94420 + Interview Questions in Chemistry in Class 12 Page 1850 InterviewSolution

92451.	2,6-Dimethylheptane on monochlorination produces. . . . . . Derivatives
Answer» 5 6 3 4 Solution :Because `CH_(3)-UNDERSET(CH_(3))underset(\|)(C)H-CH_(2)-CH_(2)-CH_(2)-underset(CH_(3))underset(\|)(C)H-CH_(3)` has derivatives.

Discussion

92452.	25ml of a solution containing HCl and H_(2) SO_(4) required 10ml of a 1N NaOH solutionfor neutralization. 20ml of the same acid mixture on being treated with an excess of AgNO_(3) gives 0.1435 g of AgCl. The normality of the HCl and the normality of the H_(2) SO_(4) are respectivley.
Answer» 0.40N and 0.05N 0.5 N and 0.35 N 0.05 N and 0.25N 0.40 N and 0.5 N ANSWER :B

Discussion

92453.	25mL of a solution containing HCl and H_2SO_4 required 10 mL of a 1 N NaOH solution for neutralization.20 mL of the same acid mixture on being treated with an excess of AgNO_3 gives 0.1435 g of AgCl.The normality of the HCl and the normality of the H_2SO_4 are respectively.
Answer» 0.40 N and 0.05 N 0.05 N and 0.35 N 0.50 N and 0.25 N 0.40 N and 0.5 N Solution :LET normality of HCl is `N_1` and `H_2SO_4` is `N_2`. `:.` Mili eq. of HCl + Mili eq. of `H_2SO_4` = Mili eq. of NAOH `25xxN_1+25xxN_2=10xx1`..(1) `N_1+N_2=0.4`…(1) BY POAC Moles of `Cl^(-)`=moles of AgCl `(20xxN_1)/1000=0.1435/143.5=10^(-3)` `N_1=0.05 N` `N_2=0.35 N`

Discussion

92454.	25 mL of 3.0 M HCl are mixed with 75 mL of 4.0 M HCl. If the volumes are additive, the molarity of the final mixture will be :
Answer» 4.0 M 3.75 M 4.25 M 3.5 M Solution :If the TWO SOLUTIONS containing the same solute are mixed, the MOLARITY of the resulting solution is given as : `M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/V_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `=((3.0M)XX(25mL)+(4.0M)xx(75mL))/((25+75)mL)` =3.75 M

Discussion

92455.	25cm^(3) of 0.2 M solution of metal chloride (MCl_(x)) reacted with 150cm^(3) of 0.1 M AgNO_(3) solution completely to form the precipitate of AgCl. What is the formula of metal chloride ?
Answer» Solution :The reaction will :`""MCl_(x)+xAgNO_(3) rarr xAgCl+(NO_(3))_(x)` `25cm^(3)" of 0.2 M "MCl_(x)" soluiton contain "25xx0.2" MILLIMOLES, i.e., 5 millimoles of "5xx10^(-3)" moles"` `150cm^(3)" of 0.1 M "AgNO_(3)" solution CONTAINS "150xx0.1" millimoles i.e., 15 millimoles or "15xx10^(-3)" mole of "AgNO_(3)` 1 mole of `MCl_(x)` reacts with x moles of `AgNO_(3)` `therefore 5xx10^(-3)` mole of `MCl_(x)` will react with `AgNO_(3)=5xx10^(-3)x` moles `"As the given amounts react completely, "5xx10^(-3)x=15xx10^(-3)or x=3` Hence, the FORMULA will be `MCl_(3)`.

Discussion

92456.	25.4 gm of iodine and 14.2 gm of chlorine are made to react completely to yeild a mixture of ICI and ICI_(3) ratio of moles of ICI &ICI_(3) formed is (Atomic mass : I = 127 , Cl = 35.5 )
Answer» `1:1` `1:2` `1:3` `2:3` ANSWER :A

Discussion

92457.	25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of lCl and lCl_(3).Calculate the ratio of moles of lCl and lCl_(3).
Answer» `1:1` `1:2` `1: 3` `2 :3` SOLUTION :N/A

Discussion

92458.	25.3 g of sodium carbonate, Na_(2)CO_(3) is dissolved in enough water to make 250 mL of solution. If sodium carbonate completely, molar concentration of sodium ions, Na^(+) and carbonate ions, CO_(3)^(2-) are respectively (Molar mass of Na_(2)CO_(3)=106g mol^(-1))
Answer» 0.477 M and 0.477 M 0.955 M and 1.910 M 1.910 M and 0.955 M 1.90 M and 1.910 M Solution :`"Molarity of the solution"=(25.3)/(106)XX(1000)/(250)=0.955M` `Na_(2)CO_(3)rarr2Na^(+)+CO_+(3)^(2-)` `therefore""[Na^(+)]=2xx0.955M=1.910M` `[CO_(3)^(2-)]=0.955M.`

