` `:. alpha^(2) = (4 xx 10^(-6))/(0.01) = 4 xx 10^(-4) "" (because = K_(b) = 4 xx 10^(-6)M)` `:. alpha = 2 xx 10^(-2) = 0.02` or 2%.

Discussion

93520.	0.16g of an organic compound, on complete combustion, produced 0.44g of CO_(2) and 0.18g of H_(2)O. Calcuate the percentage of carbon and hydrogen in the organic compounds
Answer» Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)` `=1 xx (0.44)/(44)= 0.01` Weight of C= moles of C `xx` at. Wt . of C `=0.01 xx 12 = 0.12g` Moles of H in `H_(2)O=2 xx` moles of `H_(2)O` `=2 xx (0.18)/(18)= 0.02` weight of H= moles of `H xx` at. wt. of H `= 0.02 xx 1 = 0.02g` `THEREFORE` % of `C= (0.12)/(0.16)xx 100= 75%` and % of `H= (0.02)/(0.16) xx 100= 12.5%`

Discussion

93521.	0.16 gm of organic acid required 25 ml of (N)/(10) NaOH for complete neutralization. (M. wt of acids is 128). Then the basicity of acid is
Answer» Solution :Equivalent acid = Equivalent of NAOH `(wt)/("Gewt")= 25 XX 0.1 xx 10^(-3)` `(0.16)/("Gewt") = 25 xx 0.1 xx 10^(-3)` valancy(basicety) = `("Mol.wt")/("Gewt")=2`

Discussion

93522.	0.16 gm of a dibasic acid require 25 ml of decinormal NaOH solution for complete neutralization. The moleucular weight of the acid
Answer» 32 64 128 256 Solution :`(0.16)/(X)xx(1000)/(25)=(1)/(10)implies(6.4)/(x)=(1)/(10) thereforex=64` MOL. Wt.=`64xx2=128`.

Discussion

93523.	0.16 g of methane was subjected to combustion at 27^(@)C in a bomb calorimeter. The temperature of Calorimeter system (including water) was found to rise by 0.5^(@)C. Calculate the heat of combustion of methane at (i) constant volume (ii) constant pressure. The thermal capacity of Calorimater system is 17.7" kJ K"^(-1). (R=8.313" J mol"^(-1) K^(-1))
Answer» Solution :`CH_(4)+2O_(2) rarr CO_(2) +2H_(2)O` Total energy relesed `=17.7xx0.5=8.85` Energy released by one MOLE `CH_(4)=8.85/0.16 xx16=-885" kJ/mol"` `DeltaH=DeltaU+Deltan_(G) RT rArr Deltan_(g) =-1 rArr DeltaH=-889.95" kJ/mol"`

Discussion

93524.	0.16 g of N_2H_4are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of N_2H_4that has reacted with water in this solution. K_b( N_2H_4) = 4.0 xx 10^(-6) M
Answer» SOLUTION :`N_2H_4 + H_2O to N_2H_5OH IFF N_2H_5^(+) + OH^-`2%

Discussion

93525.	0.16 g of methane was subjected to combustion at 27^@Cin a bomb calorimeter. 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, (ii) constant pressure. The thermal capacity of the calorimeter system is 17.7 kJ K^(-1) (R = 8.31 JK^(-1) "mol"^(-1))
Answer» SOLUTION :`DELTA E= (17.7 xx 0.5)/(0.16//16) KJ "mol"^(-1)` -885 kJ, -889.986 kJ

Discussion

93526.	0.16 g of a dibasic organic acid required 25vm^(3)of 0.1 M NaOH for complete neutralization. Thre molecular mass of the acidof the acid is :
Answer» 45 90 64 128 Solution :N//A

Discussion

93527.	0.15g of an organic compound gave 0.12g of silver bromide by the Carius method. The percentage of bromine in the compound is :
Answer» `42.06`<BR>`38.96` `24.08` `34.04` Solution :`UNDERSET(188)(AgBr)=underset(80)(Br)` `188g` of `AgBr` CONTAIN bromine `=80G` `0.12g` of `AgBr` contain bromine `=(80)/(188)xx0.12` `=0.051g` PERCENTAGE of bromine `=(0.051)/(0.15)xx100` `=34.04%`

