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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
How many grams of phosphoric acid would be needed to neutralise `100 g` of magnesium hydroxide? (The molecular weight are: `H_(3)PO_(4) = 98` and `Mg(OH)_(2) = 58.3)`A. `66.7 g`B. `252 g`C. `112 g`D. `168 g` |
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Answer» Correct Answer - C Meq. of `H_(3)PO_(4)=`Meq. of `Mg(OH)_(2)` `:.` Meq. of `Mg(OH)_(2)=(100)/(58.3//2)xx100` `=3430.53` `(H_(3)P_(4)`is tribasic acid) `:. (w)/(98//3)xx1000=3430.53` `:. w=112g` |
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| 2. |
The concentration of solutions can be expressed in number of ways such that Normality, Molarity, Molality, Mole fractions, Strength, `%` by weight, `%` by volume and `%` by strength. The molarity of ionic compound is usually expressed as formality beacuse we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and `%` by weight terms. Volume of water required to convert `100 mL 0.5M NaOH` solutions to `0.2M NaOH` solution is:A. `250 mL`B. `150 mL`C. `100 mL`D. `400 mL` |
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Answer» Correct Answer - C Meq.of.dil. `NaOH= "Meq.of conc."NaOH` `0.2xxV=0.5xx100` `therefore V=250 mL` `therefore H_(2)O` needed `=250-100=150 mL` |
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| 3. |
The concentration of solutions can be expressed in number of ways such that Normality, Molarity, Molality, Mole fractions, Strength, `%` by weight, `%` by volume and `%` by strength. The molarity of ionic compound is usually expressed as formality beacuse we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and `%` by weight terms. Number of oxalate ions in `100 mL` of `0.1N` oxalic acis is:A. `(N_(A))/(100)`B. `(N_(A))/(20)`C. `(N_(A))/(200)`D. `(N_(A))/(1000)` |
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Answer» Correct Answer - C Meq.of oxalate ion `=100xx0.1=10` `therefore` Milli-mole of oxalate ion `=(10)/(2)=5` ` therefore` No. of oxalate ion `=(N_(A)xx5)/(1000)=(N_(A))/(200)` |
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| 4. |
A compound has the molecular formula `X_(4)O_(6)`. If `10 g "of" X_(4)O_(6)` has `5.72 g X`, atomic mass of `X` is:A. `32 "amu"`B. `37 "amu"`C. `42 "amu"`D. `98 "amu"` |
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Answer» Correct Answer - A `(4a+96)gX_(4)O_(6) "has" 4ag X` `:. 10 g X_(4)O_(6)"has"[(4axx10)/(4a+96)]` `:. (4axx10)/(4a+96)=5.72 :. a=32` |
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| 5. |
Which is not a molecular formula?A. `C_(6)H_(12)O_(6)`B. `Ca(NO_(3))_(2)`C. `C_(2)H_(4)O_(2)`D. `N_(2)O` |
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Answer» Correct Answer - B Ionic compound are usually referred in terms of their formula. The exact mol.wt.of ionic compounds is not determined experimentially. Simply their formula weight is reported. |
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| 6. |
Number of `H^(+)` ions in `100 mL` of `0.001 M H_(2)SO_(4)` is:A. `6xx10^(20)`B. `1.2xx10^(18)`C. `12xx10^(18)`D. `1.2xx10^(20)` |
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Answer» Correct Answer - D `"Meq.of"H_(2)SO_(4)=100xx0.001=0.1` `1000 mM"of" H_(2)SO_(4)=2 "mole"H^(+)` `:. 0.1 mMH_(2)SO_(4)=(2xx0.1xx6xx10^(23))/(1000)` `=1.2xx10^(20)` |
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| 7. |
Number of positive ions in `1.45` mole of `K_(2)SO_(4)` are:A. `1.75 xx 10^24`B. `8.73 xx 10^(23)`C. `8.73 xx 10^(24)`D. `1.75 xx 10^(23)` |
