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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. |
`25.0` litre of natural gas measured at `25^(@)C` and `740 mm` of `Hg` is bubbled through `Pb_((aq))^(2+)` to give `0.535 g` of solid residue. If natural gas contains `H_(2)S`, the only component responsible for the formation of solid residue, calculate the volume `%` of `H_(2)S`, in natural gas. |
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Answer» `underset("Solid residue")(Pb^(2+)+H_(2)srarr PbS+2H^(+))` Eq.of `H_(2)S=` Eq. of `PbS` `(w)/(32//2)=(0.535)/(240//2)` `:. w_(H_(2)S)=7.58xx10^(-2)` `:.` Moles of `S=(7.85xx10^(-2))/(34)=2.23x10^(-3)` `:. V_(H_(2)s)=(nRT)/(P)` `=(2.23xx10^(-3)xx0.0821xx298xx760)/(740)` `5.6xx10^(-2)`litre `:. % Vol=(5.6xx10^(2))/(25)xx100= 0.224` |
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| 2. |
Choose the correct statement `(S)`:A. The no. of atoms present in a molecule of gas represented its atomicityB. One mole of electron weigh `0.55 mg`C. The extent of both inter and intramolecular `H`-bonding depends on the temperatureD. None of these |
| Answer» Correct Answer - A::B::C | |
| 3. |
The formula weight of an acid is `82.0.100 cm^(3)` of a solution of this acid containing `39.0 g` of the acid per litre were completely neutralised by `95.0 cm^(3)` of aqueous `NaOH` containing `40.0 g` of `NaOH` per litre. What is the basicity of the acid? |
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Answer» Normality of acis `= (39)/(82//n xx1)` (`n` is basicity of acid) Normality of `NaOH = (40)/(40) xx (1000)/(1000) = 1` Now, Meq. of acid `=` Meq. of `NaOH` `(39n)/(82) xx 100 = 1 xx 95` `:. n = 2`, i.e., acid is diabasic. |
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| 4. |
A mixture containing one mole of `BaF_(2)` and two mole of `H_(2)SO_(4)` will be neutralised by:A. 1 mole `KOH`B. 4 mole `KOH`C. 2 mole `KOH`D. 2 mole `Ca(OH)_(2)` |
| Answer» Correct Answer - C | |
| 5. |
A sample of `H_(2)SO_(4)` `("density" 1.787 g mL^(-1))` is labelleed as `80%` by weight. What is molarity of acid? What volume of acid has to be used to make `1"litre of" 0.2MH_(2)SO_(4)`? |
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Answer» `H_(2)SO_(4)` is `80%` by weight `:. Wt. of H_(2)SO_(4)= 80 g` W.t. of solution `= 100 g` `:.` Volume of solution `=(100)/(1.787)mL` `=(100)/(1.787xx1000)"litre"` `:. M_(H_(2)SO_(4))={(80)/(98xx(100)/(1.787xx1000))}=14.59` Let `V mL` of this `H_(2)SO_(4)` are used to prepare `1 "litre"` of `0.2MH_(2)SO_(4)`, then `mM` of conc. `H_(2)SO_(4)=mM` of dil. `H_(2)SO_(4)` (`mM` does not change on dilution) `Vxx14.59= 1000xx0.2` `V= 13.71 mL` |
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| 6. |
The normality of `0.3M` phosphorous acid `H_(3)PO_(3)` is:A. `0.1`B. `0.9`C. `0.3`D. `0.6` |
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Answer» Correct Answer - D `H_(3)PO_(3)(HO-overset(H)overset(|)underset(O)underset(||)P-OH)` is dibasic acid. |
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| 7. |
For the reaction, `N_(2)O_(5(g))hArr2NO_(2(g))+0.50_(2(g))`, Calculate the mole fraction of `N_(2)O_(5(g))` decomposed at a constant volume and temperature, if the initial pressure is `600 mm Hg` and the pressure at any time is `960 mm Hg`. Assume ideal gas behaviour. |