Discussion

93528.	0.15g of an organic compound (A) gave 0.33g of CO_(2) and 0.18g of H_(2)O. The molecular weight of (A) is 60. The compound (A) on dehydration gave a hydrocarbon (B) containing 85.7% of C. (B), on successive treatment with hydriodic acid and silver hydroxide gave a product (C ), isomeric with (A). Find structural formula of (A), (B) and (C )
Answer» Solution :`[((A) CH_(3)CH_(2)CH_(2)OH),((B) CH_(3)CH= CH_(2)),((C ) CH_(3)CH(OH)CH_(3))]`

Discussion

93529.	0.15 mole of pyridinium chloride has been added into 500 cc of 0.2 M pyridinesolution. Calculate the pH and hydroxyl-ion concentration in this resulting solution assuming no change in volume. K_b for pyridine = 1.5 xx 10^(-9) M .
Answer» SOLUTION :`5, 10^(-9)`M

Discussion

93530.	0.15 molalsolutionof NaCIhas freezingpoint -0.52 ^(@)C Calculatevan'tHofffactor.(K_(f)= 1.86 K kg mol^(-1) )
Answer» SOLUTION :Given :concentration`= m=0.15 M ` Freezingpoint of THESOLUTION=- 0.52 `.^(@)C` `K_(f)= 1.86K kg MOL^(-1)` `:. T_(f) =273 + (-0.52)= 272 .48 K` `T_(o)= 273 K` `:.` Depressionin freezingpoint `=DeltaT_(f(OBS))=T_(o) -T_(f)` `=273- 272.48` `= 0.52K` Theoreticaldepression infreezingpoint `= Delta T_(f_(th))` `Delta _(f_(th)) = K_(f)xx m =1.86 xx 0.15 = 0. 279 K` van't Hofffactori is `i= (Delta _(f(ob)))/(Delta_(T_(f_((th)))))=(0.52)/(0.279) = 1.864`

Discussion

93531.	0.146g of an acid requires 20 ml of 0.1 N NaOH for complete neutralization. The equivalent weight of the acid is
Answer» 45 53 40 63 Solution :20 ML of 0.1N NaoH NEUTRALIZE 20 ml of 0.1 N acid weight of acid=0.126g volume`=20ml=(20)/(1000)` litre Normality`=0.1N,` EQUIVALENT weight=? Equivalent weight`=("Weight of acid")/(NxxV)` `=(0.126xx1000)/(0.1xx20)=63`

Discussion

93532.	0.12g of organic compound containing sulphur on treatment with Conc. HNO_(3) in a carius tube and then with excess of BaCl_(2) produced 0.233g of BaSO_(4). % of 'S' in compound ...........
Answer» Solution :`%S=(32 xx "WEIGHT of" BaSO_(4) xx 100)/(233 xx "weight of ORGANIC compound")10%, 1 xx 10 =1`

Discussion

93533.	0.126g of an acid requires 20 ml of 0.1N NaOH for complete neutralization. Equivalent weight of the acid is
Answer» 45 53 40 63 Answer :D

Discussion

93534.	0.126 g of an acid requires 20 mL of 0.1 N NaOH for complete neutralisation . Eq.wt of the acid is
Answer» 45 53 40 63 Answer :d

Discussion

93535.	0.126 g of an acid is needed to completely neutralise 20 mL 0.1 N NaOH solution. The equivalent weight of the acid is
Answer» 53 40 45 63 Solution :Equivalents of acid = Equivalents of base `=(NV)/1000 =(0.1 XX 20)/1000 = 2 xx 10^(-3)` No. of gram equivalents `=("mass")/("Equivalent mass")` `2 xx 10^(-3) = 0.126/("Equivalent mass")` Equivalent mass `=0.126/(2 xx 10^(-3)) = 126/2 = 63`