Discussion

92459.	25.2 gm of a mixture of NaHCO_(3) and Na_(2)CO_(3) is heated strongly, 0.66 gm of CO_(2) gas is evolved then the % mass of Na_(2) CO_(3) present in original mixture is -
Answer» 0.1 0.2 0.45 0.9 Solution :`2NaHCO_(3) overset(Delta)(rarr) Na_(2)CO_(3) + H_(2)O + CO_(2)` `n_(CO_(2)) = (0.66)/(44) = (3)/(200)` `n_(NaHCO_(3)) = n_(CO_(2)) xx 2 = (3)/(200) xx 2 = 0.03` `2NaHCO_(3) rarr Na_(2)CO_(3) + CO_(2) + H_(2)O` `0.03` MOLE`N = (0.03)/(2)` mole Mass `= 0.03 xx 84 = 2.52gm (NaHCO_(3))` `%` by mass of `NaHCO_(3) = 10%` & `Na_(2)CO_(3) = 90%`

Discussion

92460.	250 mLof 0.1 M HCL and 250 mL of 0.1 M KOH, both being at the same temperature, are mixed throughtly and the temperature rise is found to be DeltaT_(1). If the experiment is repeated using 500 mL each of the two solutions and DeltaT_(2) is the temperature rise, then which is true?
Answer» `DeltaT_(2)gt2DeltaT_(1)` `DeltaT_(1)=2DeltaT_(2)` `DeltaT_(1)=DeltaT_(2)` None of these Solution :In the second EXPERIEMENT, the heat produced is twice that in first case but at the same TIME thermal capacity of the resulting solution (twice as that in first case) is doubled. Hence, temperature rise will be the same.

Discussion

92461.	2.505 gm of a hydrated dibasic acid required 35 ml 1N NaOH for complete reaction. When 1.01 gm of the hydrated acid is heated to constant weight 0.72 gm of the anhydrous acid is obtained. The degree of hydration of this acid will be:
Answer» 10 3 12 2 Answer :D

Discussion

92462.	250 mL of sodium carbonate solution contains 2.65 grams of Na_(2)CO_(3). If 10 mL of this solution is diluted to one litre, what is the concentration of the resultant solution? (Mol. wt. of Na_(2)CO_(3) = 106)
Answer» 0.1 M 0.001 M 0.01 M `10^(-4)` M Solution :MOLARITY of given `N_(2)CO_(3)` solution `=2.65/106 XX 1/250 xx 1000 = 0.1 M` `M_(1)V_(1) = M_(2)V_(2)` or `10 xx 0.1 = 1000 xx M_(2)` or `M_(2) = 0.001 M`

Discussion

92463.	250 ml of 6 M HCl and 650 ml of 3 M HCl were mixed together. What volume of water is to be added so that the molarity of the final solution is 3 M ?
Answer» 300 ml 1150 ml 250 ml 135 ml Solution :Moles of HCl in 250 ml of 6M `=(6xx250)/(1000)` `=1.5` moles Moles of HCl in 650 ml of `3M=(3xx650)/(1000)` `=1.95` moles Total moles `=1.5+1.95=3.45` moles Molarity of solution `=(3.45)/(900)xx1000` `=3.833M` `M_(1)V_(1)=M_(2)V_(2)` `3.833xx900=3.0xxV_(2)` or `V_(2)=(3.833xx900)/(3.0)=1150` ml VOLUME of water added `=1150-900` `=250` ml

Discussion

92464.	25.0 mL of 1.0M HCl is combined with 35.0 mL of 0.5 M NaOH. The initial temperatures of the solutions is 25^@C, the density of the solution is 1.0 g/mL, the specific heat capacity of the solution is 4.184 J//g .^(@)C, the reaction is completed in insulated beaker, and the standard enthalpy of reaction for H^(+)(aq) + OH^(-)(aq) rarr H_(2)O (l) is -56 kJ//mol . What is the final temperature of the solution?
Answer» `27^@C` `28.9^@C` `30.1^@C` `32.8^@C` Solution :Heat Liberated due to neutralisation `=56 xx (0.5 xx 3.5 xx 10^(-3)) kJ = 0.98 kJ` [Meq. `H_2O` formed = `0.5 xx 35` ] `implies ""`Heat Liberated due to neutralisation = Heat gained by the system `implies 980 = (60 xx 1) xx 4.18 xx (T - 25) [ :' "TOTAL volume of solution" = 60 ml] implies T = 28.9^(@)C`.[ Total volume of solution = 60 ml]

Discussion

92465.	2.50 g sample of sodium bicarbonate when strongly heated gave 300cm^(3)" of "CO_(2) measured at 27^(@) and 760 mm pressure. Calculate the percentage purity of the sample.
Answer» ANSWER :`81.9%`