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Answer» `{:(,N_(2)O_(5),hArr,2NO(g)+,(1)/(2)O(g)),("Initial pressure",600,,0,0),("Final pressure",600-P,,2P,P//2):}` `P prop` moles if `V,T` are constant (where mole equivalent to pressure `P` are decomposed) Thus, `600 - P + 2P + P//2 = 960` or `P = 240mm Hg` Thus, mole fraction of `N_(2)O_(5)` decomposed `= (240)/(600) = 0.4` |
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| 8. |
The density of `1 M` solution of `NaCl` is `1.0585 g mL^(-1)`. The molality of the solution isA. `1.0585`B. `1.0`C. `0.10`D. `0.0585` |
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Answer» Correct Answer - B `m=("Mole of" NaCl)/("weight of solvent in" kg)=(1)/(1)=1` `"wt. of solvent"="wt.of solution"-"wt. of" NaCl` `=1.0585xx1000-58.5` `=1058.5-58.5=100 g=1kg` |
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| 9. |
The percentage of sodium in a breakfast careal labelled as `110 mg` of sodium per `100 g` of cereal is:A. `11%`B. `0.110%`C. `0.110%`D. `110%` |
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Answer» Correct Answer - C `100 g` cereal has `Na=110xx10^(-3)=0.11 g` |
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| 10. |
The equivalent weight of iron in `Fe_(2)O` would be:A. `18.6`B. `26.66`C. `56`D. `112` |
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Answer» Correct Answer - A `E_(Fe_(2)O_(3))=M//6` and `E_(Fe)=(at.wt)/(3)=(56)/(3)` |
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| 11. |
Which quantity is (are) independent of temperature?A. Mole faractionB. MolalityC. MolarityD. `%` by weight |
| Answer» Correct Answer - A::B::D | |
| 12. |
What is the purity of conc. `H_(2)SO_(4)("density" 1.8g//mL)` if `5 mL` of it is neutralized completely with `84.6mL` of `2.0N NaOH`? |
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Answer» Meq. of acid in `5 mL=` Meq. of `NaOH` used `(w)/(49)xx100= 84.6xx2` `:.` Purity of `H_(2)SO_(4)=(8.29)/(9)xx100` `92.1% H_(2)SO_(4)` |
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| 13. |
Weight of oxygen in `Fe_(2)O_(3)` and `FeO` in the simple ratio for the same amount of iron is:A. `1:2`B. `2:1`C. `3:2`D. `1:3 |
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Answer» Correct Answer - C `112:48::56:16` `:.` Simple ratio of `O` for same amount of `Fe "is" 48:32i.e.,3:2` |
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| 14. |
In which mode of expression, the concentration of a solution remains independent of temperature?A. MolarityB. MolalityC. FormalityD. Normality |
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Answer» Correct Answer - B Weights are independent of temperature. |
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| 15. |
The haemoglobin from the red blood corpuscles of most mammals contains approximately `0.33%` of iron by weight. The molecular weight of haemoglobin as `67,200`. The number of iron atoms in each molecule of haemoglobin is (atomic weight of iron `=56`):A. `2`B. `3`C. `4`D. `5` |
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Answer» Correct Answer - C `100 g "sample"-=0.33 "iron"` `67200 g-=221.8 g "iron"` `:.` Number of iron atoms per molecule of haemoglobin `=(221.8)/(56)=4` |
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| 16. |
Calculate the weight of `NaOH` in `75` mill-equivalents. |
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Answer» Correct Answer - 3 `because Meq. =(w)/(E )xx1000` `75=(w)/(40)xx1000rArr therefore w=3g` |
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| 17. |