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Answer» Correct Answer - A `1 "mole" K_(2)SO_(4)"has" 2N K^(+)"ions"`. |
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| 8. |
`20 g` of an acid furnishes `0.5 "mole of" H_(3)O^(+)` ions in its aqueous solution. The value of `1` equivalent of the acid will be:A. `40 g`B. `20 g`C. `10 g`D. `100 g` |
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Answer» Correct Answer - A `0.5` mole of `H_(3)O^(+)=20 g`, Also `H_(3)O^(+)` is monovalent, Thus, mol.wt.`= Eq.wt.` `:. 1 "mole of" H_(3)O^(+)= 40 g` |
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| 9. |
In a compound `A_(x)B_(y)`:A. Mole of `A= "Mole of" B = "Mole of" A_(x)B_(y)`B. Eq. of `A=Eq. "of" B=Eq. "of" A_(x)B_(y)`C. `yxx"mole of" A=yxx"mole of" B=(x+y)xx"mole of" A_(x)B_(y)`D. `yxx"mole of" A= yxx"mole of"B` |
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Answer» Correct Answer - B Equal equivalent reacts together to give same equivalent of product. |
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| 10. |
A solution required `[OH^(-)]=2M`. If degree of dissociation of `Mg(OH)_(2)` is `alpha`, what analytical molarity solution of `Mg(OH)_(2)` isA. `alpha`B. `2 alpha`C. `(1)/(2alpha)`D. `(1)/(alpha)` |
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Answer» Correct Answer - D `[OH^(-)]=2M` `{:(Mg(OH)_(2)hArr,Mg^(2+)+,2OH^(-),,),(C,0,0,,),(C(1-alpha),Calpha,2Calpha,,):}` `:. 2 Calpha=2M` `:. C=(2)/(2 alpha)=(1)/(alpha)` |
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| 11. |
`10 "mole of" kSO_(2)` and `15 "mole of" O_(2)` were passed over catalyst to produce `8 "mole of" SO_(3)`. The ratio of `SO_(2)` and `SO_(3)` moles in mixture is:A. `5//4`B. `1//4`C. `1//2`D. `3//4` |
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Answer» Correct Answer - B `{:(,SO_(2)+,1//2O_(2)hArr,SO_(3),),("Initial mole",10,15,,),("Final mole",10-x,15-x/2,x,):}` `x=8` `:. SO_(2):SO_(3)::2:8` |
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| 12. |
How many moles are in `96 g O_(2)?` |
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Answer» Correct Answer - 3 `because 32 gO_(2) "has mole"=1` `:. 96 g O_(2) "has mole"=(96xx1)/(32)=3` |
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| 13. |
A sample of hard water contains `96 pp m."of" SO_(4)^(2-)` and `183 pp m "of" HCO_(3)^(-)`, with `Ca^(2+)` as the only cation. How many moles of `CaO` will be required to remove `HCO_(3)^(-)` from `1000 kg` of this water? If `1000 kg` of this water is treated with the amount of `CaO` calculated above, what will be the concentration (in ppm)of residual `Ca^(2+)` ions (Assume `CaCO_(3)` to be completely insoluble in water)? If the `Ca^(2+)` ions in one litre of the treated water are completely exchange with hydrogen ions, what will be its `pH` (One ppm means one part of the substance in one million part of water, weight`//`weight)? |
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Answer» Sample of hard water contains `96 pp m SO_(4)^(2-)` and `40 pp m Ca^(2+)` ions left in solution are `CaSO_(4),i.e., 40pp m` Now `1 "litre"` water contains `Ca^(2+)` after removal of `Ca(HCO_(3))_(2)=(40xx10^(3))/(10^(6))=40xx10^(-3)g` or `[Ca^(2+)]=(40xx10^(-3))/(40)=10^(-3)` If these `Ca^(2+)` are exchanged with `H^(+)`, then `[H^(++)]` in solution `=2xx10^(-3)` `:. pH= - log2xx10^(-3)=2.6989` |
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| 14. |