Discussion

92466.	25.0 g of FeSO_4. 7H_2O was dissolved in water containing dilute H_2SO_4, and the volume was made up to 1.0 L.25.0 mL of this solution required 20 mL of an N//10KMnO_4 solution for complete oxidation.The percentage of FeSO_4. 7H_2O in the acidic solution is
Answer» 0.78 0.98 0.89 0.79 Solution :MILI EQ. of `FeSO_4. 7H_2O` in 25 ml =Mili eq. of `KMnO_4` used =2 Milli eq. Mili eq. of `FeSO_4. 7H_2O` in 1000 ml =80 Mili eq. mass of `FeSO_4. 7H_2O` in solution =`80/1xx278xx1/1000=22.24` gm % of `FeSO_4 . 7H_2O=22.24/25xx100=88.96=89%`

Discussion

92467.	25 volumes of H_(2)O_(2) means
Answer» `25%H_(2)O_(2)` `25CM^(3)` of the SOLUTION contains 1 G of `H_(2)O_(2)` `1cm^(3)` of solution liberates `25cm^(3)` of `O_(2)` at N.T.P. `25cm^(3)` of the solution contains 1 MOLE of `H_(2)O_(2)`. Answer :C

Discussion

92468.	25 mL of household bleachsolution was mixed with30 mL of o.50 M Kl and 10 mL of 4N aceticacid. In the titration of the liberated iodine, 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point. The molarityof the household bleach solution is :
Answer» 0.48 M 0.96 M 0.24 M 0.024 M Answer :C

Discussion

92469.	25 ml of H_(2)O_(2) solution were added to excess of acidified solution of Kl and iodine so liberated required 20 ml of 0.1 N Na_(2)S_(2)O_(3) for titration . The normality of H_(2)O_(2) is
Answer» `0.02` `0.04` `0.08` `0.03` SOLUTION :N/A

Discussion

92470.	Bleaching powder and bleach solution are produced on a large scale and used in several hous-hold products. The effectiveness of bleach solution id often measured by iodometry. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point. The molarity of the household bleach solution is :
Answer» 0.24 M 0. 48 0.024 M 0.96 M Answer :A

Discussion

92471.	25 ml of H_(2)O_(2) solution was added to the excess of acidified Kl solution . The iodine so liberated required 40 ml of 0.1 N sodium thiosulphate solution. Calculate the normality of H_(2)O_(2) solution :
Answer» `0.08` `0.16` `0.02` `0.20` Solution :Let the NORMALITY of `H_(2)O_(2)` SOLN be`N_(x)`. No. of milliequivalents of `Na_(2)S_(2)O_(3) ` = No. of milliequivalents of `H_(2)O_(2)` `40 XX 0.1 = 25 xx N_(x)` ` N_(x) = (40 xx 0.1)/25 = 0.16` N

Discussion

92472.	25 ml of FeC_2O_4 dissolved in 186 gm of water calculate depression in freezing point. It 10 ml of same FeC_2O_4 titrated with 30 ml of 0.4 M KMnO_4 in acidic medium (k_f for H_2O=1.86,Assume 100% ionisation of FeC_2O_4).
Answer» Solution :`x_(FeC_2O_4)=x_(KMnO_4) " " undersetn(FeC_2O_4)toundersetn(Fe^(+2))+undersetn(C_2O_4^(-2))` `x_(Fe^(+2))+x_(C_2O_4^(-2))=x_(KMnO_4) " " n_(FeC_2O_4)=(MV)/1000` `1xxn+2xxn=(0.4xx5xx30)/1000 " " 0.02=(Mxx10)/1000` `n=(2xx10)/1000=0.02 " " M=2` n=no . of mole of SUBSTANCE X=equivalent of substance `DeltaT_f=i K_fm=2xx1.86xx((25xx2)/1000xx1000/186)=100/100=1`

Discussion

92473.	25 ml of a solution of Na_(2)CO_(3) having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of 0.84N-H_(2)SO_(4) that will be completely neutralized by 125 grams of the Na_(2)CO_(3) solution
Answer» 460 ml 540 ml 480 ml 470 ml Solution :`N_(1)V_(1)=N_(2)V_(2)` `Nxx25=(109.25xx32.9)/(36.5)impliesN=(109.5xx32.9)/(36.5xx25)` `N_(3)V_(3)=N_(4)V_(4)""(V_(3)=(m)/(d),V_(3)=(125)/(1.25))` `(109.5xx32.9)/(36.5xx25)xx100=0.84xxV impliesV=470ml`

Discussion

92474.	25 mL ofa solution of Fe^(2+)ions was titrated with a solution of the sxidising agent Cr_(2)O_(7)^(2-). 32.45 mL of 0.0153 M K_(2)Cr_(2)O_(7) solution war required .What is the prepare 1 litre of normal acid solution ?
Answer» SOLUTION :`179.4, 820.6`

Discussion

92475.	25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 mL. The molarity of barium hydroxide solution was
Answer» 0.14 0.28 0.35 0.07 Answer :D

Discussion

92476.	25 mL of a solution of Fe^(2+) ions was titrated with a solution of the oxidizing agent Cr_(2)O_(7)^(2-) . 50 mL of 0.01 M K_(2)Cr_(2)O_(7) solution was required. What is the molarity of the Fe^(2+) solution ?
Answer» ANSWER :0.12 M.