What is the strength in `g//L` of a solution of `H_(2)SO_(4), 14 mL` of which neutralized `20 mL` of `N//10 NaOH` solution? |
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Answer» Correct Answer - 7 Meq.of `H_(2)SO_(4)= "Meq.of" NaOH` `Nxx14=20xx(1)/(10) or N=(20)/(10xx14)` `because` Strength `=NxxE=(20)/(10xx14)xx49` `=7.0 g "per litre"` |
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| 18. |
Find the molality of `H_(2)SO_(4)` solution whose specific gravity is `1.98 mL^(-1)` and `93%` by volume `H_(2)SO_(4)` |
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Answer» Correct Answer - 9 `H_(2)SO_(4)` is `93%` by volume `:.` weight of `H_(2)SO_(4) = 93g` and volume of solution `= 100mL` `:.` wt.of solution `= 1.98 xx 100 = 198 g` `:.` wt. of water = wt. of solution -wt.of `H_(2)SO_(4)` `193 - 93 = 105g` Now, molarity `= ("Moles of" H_(2)SO_(4))/("Wt.of water"("in" kg))` |
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| 19. |
Calculte the degree of hardness of a sample of water containing `6 mg` of `MgSO_(4)` per kg of water. |
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Answer» Correct Answer - 5 `because 1 g-"mole"` or `120 g MgSO_(4) "has" =1 g-"mole"` or `100g CaCO_(3)` `therefore 6xx10^(-3)g MgSO_(4) "has" =(100xx6xx10^(-3))/(120)` `=5xx10^(-3)gCaCO_(3)` Thus `1000 g` of water contains `MgSO_(4)` equivalent to `5xx10^(-3)gCaCO_(3)` `therefore 10^(6)g` water contains `=(5xx10^(-3))/(1000)xx10^(6)` `5 g "of" CaCO_(3)` |
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| 20. |
Calculate the normality of `0.74 g Ca(OH)_(2)` in `10 mL` of solution. |
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Answer» Correct Answer - 2 `because "Eq.of"Ca(OH)_(2)=74//2` and volume of solution `=10xx10^(-3)L` `therefore "Normality"=(w)/(ExxV_(("in litre")))` `=(0.74xx2)/(74xx10xx10^(-3))=2N` |
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| 21. |
Which sample contains the largest number of atoms?A. `1 mg of C_(4)H_(10)`B. `1 mg of N_(2)`C. `1 mg of Na`D. `1 mL` of water |
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Answer» Correct Answer - D `1 mg C_(4)H_(10)=(14N)/(58)xx10^(-3)"atoms"` `1 mgN_(2)=(2Nxx10^(-3))/(28)"atoms"` `1 mg Na=(Nxx10^(-3))/(23)"atoms"` `1 mL=1 g H_(2)O=(3N)/(18)"atoms"` ( `:. Mg "of a substance"=N "molecules"= axxN "atoms"`, where `a` is number of atoms in one molecule) |
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| 22. |
One litre of `N//2 HCl` solution was heated in a beaker. When the volume was reduced to `600 mL`, `9.125 g` of `HCl` was lost out. The new normality of solution is a. `~~ 0.4` b. `~~ 0.8` c. `~~ 0.4` d. `~~ 0.2`A. `6.85`B. `0.685`C. `0.1043`D. `6.50` |
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Answer» Correct Answer - B Milli-mole of `HCl=1,000xx(1)/(2)=500` Milli-mole of `HCl` given out `=(3.25)/(36.5)xx10^(23)= 89.04` Milli-mole of `HCl` left `= 500-89.04=410.96` New molarity `=(410.96)/(600)=0.685 M` |
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| 23. |
Potash alum and chrome alum are examples of:A. AllotropyB. IsomerismC. IsomorphismD. Tautomerism |
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Answer» Correct Answer - C All Alums are isomorphs to each other. |
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| 24. |