Upon mixing `50.0 mL` of `0.1 M` lead nitrate solution with `50.0 mL` of `0.05 M` chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentration of the species left behind in the final solution. Which is the limiting reagent? |
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Answer» `{:(,3Pb(NO_(3))_(2)+,Cr_(2)(SO_(4))_(3) ,rarr,3PbSO_(4)darr+,2Cr(NO_(3))_(3),,,),("Meq.",45 xx 0.25xx2,25 xx 0.1 xx 6,,,,,,),("before",=22.5,=15,,0,0,,,),("reaction",,,,,,,,),("Meq. after",7.5,0,,15,15,,,),("reaction",,,,,,,,):}` `:.` Meq. Of `PbSO_(4)` precipitated `= 15` `:.` millie-mole of `PbSO_(4)` precipitated `= (15)/(2)` `:.` Mole of `PbSO_(4)` precipitated `= (15)/(2) xx (1)/(1000) = 0.0075` `because [ ] = ("Meq.")/("total volume" xx "valency" )` Also `[Pb^(2+)] = (7.5)/(70 xx 2) = 0.0536 M` `[NO_(3)^(-)] = (7.5 +15)/(70 xx 1) = 0.32 M` `[Cr^(3+)] = (15)/(70 xx 3) = 0.0714 M` |
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| 15. |
The percentage of `P_(2)O_(5)` in diammonium hydrogen phosphate is:A. `77.58`B. `46.96`C. `53.78`D. `23.48` |
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Answer» Correct Answer - C `2(NH_(4))_(2)HPO_(4)rarrP_(2)O_(5)` `:. 2xx132 g(NH_(4))_(2)HPO_(4) "has" 142 g P_(2)O_(5)` `:. 100 g (NH_(4))_(2) HPO_(4) "has" (142xx100)/(2xx132)gP_(2)O_(5)` `= 53.78 g P_(2)O_(5)` |
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| 16. |
Calculate the percentage of `BaO` in `29.0 g` mixture of `BaO` and `CaO` which just reacts with `100.8 mL` of `6.0M HCl`. |
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Answer» Let a g of `BaO` and b g of `CaO` are present. `:. a+b=29`….(i) Meq. of `BaO+Meq`. of `CaO=Meq`.of `HCl` `:. 56a+153b=2591`….(iii) By eq. (i) and (ii), `a=19.03 g, b=9.97 g` `:. %` of `BaO=(19.03)/(29)xx100= 65.62%` |
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| 17. |
How would you prepare exactly `3.0 "litre"` of `1.0M NaOH` by mixing proportions of stock solutions of `2.50M NaOH` and `0.40M NaOH`? No water is to be used. |
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Answer» Let `V mL` of `2.50M NaOH` be mixed with `(3000-V) mL` of `0.40M NaOH` Meq. Of `2.50 M NaOH`+ Meq. Of `0.4 M NaOH`=Meq. Of `1.0 M NaOH` `2.50xxV+0.4(3000-v)=3xx1000xx1(M=N)` `:. V=857.14 mL` `:. 857.14 mL` of `2.50 M NaOH` and `2142.96 mL` of `0.4 M NaOH` are to be mixed. |
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| 18. |
The acidic substance in vinegar is acetic acid `(CH_(3)COOH)`. When `6.0 g` of a certain vinegar was titrated with `0.1MNaOH`. `40.11 mL` of base had to be added to reach the equivalence point. What per cent by mass of this sample of vinegar is acetic acid? |
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Answer» Meq. of acetic acid = Meq. of `NaOH` `(w)/(60)xx1000= 0.1xx40.11` `w= 0.241 g` `:. %` of acetic acid in vinegar `=(0.241)/(6)xx100= 4.01` |
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| 19. |
The isotopic abundance of `C-12` and `C-14 "is" 98%` and `2%` respectively. What would be the number of `C-14` isotope in `12 g` carbon sample?A. `1.032xx10^(22)`B. `3.01xx10^(23)`C. `5.88xx10^(23)`D. `6.02xx10^(23)` |
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Answer» Correct Answer - A Weight of `C-14` isotope in `12 g` sample `=(2)/(100)xx12` `:.` No. of `C-14` isotopes `=(2xx12xxN)/(100xx14)=1.032xx10^(22)` |
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| 20. |
The total ionic strength (toal molarity of all ions containing `0.1M "of"CuSO_(4)` and `0.1 M "of" Al_(2)(SO_(4))_(3)` is:A. `0.2 M`B. `0.7 M`C. `0.8 M`D. `1.2 M` |