Discussion

92477.	25 ml of a dilute aqueous solution of p-hydroxy benzoic acid is titrated with NaOH ( aq), the soloution has pH = 4.57 , when 8.12 ml of 0.0200 M NaOH had been added, and pH = 7.02 after 16.24 ml had been added ( the equivalent point). Use these data to determine Ka_(1) and Ka_(2) for p-hydroxy benoic acid. HOC_(6) H_(4) COOH + H_(2) OhArr H_(3) O^(+) + HOC_(6) H_(4) COO^(-) HOC_(6) H_(4) COO^(-) + H_(2) O hArr H_(3) O^(+) + ""^(-)OC_(6)H_(5)COO^(-) , Ka_(2)
Answer» Solution :It is GIVEN that the equivalent point occurs when 16.24ml of NaOH had been added. Thus after 8.12 ml had been added, half neutralization occurs. At the half-neutralisation point, `[HOC_(6)H_(4)COO^(-)] = [ HOC_(6)H_(4)COOH^(-) ]` then, pH `= pKa_(1) + log"" ( [ HOC_(6)H_(4)COO^(-)])/([HOC_(6)H_(4)COOH])=pKa_(1)` `:. pKa_(1) = pH = 4.57 , log Ka_(1) = - 4.57` `:. Ka_(1) = 2.7 xx 10^(-5)` `HOC_(6)H_(4)COO^(-) ` is now amphoteric ion, a base as well as an acid `HOC_(6)H_(4)COO^(-) + H_(2) OhArr O^(-) C_(6) H_(4) COO^(-) + H_(3)O^(-) `( acidic) `HOC_(6)H_(4)COO^(-) + H_(2) O hArr HOC_(6)H_(4)COOH + OH^(-) `( basic ) For such cases, when equivalent point is obtained at pH = 7.02, `pH = (( p Ka_(1)+ pKa_(2))/(2))` `7.02 = ( 4.57+ pKa_(2))/( 2)` `pKa_(2) = 9.47` `:. Ka_(2) = 3.4 xx 10^(-10)`

Discussion

92478.	25 ml of "4 N HNO"_(3), 15 ml of "1 MH"_(2)SO_(4) and 20 ml of HCl of unknown concentration are mixed and made upto 1 litre. 20 ml of this solution required 26 ml of Ba(OH)_(2) solution. The Ba(OH)_(2) solution is prepared by dissolving 4.725 g of Ba(OH)_(2) per 250 ml of pure water. Calculate the normality of HCl solution. [Hint : Take hydrated form of Barium Hydroxide, i.e. Ba(OH)_(2)·8 H_(2)O]
Answer» Solution :Equivalent WEIGHT of `BA(OH)_(2).8H_(2)O=(315)/(2)=157.5` Normality of `Ba(OH)_(2).8H_(2)O` solution `=(4.725)/(157.5xx0.25)=0.1198` EQUIVALENTS of `Ba(OH)_(2).8H_(2)O` solution used `=(26xx0.1198)/(1000)` NUMBER of equivalents of acid `=20xx((25xx4)+(15xx2)+(20xxa))/(1000xx1000)=20xx(130+20a)/(1000)` Number of equivalents of `Ba(OH)_(2)·8H_(2)O` solution = Number of equivalents of acid `20xx(130+20a)/(1000)=(26xx0.1198)/(1000), a=1.287N`

Discussion

92479.	25 ml of 3.0 M HNO_3 are mixed with 75 ml of 4.0 M HNO_3 . If the volume are additive the molarity of the final mixture would be :
Answer» 3.25 M 4.0M 3.75 M 3.50 M Answer :C

Discussion

92480.	25mL of 2N HCl, 50 mL "of" 4N HNO_(3) and xmL H_(2)SO_(4) are mixed together and the total volume is made up to 1L after dilution. 50mL if this acid ixture completely reacteed with 25mL of a 1 N Na_(2)CO_(3) solution. The value of x is:
Answer» 250 ml 62.5 ml 100 ml 125 ml Answer :B

Discussion

92481.	2.5 ml of (2)/(5)M weak monoacidic base (K_(b) = 1 xx 10^(-12) at 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@)C. The concentration of h^(+) at equivalence point is (K_(w) = 1xx 10^(-14) at 25^(@)C)
Answer» `3.7 xx 10^(-13) M` `3.2 xx 10^(-7) M` `3.2 xx 10^(-2) M` `2.7 xx 10^(-2) M` Solution :`BOH + HCl RARR BCL + H_(2)O` At equivalence, `M_(1)V_(1) = M_(2)V_(2)` `(2)/(5) xx 2.5 = (2)/(15) xx V_(2), V_(2) = 7.5 ML HCl = 7.5 ml` Moles of BOH in 2.5 ml = 0.01 Moles of HCl in 7.5 ml = 0.001 `:.` Moles of salt formed = 0.001 [salt of S.A.F. W. B] Total value = 2.5 + 7.5 = 10 ml. = 0.01 Conc. of salt `= (0.001)/(0.01) = 0.1` mole/litre HYDROLYSIS of salt take place `pH = 7 - (1)/(2) (pK_(b) + log C)` `pK_(b) = -log 10^(-12) = 12 pH = 7 - (1)/(2)(12-1)` `[because log C = -1 log.1 = -1]` `pH = 1.5, -log H^(+) = 1.5 log H^(+) = -1.5, log H^(+) = 2.5` `H^(+) = 3.2 xx 10^(-2)`.