The term mole first used by Ostwald in `1896` refers for the ratio of mass of a substance in `g` and its molecular weight. `1 "mole"` of a gaseous compound occupies `22.4 "litre"` at `NTP` and contains `6.023xx10^(23)` molecules of gas. The amount of sulphur required to produce `100 "mole"` of `H_(2)SO_(4)` is:A. `3.2xx10^(3)g`B. `32.65 g`C. `32 g`D. `3.2 g` |
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Answer» Correct Answer - A `1 "mole" H_(2)SO_(4)equiv 1 "mole" S equiv 32 g S` |
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| 25. |
The term mole first used by Ostwald in `1896` refers for the ratio of mass of a substance in `g` and its molecular weight. `1 "mole"` of a gaseous compound occupies `22.4 "litre"` at `NTP` and contains `6.023xx10^(23)` molecules of gas. weight of `1 "atom"` of hydrogen is:A. `1.66xx10^(-24) "amu"`B. `3.32xx10^(-24)g`C. `1.66xx10^(-24)g`D. `3.32xx10^(-24) "amu"` |
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Answer» Correct Answer - C `N "atoms of" H=1g` |
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| 26. |
The term mole first used by Ostwald in `1896` refers for the ratio of mass of a substance in `g` and its molecular weight. `1 "mole"` of a gaseous compound occupies `22.4 "litre"` at `NTP` and contains `6.023xx10^(23)` molecules of gas. The volume of air needed to burning `12 g` carbon completely at `STP` is:A. `22.4 "litre"`B. `112 "litre"`C. `44.8 "litre"`D. `50 "litre"` |
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Answer» Correct Answer - B `C+O_(2)rarrCO_(2)` `12 g` carbon needs `22.4 "litre air" 22.4 "litre" O_(2)` or `(22.4xx100)/(20)"litre air"=112 "litre air"` |
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| 27. |
What is the molecular weight of a substance, each molecule of which contains `9` carbon atoms, `13` hydrogen atoms and `2.33 xx 10^(-23) g` of other component ? |
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Answer» The molecule has `C, H` and other component. w.t of `9 C` atoms `=12xx9 =108` amu w.t of `13 H` atoms `=13xx1= 13` amu w.t of other component `=(2.33xx10^(-23))/(1.66xx10^(-24))= 14.04` amu `:.` Total weight of one molecule `= 108+13+14.04` `=135.04` amu `:.` Mol.wt. of substance `= 135.04` |
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| 28. |
Which of the following will weigh maximum amount? (a). `20 g` iron, (b) `1.2 g` atom of `N`, (c ) `1xx10^(23)` atoms of carbon, (d) `1.12 litre` of `O_(2) at STP`. |
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Answer» (a) Mass of iron `= 20 g` (b) Mass of `1.2 g`-atom of `N = 14xx1.2= 16.8 g` (c ) Mass of `1xx10^(23)` atoms of `C=(12xx1xx10^(23))/(6.023xx10^(23))` `1.99 g` (d) Mass of `1.12` litre of `O_(2) at STP=(32xx1.2)/(22.4)= 1.6 g` Thus `20 g` iron has maximum weight. |
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| 29. |
`11.2 "litre"` of a gas `STP` weighs `14 g`. The gases would be:A. `N_(2)`B. `CO`C. `N_(2)O`D. `B_(2)H_(6)` |
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Answer» Correct Answer - A::B::D Molecular weight of gas `=(14xx22.4)/(11.2)=28` i.e., `CO,N_(2),B_(2)H_(6)` |
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| 30. |
Total number of electrons present in `11.2 litre` of `NH_(3)` at `STP` are:A. `6.02 xx 10^(23)`B. `3.01 xx 10^(23)`C. `3.01 xx 10^(24)`D. `5.1 xx 10^(24)` |
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Answer» Correct Answer - B `22.4 "litre" NH_(3)=1 "mole" NH_(3)=10N"electron"` |