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Answer» Correct Answer - B One mole of `CuSO_(4)` gives `1 "mole of" Cu^(2+)` and `1 "mole of" SO_(4)^(2-)`. Also `1 "mole of" Al_(2)(SO_(4))_(3)` gives `2 "mole of" Al^(3+)` and `3"mole of" SO_(4)^(2-)`. Thus total moles of all the ions present in solution having `0.1 M of CuSO_(4)` and `0.1M of Al_(2)(SO_(4))_(3)` is `0.1+0.1+0.2+0.3=0.7` |
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| 21. |
`20 mL` of `0.2 M Al_(2) (SO_(4))_(3)` is mixed with `20 mL` of `0.6 M BaCl_(2)`. Calculate the concentration of each ion in solution.A. `0.6N, 0.6 N`B. `0.2 N, 0.6 N`C. `0.6 N, 0.2 N`D. `0.2N, 0.2 N` |
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Answer» Correct Answer - A `{:(,Al_(2)(SO_(4))_(3)+,BaCl_(2)rarr,BaSO_(4)darr+,AlCl_(3)),(Meq.at,20xx0.2xx6,20xx0.6xx2,0,0),(t=0,=24,=24,,),("Meq.after reaction",0,0,24,24):}` , `therefore [Al^(3+)]=(24)/(40)=0.6N` `[Cl^(-)]=(24)/(40)=0.6N` |
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| 22. |
Which one are correct about the solution that contains `3.42 "ppm" Al_(2)(SO_(4))_(3)` and `1.42 "ppm" Na_(2)SO_(4)`?A. `[SO_(4)^(2-)]=[Na^(+)]=[Al^(3+)]`B. `[Na^(+)]+[Al^(3+)]=[SO_(4)^(2-)]`C. `[SO_(4)^(2-)]=[Na^(+)]`D. `[Al^(3+)]=[Na^(+)]` |
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Answer» Correct Answer - B::D `3.42 "ppm"Al(SO_(4))_(3)` `-=(96xx3xx3.42)/(342)"ppm"SO_(4)^(2-)=2.88"ppm"SO_(4)^(2-)` `-=(27xx2xx3.42)/(342)"ppm"Al^(3+)=0.54"ppm"Al^(3+)` `1.42"ppm"Na_(2)SO_(4)=(96xx1.42)/(142)"ppm"SO_(4)^(2-)` `=0.96"ppm"SO_(4)^(2-)=(46xx1.42)/(142)"ppm"Na^(+)=0.46"ppm"Na^(+)` `:. [Al^(3+)]=(0.54xx10^(3))/(27xx10^(6))=2.0xx10^(-5)M` `[SO_(4)^(2-)]=((2.88+0.96)xx10^(3))/(96xx10^(6))=2.0xx10^(-5)M` `[Na^(+)]=(0.46xx10^(3))/(23xx10^(6))=2.0xx10^(-5)M` |
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| 23. |
A mixture containing only `Na_(2) CO_(3)` and `K_(2) CO_(3)` and weighing `1.22 g` was dissolved in water to form `100 mL` of solution: `20 mL` of this solution required `40 mL` of `0.1 N HCl` for neutralisation. a. Calculate the weight of `K_(2) CO_(3)` in the mixture. b. If another `20 mL` of the same solution is treated with excess of `BaCl_(2)`, what will be the weight of precipitate thus obtained? (Molarcular of `Na_(2)CO_(3) = 106`, `K_(2) CO_(3) = 138, BaCO_(3) = 197.4`) |
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Answer» Weight of `Na_(2)CO_(3) = a g` Weight of `K_(2)CO_(3) = b g` `:. a+b = 1.20` …(1) For neutralization reaction of `100 mL` solution Meq. Of `Na_(2)CO_(3) + "Meq. of" K_(2)CO_(3) = "Meq.of" HCl` `(a)/(106//2) xx 1000 + (b)/(138//2) xx 1000 = (40 xx 0.1 xx 100)/(20)` `:. 69a+53b = 73.14` ...(2) By eqs. (1) and (2), `a = 0.5962 g, b = 0.6038 g` Further, solution of `Na_(2)CO_(3) + K_(2)CO_(3)` gives ppt. of `BaCO_(3)` with `BaCl_(2)`. `"Meq. of" BaCO_(3) = "Meq. of" Na_(2)CO_(3)+"Meq. of"K_(2)CO_(3)` (in `20 mL`) `= "Meq. of HCl "for" 20mL "mixture" = 40 xx 0.1 = 4` `:. (w)/(197//2) xx 1000 = 4` `:.` Wt. of `BaCO_(3) = 0.394 g` |
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| 24. |
`Al_(2)(SO_(4))_(3).XH_(2)O` has `8.1%` aluminium by mass. The value of `X` is:A. `4`B. `10`C. `16`D. `18` |
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Answer» Correct Answer - D `:. 100 g` atom has `Al=8.1` `:. 342+18x g` compound has `54 g Al` `:. 100 g` compound has `(54xx100)/(342+18 x)` `=8.1` `:. X=18` |
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| 25. |