Discussion

92482.	2.5 mL of 2/5 M weak monoacidic base (K_(b)=1xx10^(-12)" at "25^(@)C) is titrated with 2/15 M HCl in water at 25^(@). The concentration of H^(+) at equivalence point is (K_(w)=1xx10^(-14)" at "25^(@)C)
Answer» `3.7xx10^(-13)M` `3.2xx10^(-7)M` `3.2xx10^(-2)M` `2.7xx10^(-2)M` Solution :The base is very weak, the CATIONS of the SALT formed at the equivalence point shall undergo hydrolysis to a greater extent and so the degree of hydrolysis, h cannot be neglected in comparison to 1. Now apply `Kh=(Ch^(2))/(1-h)=K_(w)/K_(b),C=0.1M,[H^(+)]=Ch`

Discussion

92483.	25 mL of 0.017 " M " H_(2)SO_(3)^(-)in strongly acidic required the addition of 16.9mL of 0.01M MnO_(4)^(-) for itscomplete oxidation t SO_(4)^(2-)or HsO_(4)^(-) .in neutralsolution it required28.6 mL . Assign oxidation numbers to Mn in eachof the products .
Answer» SOLUTION :` {:( HSO_(3)^(-),overset(" change in ON = 2") to,SO_(4)^(2-),or HSO_(4)^(-)),(+4,,+6,"+6"):}` ` :. 0.017 " M " HSO_(3)^(-) -= 2 xx 0.017 N "" ..(Eqn.6i)` ` = 0.034 N ` In the first case suppose the On of Mn in the PRODUCTIS X ` :. " " 0.01 " M " MnO_(4)^(-) = 0.01 (7-X) " N " MnO_(4)^(-) "" ...(Eqn.6i)` m.e of `HSO_(3)^(-) = " m.e of " MnO_(4)^(-)` `0.034 xx 25 = 0.01 ( 7 - X ) 16.9` ` :. 7 - X = (0.034 xx25)/(16.9 xx0.01)= 5.00` or X = 2 . Now in the second titration , suppose the oN of Mn in the product is Y . ` :. 0.01" M " MnO_(4)^(-) = 0.01 (7 - Y) N MnO_(4)^(-)` ` 0.034 xx 25 = 0.01 (7-Y) xx 28.6` ` 7 - Y = = (0.034 xx 25)/(0.01 xx 28.6) = 3 ` ` :. Y = 4 `

Discussion

92484.	2.5 mL of (2)/(5) M weak monoacidic base ( K_(b) = 1 xx 10^(-12)at 25^(@)C ) is titrated with (2)/( 15) M HCl in water at 25^(@)C . The concentration of H^(+) at equivalence point is ( K_(w) = 1 xx 10^(-14)at 25^(@)C )
Answer» `3.7 xx 10^(-13) M` `3.2 xx 10^(-7) M` `3.2 xx 10^(-2) M` NONE of these Answer :B

Discussion

92485.	25 ml from a stock solution containing NaHCO_(3) andNa_(2)CO_(3) was diluted to 250 ml with CO_(2) free distilled water. 25 ml of the diluted solution when titrated with 0.12 M HCl required 8 ml., when phenolphthalein was used as an indicator. Na_(2)CO_(3) + HCl overset("HPh")(rarr)NaHCO_(3) When 20 ml of diluted solution was titrated with same acid it required 18 ml when methlyorange was used as an indicator. Na_(2)CO_(3)+2HCloverset("MeOH")(rarr)2NaCl+H_(2)O +CO_(2) NaHCO_(3) +HCl overset( MeOH)(rarr) NaCl +H_(2)O +CO_(2) Millimoles of NaHCO_(3) present in stock solution.
Answer» 0.624 2.16 1.536 7.8 Answer :D

Discussion

92486.	25 ml from a stock solution containing NaHCO_(3) andNa_(2)CO_(3) was diluted to 250 ml with CO_(2) free distilled water. 25 ml of the diluted solution when titrated with 0.12 M HCl required 8 ml., when phenolphthalein was used as an indicator. Na_(2)CO_(3) + HCl overset("HPh")(rarr)NaHCO_(3) When 20 ml of diluted solution was titrated with same acid it required 18 ml when methlyorange was used as an indicator. Na_(2)CO_(3)+2HCloverset("MeOH")(rarr)2NaCl+H_(2)O +CO_(2) NaHCO_(3) +HCl overset( MeOH)(rarr) NaCl +H_(2)O +CO_(2) Amount of NaOH that should be added to convert all bicarbonate into carbonate in 100 ml stock solution
Answer» 1.248gm 0.312 GM `3.12 XX 10^(-2) `gm `7.8 xx 10^(_3)` gm ANSWER :A