Discussion

93536.	0.123g of an organic compound produced 0.099g of CO_(2) and 0.0507g of H_(2)O. 0.185g of the same compound produced 0.319g of AgBr. Find the percentage of carbon, hydrogen and bromine in the compound
Answer» SOLUTION :`21.96%, 4.48%, 73.36%`

Discussion

93537.	0.1 mole of N_(2)O_(4(g)) was sealed in a tube under one atmospheic conditions at 25^(@)C. Calculate the number of moles of NO_(2(g)) present, if the equlibrium N_(2)O_(4(g))hArr2NO_(2(g))(k_(p)=0.14) is reache3d after some time
Answer» <P>`1.8xx10^(2)` `2.8xx10^(2)` `0.034` `2.8xx10^(-2)` Solution :`underset(0.1)(N_(2)O_(4))hArrunderset(0)(2NO_(2))` `(.1-alpha)""2ALPHA` `thereforeP prop0.1` If V and T are constant `(Pprop0.1+alpha)P=(0.1+alpha)//0.1` `K_(p)=([2alpha]^(2))/([0.1-alpha])xx[(P)/(0.1+alpha)]or K_(p)=(40alpha^(2))/([0.1-alpha])=0.14` `alpha=0.017` `NO_(2)=0.017xx2=0.034` MOLE

Discussion

93538.	0.1 mole of sugar is dissolved in 250 g of water. The freezing point of the solution is [K_(f) "for" H_(2)O = 1.86^(@)C "molal"^(-1)]
Answer» `-0.460^(@)C` `0.460^(@)C` `-0.744^(@)C` `-186^(@)C` ANSWER :C

Discussion

93539.	0.1 mole of XeF_(6) is treated with 1.8 g of water.The product obtained is
Answer» `XeO_(3)` `XeO_(2)F_(2)` `XeOF_(4)` `XE + XeO_(3)` Solution :(c) : `n_(H_(2)O) = (1.8)/(18) = 0.1` mole, `n_(XeF_(6)) = 0.1` mole `XeF_(6) + H_(2)O to XeOF_(4) + 2HF`

Discussion

93540.	0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is added to 0.08 mole of HCl and the solution is diluted to one litre. The resulting [H^(+)] is
Answer» `1.6xx10^(-11)` `8XX10^(-11)` `5XX10^(-5)` `8xx10^(-2)` Solution :The resulting solution is a buffer solution.

Discussion

93541.	0.1 mole of CH_(3)NH_(2)(K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H^(+) concentration in the solution
Answer» `8 xx 10^(-2)M` `8 xx 10^(-11) M` `1.6 xx 10^(-11) M` `8 xx 10^(-5) M` Solution :`{:(CH_(3)NH_(2),+,HCl,RARR,CH_(3)NH_(3)^(+)Cl^(-)),(0.1,,0.08,,0),(0.02,,0,,0.08):}` As it is a BASIC buffer solution. `pOH = pK_(b) + log.(0.08)/(0.02) = - log5 xx 10^(-4) + log 4` `= 3.30 + 0.602 = 3.902` `PH = 14 - 3.92 = 10.09`, `[H^(+)] = 7.99 xx 10^(-11) = 8 xx 10^(-11) M`

Discussion

93542.	0.1 mole of CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole for HCl and diluted to one litre. What will be the H^(+) concentration in the solution
Answer» `8 xx 10^(-2)M` `8 xx 10^(-11)M` `1.6 xx 10^(-11)M` `8 xx 10^(-5) M` SOLUTION :`{:(CH_(3)NH_(2)+,HCL, rarr, CH_(3)NH_(3)^(+)Cl^(-)),(0.1, 0.08,, 0),(0.02, 0,, 0.08):}` (Basic buffer solutions) `[OH^(-)] = K_(b) xx ("Base")/("Salt") = 5 xx 10^(-4) xx (0.02)/(0.08) = 1.25 xx 10^(-4)` `:. [H^(+)] = (10^(-14))/([OH^(-)]) = (10^(-14))/(1.25 xx 10^(-4)) = 8 xx 10^(-11) M`.