`0.7 g "of" Na_(2)CO_(3).xH_(2)O` were dissolved in water and the volume was made to `100 mL, 20 mL` of this solution required `19.8 mL "of" N//10 HCl` for complete neutralization. The value of `x` is:A. `7`B. `3`C. `2`D. `5` |
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Answer» Correct Answer - C Meq. of `Na_(2)CO_(3).xH_(2)O "in" 20 mL=19.8xx(1)/(10)` `:.` Meq.of `Na_(2)CO_(3).xH_(2)O "in" 100 mL=19.8xx(1)/(10)xx5` `:. (w)/(E )xx1000=19.8xx(1)/(10)xx5` or `(0.7)/(M//2)xx1000=(19.8)/(2)` `:. M= 141.41` `23xx2+12+3xx16+18x=141.41` `:. x=2` |
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| 26. |
A solution contains `Na_(2)CO_(3)` and `NaHCO_(3). 10 mL` of the solution required `2.5 mL "of" 0.1M H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further `2.5 mL "of" 0.2M H_(2)SO_(4)`was required. The amount of `Na_(2)CO_(3)` and `NaHCO_(3)` in `1 "litre"` of the solution is:A. `5.3 g` and `4.2 g`B. `3.3 g` and `6.2 g`C. `4.2 g` and `5.3 g`D. `6.2 g` and `3.3 g` |
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Answer» Correct Answer - A For phenolphthalein: `(1)/(2)`Meq.of `Na_(2)CO_(3)= 2.5xx0.1xx2=0.5` For methyl orange: `(1)/(2)`Meq.of `Na_(2)CO_(3)+`Meq.of `NaHCO_(3)` `=2.5xx0.2xx2=1.0` `:.` Meq. of `NaHCO_(3)=0.5` and Meq. of `Na_(2)CO_(3)=1.0` `:. (w)/(84)xx10000=0.5` `(w)/(160//2)xx1000=1` `:. w=0.042 g "in" 10 mL` `=4.2 "in" 1 L` `:. w=0.053 g "in" 10 mL` `=5.3 "in" L` |
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| 27. |
Calculate the number of moles of water in `610 g BaCl_(2).2H_(2)O` |
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Answer» Correct Answer - 5 Molecular weight of `BaCl_(2).2H_(2)O=244 g` `because 244 g BaCl_(2).2H_(2)Oequiv36gH_(2)O=2 "mole of" H_(2)O` `therefore 610 g BaCl_(2).2H_(2)Oequiv(2xx610)/(244)` `=5 "mole of" H_(2)O` |
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| 28. |
The measured density at `NTP` of `He` is `0.1784 g L^(-1)`. Calculate the weight of `1 "mole"` of `He`. |
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Answer» Correct Answer - 4 `because 1 "mole"` of `He` occupies `22.4"litre"`volume `therefore1 "litre"` volume weighs `=0.1784 g` `therefore 22.4"litre"` volume weighs `=0.1784xx22.4=4 g mol^(-1)` |
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| 29. |
A branded tooth paste contains `0.754 g` sodium in form of sodium monofluoroortho phosphate `(Na_(3)PO_(4)F)` in `100 mL` solution. Calculate the amount of `Na_(3)PO_(4)F` present in `100 mL` of solution. |
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Answer» Correct Answer - 2 `"Mol.wt.of" Na_(3)PO_(4)F=183` `1 "mole of "Na_(3)PO_(4)Fequiv3"mole" Na=1 "mole" F` `therefore "Mole of" Na=(0.754)/(23)=0.033` `therefore "Mole of" Na_(3)PO_(4)F=(0.033)/(3)=0.011` `therefore` Amount of `Na_(3)PO_(4)F=0.011xx183=2.0g` |
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| 30. |
`Fe(SO_(4))_(3)` is empirical formula of a crystalline compound to iron. It is used in water and sewage treatment to aid in the removal of suspended impurities. Calculate the mass percentage of iron, sulphur and oxygen in this compound. |
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Answer» Mol. Wt of `Fe_(2)(SO_(4))_(3)= 2xxat.wt. of Fe+` `3xxat.wt. of S+12xxat.wt."of oxygen"` `=2xx56+3xx32+12xx16` `=112+96+192=400` `:. % of Fe=(112)/(400)xx100= 28%` `:. % of S=(96)/(400)xx100= 24%` `:. % of O =(192)/(400)xx100=48%` |
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| 31. |
A hydrate of iron (III) thiocynate `Fe(SCN)_(3)`, was found to contain `19% H_(2)O`. What is the formula of the hydrate? |