Discussion

92487.	25 ml from a stock solution containing NaHCO_(3) andNa_(2)CO_(3) was diluted to 250 ml with CO_(2) free distilled water. 25 ml of the diluted solution when titrated with 0.12 M HCl required 8 ml., when phenolphthalein was used as an indicator. Na_(2)CO_(3) + HCl overset("HPh")(rarr)NaHCO_(3) When 20 ml of diluted solution was titrated with same acid it required 18 ml when methlyorange was used as an indicator. Na_(2)CO_(3)+2HCloverset("MeOH")(rarr)2NaCl+H_(2)O +CO_(2) NaHCO_(3) +HCl overset( MeOH)(rarr) NaCl +H_(2)O +CO_(2) Concentration of NaHCO_(3) in gm//lit.
Answer» 0.312 2.62 3.12 26.208 Answer :D

Discussion

92488.	25 mL Ba(OH)_(2) solution is neutralized by 35 mL 0.1 M HCl, what will be molarity of Ba(OH)_(2) solution ?
Answer» `0.42` `0.21` `0.07` `0.14` ANSWER :C

Discussion

92489.	2.5 litre of 1M NaOH solution is mixed with a 3.0 litres of 0.5 M NaOH solution. The molarity of the resulting solution is :
Answer» 0.80 M 1.0 M 0.73 M 0.50 M. Solution :`M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `M_(3)=(1xx2.5+0.5xx3)/(2.5+3)` `(4.0)/(5.5)=0.73M`

Discussion

92490.	2.5 L of a sample of a gas at 27^(@)C and 1 bar pressure is compressed to a volume of 500 mL keeping the temperature constant, find the % increase in the pressure.
Answer» Solution :`P_(1)V_(1)=P_(2)V_(2)rArr1xx2.5=P_(2)xx1//2rArrP_(2)=5` bar % INCREASE `=("Increase in pressure")/("Initial pressure")xx100=(4)/(1)xx100=400%`

Discussion

92491.	2.5g sample of copper is dissolved in excess of H_(2)SO_(4) to prepare100 mL of 0.02 MCuSO_(4) (aq) . 10 mL of 0.02 M solution of CuSO_(4) (aq) is mixed with excess of Kl to show the following changes CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2) 2Cu l_(2) to Cu_(2)l_(2) + l_(2) The liberated iodine is titrated with hypo (na_(2)S_(2)O_(3)) andrequires V mL of 0.1 M hypo solution for its complete reduction . The amount of l_(2) liberated in the reaction of 10 mL of 0.02 M solution with Kl (excess ) is :
Answer» `0.051` G `0.0254` g `0.102` g `0.204` g SOLUTION :N/A

Discussion

92492.	2.5g sample of copper is dissolved in excess of H_(2)SO_(4) to prepare100 mL of 0.02 MCuSO_(4) (aq) . 10 mL of 0.02 M solution of CuSO_(4) (aq) is mixed with excess of Kl to show the following changes CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2) 2Cu l_(2) to Cu_(2)l_(2) + l_(2) The liberated iodine is titrated with hypo (na_(2)S_(2)O_(3)) andrequires V mL of 0.1 M hypo solution for its complete reduction . Percentage purity of sample is :
Answer» `10.16` `5.08` `2.54` `1.27` SOLUTION :N/A

Discussion

92493.	2.5g sample of copper is dissolved in excess of H_(2)SO_(4) to prepare100 mL of 0.02 MCuSO_(4) (aq) . 10 mL of 0.02 M solution of CuSO_(4) (aq) is mixed with excess of Kl to show the following changes CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2)2Cu l_(2) to Cu_(2)l_(2) + l_(2) The liberated iodine is titrated with hypo (na_(2)S_(2)O_(3)) andrequires V mL of 0.1 M hypo solution for its complete reduction . The volume (V) of hypo required is :
Answer» 2 ML 20 mL 1 mL 10 mL Solution :N/A

Discussion

92494.	2.5 g of pure calcium carbonate, when strongly heated left a residue of 1.400 g. The evolved gas was found to occupy 624 mL at 27^@ C and 755 mm pressure. Calculate the molecular weight of the gas.
Answer» ANSWER :43.67

Discussion

92495.	25 g of MCl_(4) contains 0.5 mol chlorine then its molecular mass is:
Answer» 100g `"mol"^(-1)` 200g `"mol"^(-1)` `150g "mol"^(-1)` `400g "mol"^(-1)` SOLUTION :1 mol of `MCl_(4)` contains 4 mol of CHLORINE `because`0.5 mol chlorine is present in 25 g of `MCl_(4)` `THEREFORE` 4 mol chlorine will be present in `(25)/(0.5)xx4` i.e. 200g of `MCl_(4)`.