Discussion

93543.	0.1 mol of sugar was dissolved in 1 kg of water. The freezing point of the solution was found to be 272.814 K. What conclusion would you draw about the molecular state of sugar? K_(f) for water is "1.86 K kg mol"^(-1).
Answer» Solution :Proceed as in Solved Problem 1 or calculate the MOLECULAR MASS of sugar from the given the data which COMES out ot be 342, i.e., same as the NORMAL molecular mass of sugar.

Discussion

93544.	0.1 molal aqueous solution of sodium bromide freezes at -0.335^(@)C at atmospheric pressure. K_(f) for water is 1.86^(@)C. The percentage of dissociation of the salt in solution is
Answer» 90 80 58 98 Solution :`DELTA T_(f) = i K_(f) m or 0.335 = i xx 1.86 xx 0.1 or i= 1.80 ` `alpha = (i-1)/(n-1) = (i-1)/(2-1)` For NABR `i = 1 + alpha or alpha = 0.80 = 80%`

Discussion

93545.	0.1 mole of a carbohydrate with empirical formula CH_(2)O contains 1 g of hydrogen. What is its molecular formula?
Answer» `C_(5)H_(10)O_(5)` `C_(6)H_(12)O_(6)` `C_(4)H_(8)O_(4)` `C_(3)H_(6)O_(3)` Solution :0.1 mole of carbohydrate with E.F. `CH_(2)O` contains 1 gof hydrogen. `therefore "1 mole of carbohydrate will contain hydrogen"` = 10 g = 10 g atoms In `CH_(2)O`, g atomic RATIO of `C:H:O=1:2:1` `therefore" With 10 g ATOM of H, g atoms of C combined"` `="5 and g atoms of O combined = 5. Hence, actual formula (molecular formula) will be "C_(5)H_(10)O_(5)`.

Discussion

93546.	0.1 mole Ni is obtained on electrolysis of NiSO_(4) solution in neural solution, how much faraday electricity should be pass through the cell ?
Answer» 1F 0.1 F 0.2 F 0.01 F Solution :`NiSO_(4) to NI^(2+)+SO_(4)^(2-)` `Ni^(2+)+2e^(-) to Ni` So, 1 mole Ni is OBTAINED by 2F Then 0.1 mole Ni is obtained by `=(0.1" mole"xx2F)/(1" mole")=0.2F`

Discussion

93547.	0.1 mol of MnO_(4)^(-) in acidic medium can oxidize
Answer» `0.5 " MOL of " Fe^(2+)` `0.166 " mol of " FeC_(2)O_(4)` `0.25 " mol of " C_(2)O_(4)^(2-)` `0.60 " mol of " Cr_(2)O_(7)^(2-)` SOLUTION :N/A

Discussion

93548.	0.1 mol of CH_(3)NH_(2)(K_(b)=5.0xx10^(-4)) is mixed with 0.08 mol of HCl and diluted to one litre. What will be the H^(+) ion concentration in the solution ?
Answer» `8xx10^(-2)M` `8xx10^(-11)M` `1.6xx10^(-11)M` `8xx10^(-5)M` SOLUTION :`underset(0.02)underset(0.1)(CH_(3)NH_(2))+underset(0)underset(0.08)(HCl)RARR underset(0.08)underset(0)(CH_(3)NH_(s)^(+)Cl^(-))` For basic buffer `pOH=pK_(b)+"LOG"(0.08)/(0.02)` `pK_(b)=-log(5xx10^(-4))=3.30` `therefore pOH=pK_(b)+0.602` `=3.30+0.602=3.902` `pH=-log(3.902)=10.09` `[H^(+)]=-log(10.09)=7.99xx10^(-11)` or`=8xx10^(-11)M`

Discussion

93549.	0.1 mol HCl is dissolved in distilled water of volume V, then, at underset(Vtooo)(lim) (pH)_("solution") is equal to :
Answer» zero 1 7 14 Answer :C

Discussion

93550.	0.1 M solution of two electrolytes P and Q have specific conductance 4 times 10^(-4)" S "cm^(-1) " and " 6 times 10^(-6)" S "cm^(-1) respectively. Which among the following will have greater resistance to the flow of current? Give reason.
Answer» Solution :Specific CONDUCTANCE `K=C(l/a)` `""k=1/R(l/a)` That is `k ALPHA 1/R` Therefore Q will have greater resistance.