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Answer» Let the hydrate be `Fe(SCN)_(3). mH_(2)O` Molecular weight of hydrate `=56+3xx(32+12+14)+18 m= 230+18m` `:. % of H_(2)O=(18mxx10)/(230+18m)= 19` or `m=3` `:.` Formula is `Fe(SCN)_(3).3H_(2)O`. |
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| 32. |
Calulate the weight of lime `(CaO)` obtained by heating `300 kg` of `90%` pure limestone `(CaCO_(3))`. |
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Answer» `100 kg` impure sample has pure `CaCO_(3)= 90 kg` `:. 300 kg` impure sample has pure `CaCO_(3)` `=(90xx300)/(100)= 270 kg` `CaCO_(3)rarrCaO+CO_(2)` `:. 100 kg CaCO_(3)` gives `CaO= 56 kg` `:. 270 kg CaCO_(3)` gives `CaO=(56xx270)/(100)= 151.120 kg` |
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| 33. |
A solution of specific gravity `1.6 g mL^(-1)` is `67%` by weight. What will be the `%` by weight of the solution of same acid if it is diluted to specific gravity `1.2 g mL^(-1)`? |
| Answer» Correct Answer - `29.78%` ; | |
| 34. |
Igniting `MnO_(2)` in air converts it quantitatively to `Mn_(3)O_(4)`. A sample of pyrolusite is of the following composition: `MnO_(2) = 80%`, `SiO_(2)` and other inert constituents = 15%, and rest bearing `H_(2) O`. The sample is ignited to constant weight. What is the percent of `Mn` in the ingnited sample? |
| Answer» Correct Answer - `59.37%` ; | |
| 35. |
What weight of `AgCl` would be precipitated if `10 mL HCl` gas `12^(@)C` and `750 mm` pressure were passed into excess of silver nitrate? |
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Answer» Meq. Of `AgCl =` Meq. of `HCl =` milli-mole of `HCl_((g))` `= (PV)/(RT) xx 10^(3)` `(w)/(143.5) xx 1000 = (750)/(760) xx (10)/(1000) xx (10^(3))/(0.0821 xx 285) = 0.422` `:. w_(AgCl) = (0.422 xx 143.5)/(1000) = 0.0605 g` |
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| 36. |
What weight of `AgCl` will be precipitated when a solution containing `4.77 g NaCl` is added to a solution of `5.77 g` of `AgNO_(3)`.A. `4.88 g`B. `5.77 g`C. `4.77 g`D. None of these |
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Answer» Correct Answer - A `{:(,AgNO_(3)+,NaClrarr,AgCl+,NaNO_(3)),(,(5.77)/(170),(4.77)/(58.5),0,0),("Initial mole",0.034,0.082,0,0),("Mole after mixing",0,0.048,0.034,0.034):}` `therefore` Weight of `AgCl=0.034xx143.5=4.88g` |
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| 37. |
Haemoglobin contains `0.312%` iron by weight. The molecular weight of haemoglobin in `89600`. Find the number of iron atoms per molecular of haemoglobin. |
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Answer» Correct Answer - 5 `because 100 g` haemoglobin has iron `= 0.312 g` `:. 89600 g` haemoglobin has iron `= (0.312 xx 89600)/(100) = 279.55` `:. 1 "mole"` or `N` molecules of haemoglobin has `= (279.55)/(56) ~~ 5g-"atom of Fe"` |
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| 38. |
When the same amount of zinc is treated separately with excess of `H_(2)SO_(4)` and excess of `NaOH`, the ratio of volumes of `H_(2)` evolved is:A. `1:1`B. `1:2`C. `2:1`D. `9:4` |
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Answer» Correct Answer - A `Zn+H_(2)SO_(4)rarrZnSO_(4)+H_(2)`, `Zn+2NaOHrarrNaZnO_(2)+H_(2)` |
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| 39. |
A sample of `Mg` was burnt in air to give a mixure of `MgO` and `Mg_(3)N_(2)`. The ash was dissolved in `60 Meq`. of `HCl` and the resulting solution was back titrated with `NaOH`. `12 Meq`. Of `NaOH` was then added and the solution distrilled. The ammonia released was then trapped in `10 Meq`. of second acid solution. Back titration of this solution required `6 Meq`. of the base Calculate the percentage of `Mg` burnt to the nitride. |