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| 31. |
At `STP 5.6 "litre"` of a gas weigh `60 g`. The vapour density of gas is:A. `60`B. `120`C. `30`D. `240` |
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Answer» Correct Answer - B `5.6 "litre"=60 g` `:. 22.4 "litre"=240 g` ormol.wt.of gas`= 240` `:.` vapour density`=120` |
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| 32. |
`4.4 g of CO_(2)` and `2.24 "litre" of H_(2)` at `STP` are mixed in a container. The total number of molecules present in the container will be:A. `6.022xx10^(23)`B. `1.2044xx10^(23)`C. `2 "mole"`D. `6.023xx10^(24)` |
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Answer» Correct Answer - B `44 gCO_(2)=N "molecules"` `:. 4.4 g CO_(2) = (N)/(10)` molecules `22.4` litre `H_(2)` at STP `=N` molecules `:. 2.24` litre `H_(2)` at STP `=(N)/(10)` molecules Thus total molecules `= (N)/(10) + (N)/(10) = (N)/(5)` |
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| 33. |
`1 g` of calcium was burnt in excess of `O_(2)` and the oxide was dissolved in water to make up `1 L` solution. Calculate the normality of alkaline soluiton.A. `0.05, 0.025`B. `0.1, 0.05`C. `0.1, 0.2`D. `0.01, 0.02` |
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Answer» Correct Answer - A `"Eq.of"Ca=Eq.of CaO=(1)/(20)` `N_(CaO)=(1)/(20xx1)=0.05` `M_(CaO)=(0.05)/(2)=0.025` |
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| 34. |
Number of ions present in `2.0 "litre"` of a solution of `0.8 M K_(4)Fe(CN)_(6)` is:A. `4.8xx10^(22)`B. `4.8xx10^(24)`C. `9.6xx10^(24)`D. `9.6xx10^(22)` |
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Answer» Correct Answer - B Mole of `K_(4)Fe(CN)_(6)=2xx0.8=1.6` Also `1 "mole of" K_(4)Fe(CN)_(6)` gives `4K^(+)` and `1 Fe(CN)_(6)^(4-) "ion"`. Thus total ions in `1.6 "mole"` `=1.6xx5xxN=48.184xx10^(23)` |
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| 35. |
Mole fraction of `I_(2)` in `C_(6) H_(6)` is 0.2. Calculate molality of `I_(2)` in `C_(6) H_(6)`. `(Mw of C_(6) H_(6) = 78 g mol^(-1))`A. `3.2`B. `6.40`C. `1.6`D. `2.30` |
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Answer» Correct Answer - A `(n)/(n+N) = 0.2 :. (N)/(n+N) = 0.8` or `(n)/(N) = (1)/(4)` or `(n_(I_(2)))/(w_(C_(6)H_(6))) xx M_(C_(6)H_(6)) = (1)/(4)` `:.` Molality `=(n_(I_(2)))/(w_(C_(6)H_(6))) xx 1000` `= (1)/(4) xx (1000)/(M_(C_(6)H_(6)))` or Molality `= (1)/(4) xx (1000)/(78)` `= 3.2` |
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| 36. |
A `40 mL` mixture of methane and ethylene when exploded with certain volume of oxygen which is just sufficient for combustion produced `60 mL` of `CO_(2)` gas. Calculate the ratio between the volumes of `CH_(4)` and `C_(2)H_(4)` in the mixture. Wgat volume of oxygen is required if the ratio between the volumes of `C_(2)H_(4)` and `CH_(4)` is first reversed and then doubled? What volume of `CO_(2)` is produced? Assume, all the volumes being measured under identical conditions. |
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Answer» Let the volumes of `CH_(4)` and `C_(2)H_(4)` in mixture be `a, b mL` respectively and `O_(2)` is just Sufficient to burn these gases. `{:(CH_(4)+,2O_(2),rarr,CO_(2),2H_(2)O),(a,2a,,0,0),(0,0,,a,),(C_(2)H_(4)+,3O_(2),rarr,2CO_(2)+,2H_(2)O),(b,3b,,0,0),(0,0,,2a,):}` Given that `a+b=40` and `3a=60` `:. a=20` and `b=20` `:. ("Volume of"C_(2)H_(4))/("Volume of"CH_(4))=(2)/(1)` then volume of `C_(2)H_(4)=(40xx2)/(3)= 26.67 mL` and volume of `CH_(4)= 40-26.67 mL` and volume of `CH_(4)= 40-26.67=13.33 mL` Thus volume of `CO_(2)` formed in this case `= 13.33+2xx26.67= 66.64mL` and volume of `O_(2)` required in this case `= 2xx13.33+3xx26.67= 106.67mL` |