Discussion

Explore topic-wise InterviewSolutions in .

0.2 gm of fine animal charcoal is mixed with ahlf litre of acetic acid (-SM) solutioin and shaken for 30 minutes

0.2 g of a sample of H_(2)O_(2) required 10 mL of N KMnO_(4) in a titration in the presence of H_(2)SO_(4) .Purity of H_(2)O_(2) is

0.2g of an organic compoundcontaningC, Hand O on combustion gave 0.147 g of CO_(2)and0.12g of water . Thepercentage content ofoxygen in the compoundis :

0.2 g of an organic compound containing C, Hand O ,on combustion yielded 0.147 g Co_2 and 0.12 g water .The percentage of oxygen in it is :

(0.1M "urea")/((A)),(0.1M NaCl)/((B)),(0.1MBaCl_(2))/((C))

0.1M solution of which of the following compounds will have the lowest freezing point ?

0.1m solution of KCI and BaCI_(2) are preparred. The freezing point of KCI solution is found to be- 4.0^(@)C.What will be the freezing point of BaCI_(2) solutionassuming that both KCI and BaCI_(2) solution are sompletely ionised in solution ?

0.1m K_(2)SO_(4) ionises to 10 x% to boil at same temperature as 0.2 glucose. What is x ?

0.1M CH_(3)COOH(aq.) Is mixed with 0.1M NaOH(aq.). The pH of solution on mixing will be (K_(a) ofCH_(3)COOH is 1.8xx10^(-5)

0.1g of an unknown substance was dissolved in 5g of camphor and it was found that the melting point of camphor was depressed by 5.3^(@)C. If K_(f) is 39.7, find the weight of 1 mole of the solute.

0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrichemical equivalent of copper ?

0.1914g of an organic acid is dissolved in approx. 20 ml of water. 25 ml of 0.12 N NaOH required for the complete neutralization of the acid solution. The equivalent weight of the acid is

0.1890g of an organic compound gave 0.2870g of silver chloride by Carius method. Find the percentage of chlorine in the compound

0.189 g of an organic substance gave in a carius determination 0.287 g of silver chloride. What is the % of Cl in the given compound?

0.188gm of silver bromide is obtained from x gm of an organic compound and percentage of bromine in the compound is 4. Then the x value is

0.188 g of silver bromide is obtained from 4 g of an organic compound. Calculate the percentate of bromine in the compound

0.170 g of a volatile liquid displaced 69.1 mL air in Victor Meyor's method. The atmospheric temperature was 15^@ C and pressure was 755 mm. (Aqueous tension at 15^@C = 12.7 mm ). Find the molecular weight of the liquid.

0.16g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of N_(2)H_(4) that has reacted with water in this solution. The K_(b) for N_(2)H_(4) is reacted with water in this solution. The K_(b) for N_(2)H_(4) is 4.0 xx 10^(-6) M

0.16g of an organic compound, on complete combustion, produced 0.44g of CO_(2) and 0.18g of H_(2)O. Calcuate the percentage of carbon and hydrogen in the organic compounds

0.16 gm of organic acid required 25 ml of (N)/(10) NaOH for complete neutralization. (M. wt of acids is 128). Then the basicity of acid is

0.16 gm of a dibasic acid require 25 ml of decinormal NaOH solution for complete neutralization. The moleucular weight of the acid