Discussion

92496.	25 g of each of the following gases are taken at 27^(@)C and 600 mm pressure.Which of these will have the least volume?
Answer» HBr HCl HF HI Solution :Ideal gas EQUATION is PV=nRT If pressure and TEMPRATURE is same for all the gases then V`prop`n(From above equation) n=no. of moles of the gas `=(WT.)/(M.wt.)` From the options,HI has more molecular weight than remaining gases,so it has LEAST VOLUME.

Discussion

92497.	25 g of a solute of molar mass 250 g mol^(-1) is dissolved in 100 ml of water to obtain a solution whose density is 1.25 g ml^(-1). The molarity and molality of the solution are respectively
Answer» 0.75 and 1 0.8 and 1 1 and 0.8 1 and 0.75 SOLUTION :No. of moles of solute `=(25)/(250)=0.1` moles Molarity `=("No. of moles of solute")/("Volume of solvent in litre")=(0.1)/((100)/(1000))=1M` Molality `=("no. of moles of solute")/("mass of solvent in KG")` Mass `= d xx V` `=1.25xx100=125 g` Molality `=(0.1)/(125//1000)=0.8m`.

Discussion

92498.	2.5 Faraday of electricity are passed through a solution of a solution of CuSO_4. The Number of gram equivalents of copper deposited on the cathode are :
Answer» 1 2 2.5 1.25 Answer :C

Discussion

92499.	25 cm^(3) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acidsolution is
Answer» 0.064 0.045 0.015 0.032 Solution :`undersetoverset(\|)(COOH)(C )OOH + 2NAOH to undersetoverset(\|)(COONA)(C ) OONa + 2H_2O` MOLES of 0.064g of NaOH ` = (0.064)/(40) = 0.0016` mol 2 moles of NaOH react with 1 mol of oxalic acid ` therefore ` 0.0016 mol of NaOH react with oxalic acid ` = (0.0016)/(2) = 8 xx 10^(-4) ` mol volume of solution = 25 mL Molarity of oxalic acid solution` = (8 xx 10^(-4) xx 1000)/(25)` 0.032 M

Discussion

92500.	24^(@)F can be given in Kelvin as
Answer» 26.87 K 268.7 K 2687 K 2.687 K ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in .

2,6-Dimethylheptane on monochlorination produces. . . . . . Derivatives

25ml of a solution containing HCl and H_(2) SO_(4) required 10ml of a 1N NaOH solutionfor neutralization. 20ml of the same acid mixture on being treated with an excess of AgNO_(3) gives 0.1435 g of AgCl. The normality of the HCl and the normality of the H_(2) SO_(4) are respectivley.

25mL of a solution containing HCl and H_2SO_4 required 10 mL of a 1 N NaOH solution for neutralization.20 mL of the same acid mixture on being treated with an excess of AgNO_3 gives 0.1435 g of AgCl.The normality of the HCl and the normality of the H_2SO_4 are respectively.

25 mL of 3.0 M HCl are mixed with 75 mL of 4.0 M HCl. If the volumes are additive, the molarity of the final mixture will be :

25cm^(3) of 0.2 M solution of metal chloride (MCl_(x)) reacted with 150cm^(3) of 0.1 M AgNO_(3) solution completely to form the precipitate of AgCl. What is the formula of metal chloride ?

25.4 gm of iodine and 14.2 gm of chlorine are made to react completely to yeild a mixture of ICI and ICI_(3) ratio of moles of ICI &ICI_(3) formed is (Atomic mass : I = 127 , Cl = 35.5 )

25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of lCl and lCl_(3).Calculate the ratio of moles of lCl and lCl_(3).

25.3 g of sodium carbonate, Na_(2)CO_(3) is dissolved in enough water to make 250 mL of solution. If sodium carbonate completely, molar concentration of sodium ions, Na^(+) and carbonate ions, CO_(3)^(2-) are respectively (Molar mass of Na_(2)CO_(3)=106g mol^(-1))

25.2 gm of a mixture of NaHCO_(3) and Na_(2)CO_(3) is heated strongly, 0.66 gm of CO_(2) gas is evolved then the % mass of Na_(2) CO_(3) present in original mixture is -

250 mLof 0.1 M HCL and 250 mL of 0.1 M KOH, both being at the same temperature, are mixed throughtly and the temperature rise is found to be DeltaT_(1). If the experiment is repeated using 500 mL each of the two solutions and DeltaT_(2) is the temperature rise, then which is true?

2.505 gm of a hydrated dibasic acid required 35 ml 1N NaOH for complete reaction. When 1.01 gm of the hydrated acid is heated to constant weight 0.72 gm of the anhydrous acid is obtained. The degree of hydration of this acid will be:

250 mL of sodium carbonate solution contains 2.65 grams of Na_(2)CO_(3). If 10 mL of this solution is diluted to one litre, what is the concentration of the resultant solution? (Mol. wt. of Na_(2)CO_(3) = 106)

250 ml of 6 M HCl and 650 ml of 3 M HCl were mixed together. What volume of water is to be added so that the molarity of the final solution is 3 M ?