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Answer» Let total milli-mole of `Mg` used for `MgO` and `Mg_(3)N_(2)` be `a` and `b` respectively. `{:(,2Mg+,O_(2),rarr,2MgO),("Before reaction",a,,,0),("After reaction",0,,,a),(,3Mg+,N_(2),rarr,Mg_(3)N_(2)),("Before reaction",b,,,0),("After reaction",0,,,b//3):}` Now `(a+b//30)` milli-mole of `MgO` and `Mg_(3)N_(2)` are present in the mixture. `MgO + 2HCl rarr MgCl_(2)+H_(2)O`, `Mg_(3)N_(2)+8HCl rarr 3MgCl_(2) + 2NH_(4)Cl` or the solution contains `a` milli-mole of `MgCl_(2)` from `MgO` and `b` milli-mole of `MgCl_(2)` from `Mg(3)N_(2)` and `(2b)/(3)` milli-mole of `NH_(4)Cl`. Also millie-mole of `HCl` used for this purpose `{:(=2a,,+,,(8b)/(3)),("for" MgO,,,,"for"Mg_(3)N_(2)):}` Now milli-mole of `HCl` or Meq. of `HCl` (monobasic acid) `= 60-12 = 48` `2a + (8b)/(3) = 48` Further, millie-mole of `NH_(4)Cl` formed `=` milli-mole of `NH_(3)` liberated `=` milli-mole of `HCl` used for absorbing `NH_(3)` `:. (2b)/(3) = 4` or `b = 6` ...(2) From eqn. (1), `2a + (8 xx 6)/(3) = 48` or `a = 16` Thus, `%` of `Mg` used for `Mg_(3)N_(2) = (6)/((6+16)) xx 100 = 27.27%` |
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| 40. |
When dissolved in dilute `H_(2)SO_(4), 0.275 g` of metal evolved `119.7 mL` of `H_(2)` at `20^(@)C` and `780.4 mm` pressure. `H_(2)` was collected over water. Aqueous tension is `17.4` mm at `20^(@)C`. Calculate equivalent weight of metal. |
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Answer» Mole of `H_(2(n))=(PV)/(RT)` `=((780.4-17.4)xx119.7)/(760xx1000xx0.0821xx293)=5xx10^(-3)` Let eq.wt. of metal be `E`. Eq. of metal =Eq. of `H_(2)` `0.275/E`= mole of `H_(2)xx2=2xx5xx10^(-3)` `E=27.52` |
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| 41. |
A sample of `Mg` metal containing some `MgO` as impurity was dissolved in `125 mL` of `0.1N H_(2)SO_(4)`. The volume of `H_(2)` evolved at `27.5^(circ)C` and `1 atm` was `120.0 mL`. The resulting solution was found to be `0.02N` with respect to `H_(2)SO_(4)`. Calculate the weight of sample dissolved and the `%` by weight of pure `Mg` metal in sample. Neglect any change in volume. |
| Answer» Correct Answer - `122.16 g, Mg=95.57%`; | |
| 42. |
`25 mL` of `0.2M` phosphorus acid `(H_(3)PO_(3))` neutralises exact `80 mL` of a solution containing `10 g NaOH ( 50% "pure") per dm^(3)`. Report basicity of acid and write balanced chemical equation for neutralisation. |
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Answer» Meq. of acid`=Meq. of NaOH` `25xx0.2xxn=(10)/(40)xx(50)/(100)xx80` `:. N=2` `:.` Basicity of acid`=2` Neutralisaton reaction: `2NaOH+H_(3)PO_(3)rarrNa_(2)HPO_(3)+2H_(2)O` |
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| 43. |
A `100.0 mL` solution containing `HCl` and `HBr` was titrated with `0.1235 M NaOH`. The volume of base required to neutralise the acid was `47.14 mL`. Aqueous `AgNO_(3)` was then added to precipitate `Cl^(-)` and `Br^(-)` ions as `AgCl` and `AgBr`. The mass of silver halides obrained was `0.9974 g`. What were the molarities of `HCl` and `HBr` in solution? |
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Answer» Let `a M HCl` and `bMHBr` be molarities. Meq. of `HCl+Meq. of HBr=Meq. of NaOH` `axx100+bxx100= 0.1235xx47.14=5.82` `a+b= 5.82xx10^(-2)`…..(i) Also Meq. of `HCl+` Meq. Of `HBr=`Meq. of halide `AgCl+Meq`. of halide `AgBr` `=axx100+bxx100` `:. "Total weight"=(axx100xx143.5)/(1000)+(bxx100xx188)/(1000)` `=0.9974` or `143.5a+188b= 9.9974`....(ii) By eqn. (i) and (ii), `a=0.036M, b=0.022M` |