0.16 g of N_2H_4are dissolved in water and the total volume made upto 500 mL. Calculate the percentage of N_2H_4that has reacted with water in this solution. K_b( N_2H_4) = 4.0 xx 10^(-6) M

0.16 g of a dibasic organic acid required 25vm^(3)of 0.1 M NaOH for complete neutralization. Thre molecular mass of the acidof the acid is :

0.15g of an organic compound gave 0.12g of silver bromide by the Carius method. The percentage of bromine in the compound is :

0.15 mole of pyridinium chloride has been added into 500 cc of 0.2 M pyridinesolution. Calculate the pH and hydroxyl-ion concentration in this resulting solution assuming no change in volume. K_b for pyridine = 1.5 xx 10^(-9) M .

0.15 molalsolutionof NaCIhas freezingpoint -0.52 ^(@)C Calculatevan'tHofffactor.(K_(f)= 1.86 K kg mol^(-1) )

0.146g of an acid requires 20 ml of 0.1 N NaOH for complete neutralization. The equivalent weight of the acid is

0.12g of organic compound containing sulphur on treatment with Conc. HNO_(3) in a carius tube and then with excess of BaCl_(2) produced 0.233g of BaSO_(4). % of 'S' in compound ...........

0.126g of an acid requires 20 ml of 0.1N NaOH for complete neutralization. Equivalent weight of the acid is

0.126 g of an acid requires 20 mL of 0.1 N NaOH for complete neutralisation . Eq.wt of the acid is

0.126 g of an acid is needed to completely neutralise 20 mL 0.1 N NaOH solution. The equivalent weight of the acid is

0.123g of an organic compound produced 0.099g of CO_(2) and 0.0507g of H_(2)O. 0.185g of the same compound produced 0.319g of AgBr. Find the percentage of carbon, hydrogen and bromine in the compound

0.1 mole of N_(2)O_(4(g)) was sealed in a tube under one atmospheic conditions at 25^(@)C. Calculate the number of moles of NO_(2(g)) present, if the equlibrium N_(2)O_(4(g))hArr2NO_(2(g))(k_(p)=0.14) is reache3d after some time

0.1 mole of sugar is dissolved in 250 g of water. The freezing point of the solution is [K_(f) "for" H_(2)O = 1.86^(@)C "molal"^(-1)]

0.1 mole of XeF_(6) is treated with 1.8 g of water.The product obtained is

0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is added to 0.08 mole of HCl and the solution is diluted to one litre. The resulting [H^(+)] is

0.1 mole of CH_(3)NH_(2)(K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H^(+) concentration in the solution

0.1 mole of CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole for HCl and diluted to one litre. What will be the H^(+) concentration in the solution

0.1 mol of sugar was dissolved in 1 kg of water. The freezing point of the solution was found to be 272.814 K. What conclusion would you draw about the molecular state of sugar? K_(f) for water is "1.86 K kg mol"^(-1).

0.1 molal aqueous solution of sodium bromide freezes at -0.335^(@)C at atmospheric pressure. K_(f) for water is 1.86^(@)C. The percentage of dissociation of the salt in solution is

0.1 mole of a carbohydrate with empirical formula CH_(2)O contains 1 g of hydrogen. What is its molecular formula?

0.1 mole Ni is obtained on electrolysis of NiSO_(4) solution in neural solution, how much faraday electricity should be pass through the cell ?

0.1 mol of MnO_(4)^(-) in acidic medium can oxidize

0.1 mol of CH_(3)NH_(2)(K_(b)=5.0xx10^(-4)) is mixed with 0.08 mol of HCl and diluted to one litre. What will be the H^(+) ion concentration in the solution ?

0.1 mol HCl is dissolved in distilled water of volume V, then, at underset(Vtooo)(lim) (pH)_("solution") is equal to :

0.1 M solution of two electrolytes P and Q have specific conductance 4 times 10^(-4)" S "cm^(-1) " and " 6 times 10^(-6)" S "cm^(-1) respectively. Which among the following will have greater resistance to the flow of current? Give reason.