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| 37. |
A gaseous alkane was exploded with oxygen. The volume of `O_(2)` for complete combustion to `CO_(2)` formed was in the ratio of `7:4`. The molecular formula of alkane is: |
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Answer» Correct Answer - 2 `C_(n)H_(2n+2)+[n+(n+1)/(2)]O_(2)rarrnCO_(2)+(n+1)H_(2)O` Given, `("vol.of"O_(2)"used")/("vol.of" CO_(2)"formed")=(7)/(4)` `therefore (n+(n+1)//2)/(n)=(7)/(4) or n=2` |
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| 38. |
A gaseous alkane was exploded with oxygen. The volume of `O_(2)` for complete combustion to `CO_(2)` formed was in the ratio of `7:4`. The molecular formula of alkane is:A. `CH_(4)`B. `C_(2)H_(6)`C. `C_(3)H_(6)`D. `C_(4)H_(10)` |
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Answer» Correct Answer - B `C_(n)H_(2n+2)+[n+(n+1)/(2)]O_(2)rarrnCO_(2)+(n+1)H_(2)O` `("Volume of" O_(2) "used")/("Volume of" CO_(2) "formed")=(7)/(4)` `(n+(n+2)//2)/(n)=(7)/(4)` `:. n=2` and alkane is `C_(2)H_(6)`. |
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| 39. |
What is the weight of `3.01xx10^(23)` molecules of ammonia? |
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Answer» `6.023xx10^(23)` molecules of `NH_(3)` has weight `= 17 g` `:. 3.01xx10^(23)` molecules of `NH_(3)` has weight `=(17xx3.01xx10^(23))/(6.023xx10^(23))= 8.50 g` |
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| 40. |
The total number of `AlF_(3)` molecule in a sample of `AlF_(3)` containing `3.01xx10^(23)` ions of `F^(-)` is:A. `9.0xx10^(24)`B. `3.0xx10^(24)`C. `7.5xx10^(23)`D. `10^(23)` |
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Answer» Correct Answer - D `3 "ion" F^(-)=1 "molecule" Al` `:. 3xx10^(23)"ion" F^(-)=10^(23)"molecule" AlF_(3)` |
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| 41. |
`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`? |
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Answer» Correct Answer - 2 Redox changes are: For `KMnO_(4):Mn^(7+)+5erarrMn^(2+)` For `A^(n+): A^(n+)rarrA^(5+)+(5-n)e` Now meq.of `A^(n+)="meq.of" KMnO_(4)` `( becausemeq.="mole"xx"valencyfactor"xx1000)` `2.68xx10^(-3)xx(5-n)xx1000=1.61xx10^(-3)xx5xx1000` `therefore n=1.99approx 2` |
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| 42. |
`11.2 g` carbon reacts completely with `19.63` litre of `O_(2)` at `NTP`. The cooled gases are passed through `2"litre"` of `2.5N NaOH` and `Na_(2)CO_(3)` in solution. `CO` does not react with `NaOH` under these conditions. |
| Answer» Correct Answer - `NaOH=1.68N, Na_(2)CO_(3)=0.82N`; | |
| 43. |
One litre of `CO_(2)` is passed over hot coke. The volume becomes `1.4 L`. Find the composition of products, assuming measurement at `NTP`.A. `0.6 "litre" CO`B. `0.8 "litre" CO_(2)`C. `0.6"litre" CO_(2)`D. none of these |
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Answer» Correct Answer - C `{:(CO_(2)+,Crarr,2CO,,,),(1,0,,,),((1-x),,2x,,):}` `:. 1-x+2x=1.4, :. x=0.4` |
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| 44. |