2.50 g sample of sodium bicarbonate when strongly heated gave 300cm^(3)" of "CO_(2) measured at 27^(@) and 760 mm pressure. Calculate the percentage purity of the sample.

25.0 g of FeSO_4. 7H_2O was dissolved in water containing dilute H_2SO_4, and the volume was made up to 1.0 L.25.0 mL of this solution required 20 mL of an N//10KMnO_4 solution for complete oxidation.The percentage of FeSO_4. 7H_2O in the acidic solution is

25 volumes of H_(2)O_(2) means

25 mL of household bleachsolution was mixed with30 mL of o.50 M Kl and 10 mL of 4N aceticacid. In the titration of the liberated iodine, 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point. The molarityof the household bleach solution is :

25 ml of H_(2)O_(2) solution were added to excess of acidified solution of Kl and iodine so liberated required 20 ml of 0.1 N Na_(2)S_(2)O_(3) for titration . The normality of H_(2)O_(2) is

25 ml of H_(2)O_(2) solution was added to the excess of acidified Kl solution . The iodine so liberated required 40 ml of 0.1 N sodium thiosulphate solution. Calculate the normality of H_(2)O_(2) solution :

25 ml of FeC_2O_4 dissolved in 186 gm of water calculate depression in freezing point. It 10 ml of same FeC_2O_4 titrated with 30 ml of 0.4 M KMnO_4 in acidic medium (k_f for H_2O=1.86,Assume 100% ionisation of FeC_2O_4).

25 mL ofa solution of Fe^(2+)ions was titrated with a solution of the sxidising agent Cr_(2)O_(7)^(2-). 32.45 mL of 0.0153 M K_(2)Cr_(2)O_(7) solution war required .What is the prepare 1 litre of normal acid solution ?

25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 mL. The molarity of barium hydroxide solution was

25 mL of a solution of Fe^(2+) ions was titrated with a solution of the oxidizing agent Cr_(2)O_(7)^(2-) . 50 mL of 0.01 M K_(2)Cr_(2)O_(7) solution was required. What is the molarity of the Fe^(2+) solution ?

25 ml of 3.0 M HNO_3 are mixed with 75 ml of 4.0 M HNO_3 . If the volume are additive the molarity of the final mixture would be :

25mL of 2N HCl, 50 mL "of" 4N HNO_(3) and xmL H_(2)SO_(4) are mixed together and the total volume is made up to 1L after dilution. 50mL if this acid ixture completely reacteed with 25mL of a 1 N Na_(2)CO_(3) solution. The value of x is:

2.5 ml of (2)/(5)M weak monoacidic base (K_(b) = 1 xx 10^(-12) at 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@)C. The concentration of h^(+) at equivalence point is (K_(w) = 1xx 10^(-14) at 25^(@)C)

2.5 mL of 2/5 M weak monoacidic base (K_(b)=1xx10^(-12)" at "25^(@)C) is titrated with 2/15 M HCl in water at 25^(@). The concentration of H^(+) at equivalence point is (K_(w)=1xx10^(-14)" at "25^(@)C)

25 mL of 0.017 " M " H_(2)SO_(3)^(-)in strongly acidic required the addition of 16.9mL of 0.01M MnO_(4)^(-) for itscomplete oxidation t SO_(4)^(2-)or HsO_(4)^(-) .in neutralsolution it required28.6 mL . Assign oxidation numbers to Mn in eachof the products .

2.5 mL of (2)/(5) M weak monoacidic base ( K_(b) = 1 xx 10^(-12)at 25^(@)C ) is titrated with (2)/( 15) M HCl in water at 25^(@)C . The concentration of H^(+) at equivalence point is ( K_(w) = 1 xx 10^(-14)at 25^(@)C )

25 mL Ba(OH)_(2) solution is neutralized by 35 mL 0.1 M HCl, what will be molarity of Ba(OH)_(2) solution ?

2.5 litre of 1M NaOH solution is mixed with a 3.0 litres of 0.5 M NaOH solution. The molarity of the resulting solution is :

2.5 L of a sample of a gas at 27^(@)C and 1 bar pressure is compressed to a volume of 500 mL keeping the temperature constant, find the % increase in the pressure.

2.5 g of pure calcium carbonate, when strongly heated left a residue of 1.400 g. The evolved gas was found to occupy 624 mL at 27^@ C and 755 mm pressure. Calculate the molecular weight of the gas.

25 g of MCl_(4) contains 0.5 mol chlorine then its molecular mass is:

25 g of each of the following gases are taken at 27^(@)C and 600 mm pressure.Which of these will have the least volume?

25 g of a solute of molar mass 250 g mol^(-1) is dissolved in 100 ml of water to obtain a solution whose density is 1.25 g ml^(-1). The molarity and molality of the solution are respectively

2.5 Faraday of electricity are passed through a solution of a solution of CuSO_4. The Number of gram equivalents of copper deposited on the cathode are :

25 cm^(3) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acidsolution is

24^(@)F can be given in Kelvin as