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| 44. |
A piece of `Al` wieghing `2.7 g` is titrated with `75.0 mL` of `H_(2) SO_(4)` (specific gravity `1.8 mL^(-1)` and 24.7% `H_(2) SO_(4)` by weight). After the metal is completely dissolved, the solution is diluted to `400 mL`. Calculate the molarity of free `H_(2) SO_(4)` solution. |
| Answer» Correct Answer - `0.183`; | |
| 45. |
`25 mL` of `0.107 M H_(3)PO_(4)` was titrated with `0.115 M` solution of `NaOH` to the end point identified by indicator bromocresol green.This required `23.1 mL`. The titration was repeated using phenolphthalein as indicator. This time `25 mL` of `0.107 M H_(3)PO_(4)` reuired `46.2 mL` of the `0.115 M NaOH`. What is the coefficient of `n` in this equation for each reaction? `H_(3)PO_(4)+n OH^(-)rarr nH_(2)O+[H_(3-n)PO_(4)]^(n-)` |
| Answer» Correct Answer - `n=2;` | |
| 46. |
Statement `1` equivalent of `K_(2)Cr_(2)O_(7)` has `1` equivalent of `K, Cr` and `O` each. Explanation Equivalent and milli-equivalent reacts in equal number to give same eq.or meq. of product.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is wrong.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S. |
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Answer» Correct Answer - C Explanation is correct reason for statements. |
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| 47. |
A sample of peanut oil weighing `1.5763 g` is added to `25 mL "of" 0.4210 M KOH`. After saponification is complete `8.46 mL "of" 0.2732 M H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:A. `209.6`B. `108.9`C. `98.9`D. `218.9` |
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Answer» Correct Answer - A Meq.of `KOH` added `=25xx0.4210=10.525` Meq. of `KOH` left `=8.46xx0.2732xx2=4.623` `:.` Meq. of `KOH` used by oil`=10.525-4.623=5.902` or `(w)/(56)xx1000=5.902` or `w_(KOH)=0.3305g` `:.` Saponification no. `=wt.of KOH` used in `mg per g` of oil `=(0.3305)/(1.5763)xx1000=209.6` |
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| 48. |
`H_(2)C_(2)O_(4)` acts as an acid as well as an oxidising agent. The correct statements `(s)` about `H_(2)C_(2)O_(4)` is (are):A. It forms two series of saltsB. Equivalent weight of `H_(2)C_(2)O_(4)` as an acid for completer neutralisation and as oxidant are sameC. `100 mL` of `0.1M` solution of `KMnO_(4)` (acid) will be completely reduced by `50 mL` of `1M H_(2)C_(2)O_(4)`D. `100 mL` of `0.1N` solution of `Ca(OH)_(2)` will be completely neutralised by `50 mL` of `0.2M H_(2)C_(2)O_(4)` |
| Answer» Correct Answer - A::B::D | |
| 49. |
The reaction `H_(3)PO_(4)+Ca(OH)_(2)rarrCa(HPO_(4))_(2)+2H_(2)O` Which statements `(s)` is (are)true?A. Equivalent weight of `H_(3)PO_(4)` is `49`B. For complete neutralization `3//2`mole of `Ca(OH)_(2)` are neededC. Resulting mixture is neutralised by 1 mole of `KOH`D. Equivalent weight of `H_(3)PO_(4) is 98` |
| Answer» Correct Answer - A::B::C | |
| 50. |
`1.0 g` of pure calcium carbonate was found to require `50 mL` of dilute `HCl` for complete reactions. The strength of the `HCl` solution is given by:A. `4N`B. `2N`C. `0.4N`D. `0.2N` |
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Answer» Correct Answer - B Meq. of `HCl="Meq of" CaCO_(3)` `:. Nxx50=(1)/(50)xx1000 or N=0.4` |
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