A sample of gaseous hydrocarbon occupying `1.12 "litre"` at `NTP`, when completely burnt in air produced `2.2 g CO_(2)` and `1.8 g H_(2)O`. Calculate the weight of hydrocarbon taken and the volume of `O_(2)` at `NTP` required for its combustion. |
| Answer» Correct Answer - `0.8 g, 44.8 "litre" O_(2)` ; | |
| 45. |
`0.63 g` of diabasic acid was dissolved in water. The volume of the solution was made `100 mL`. `20 mL` of this acid solution required `10 mL` of `N//5 NaOH` solution. The molecular mass of acid is:A. `63`B. `126`C. `252`D. `128` |
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Answer» Correct Answer - B Meq. of acid in `20 mL=`Meq.of`NaOH` `10xx(1)/(5)=2` `:.` Meq. of acid in `100mL=(2xx1000)/(20)=10` or `(w)/(M//2)xx1000=10` `(0.63)/(M//2)xx1000=10` or `M=126` |
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| 46. |
The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` is:A. `90`B. `80.3`C. `40.13`D. `9` |
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Answer» Correct Answer - B `m=(axx1000)/(98xx910)=9` `:. a= 802.62 g//"litre"` `:. %` by weight `=80.26 g=X` |
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| 47. |
A mixture of `Xe` and `F_(2)` was heated and the white solid so formed reacted with `H_(2)` to give `81 mL of Xe` at `STP` and `HF`. The `HF` formed required `68.43 mL` of `0.3172 M NaOH` for complete neutralisation. Determine empiriacal formula of white solid. |
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Answer» `2Xe+nF_(2)rarr2XeF_(n)` `2XeF_(n)+nH_(2)rarr2Xe+2nHF` Mole of `Xe` formed `=(81)/(22400)=3.6xx10^(-3)` Meq. Of `HIF` formed `=Meq."of" NaOH` `=68.43xx0.3172` `:. m` of `HF` formed `=68.43xx0.3172` Mole of `HF` formed`=(68.43xx0.3172)/(1000)` `=21.7xx10^(-3)` `("Mole of"Xe)/("Mole of"HF)=(2)/(2n)` `:. (3.6xx10^(-3))/(21.7xx10^(-3))=(2)/(2n)` `:. n=6` Thus, compound is `XeF_(6)`. |
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| 48. |
`RH_(2)` ( ion exchange resin) can replace `Ca^(2+)`d in hard water as. `RH_(2)+Ca^(2+)rarrRCa+2H^(+)` `1 "litre"` of hard water passing through `RH_(2)` has `pH2`. Hence hardness in `pp m "of" Ca^(2+)` is:A. `200`B. `100`C. `50`D. `125` |
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Answer» Correct Answer - A `[H^(+)]=10^(-2)` `:. [Ca^(2+)]=(10^(-2))/(2)=(40xx10^(-2))/(2)=0.2 g//"litre"` `:. "Hardness"=(0.2xx10^(6))/(10^(3))=200pp m` |
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| 49. |
A mixture contains `NaCl` and unknown chloride `MCl`. (a) `1 g` of this is dissolved in water, excess of acidified `AgNO_(3)` solution is added to it, so that `2.567 g` of white `ppt`. Is obtained. (b) `1 g` of original mixture is heated to `300^(circ)C`. Some vapours come out which are absorbed in `AgNO_(3)` ( acidified) solution. `1.341 g` of white precipitate is obtained. Find the mol.wt. of unkonwn chloride. |
| Answer» Correct Answer - `53.50`; | |
| 50. |
Calculate the mass of chloride ion in `1` litre of: (a) `10%` by weight of `NaCl` solution having density `1.07 g// mL`, and (b) `10%` by weight of `AlCl_(3)` solution having density `1.10 g//mL`. |
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Answer» (a) `100 g NaCl` solution `-= 10 g NaCl` `(100)/(1.07)mL NaCl` solution `-=10 g NaCl` `:. 1000mLNaCl` solution `=(10xx1000xx1.07)/(100)g NaCl` `=(10xx1000xx1.07)/(100)xx(35.5)/(58.5)= 64.93 Cl^(-1) "ions"` (b) Similarly of `BaCl_(2)` wt. of `Cl^(-)` ion `=(10xx1000xx1.10)/(100)xx(2xx35.5)/(208)= 37.55 g` |
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