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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction `4 Ag + 8KCN +O_(2)+2H_(2)O rarr 4K[Ag(CN)_(2)]+4KOH`A. The amount of KCN required to dissolve 100g of pure Ag is 120gB. The amount of oxygen used in this process is `0.742g`C. The mount of oxygen used in this process is `7.40g`D. The volume of oxygen used at STP is `5.20` lit. |
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Answer» Correct Answer - A::C::D `4Ag +8KCN +O_(2)+2H_(2)Orarr 4K [Ag(CN)_(2)] +4KOH` `rArr 4 xx 108g` of Ag reacts with `8 xx 65g` of KCN 100g of Ag reacts with `(8 xx 65)/(4 xx 108) xx 100 = 120` Hence to dissolve 100g of Ag, the amount of `KCN` required `=120g` Hence, statement (A) is correct `rArr 4 xx 108g` of Ag require `32g` of `O_(2)` `rArr 100g` of Ag require `=(32 xx 100)/(4 xx 108) = 7.14g` Hence, choice (C) is correct Hence, volume of `O_(2)` required `= (7.4)/(32) xx 22.4 = 5.20 =` litre Hence (A),(C),(D) are correct while (B) is incorrect. |
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| 2. |
1 g of a sample of NaOH was dissolved in 50 " mL of " 0.33 M alkaline solution of `KMnO_(4)` and refluxed till all the cyanide was converted into `OCN^(ɵ)`. The reaction mixture was cooled and its 5 mL portion was acidified by adding `H_(2)SO_(4)` in excess and then titrated to end point against 19.0 " mL of " 0.1 M `FeSO_(4)` solution. The percentage purity of NaCN sample isA. `55.95%`B. `65.95%`C. `75.95%`D. `85.95%` |
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Answer» Correct Answer - C `MnO_(4)^(-) +3e rarr MnO_(2)` (n-factor =3) `(0.33 xx 3 xx 50) - (0.1 xx 1xx 19) (50)/(5) = 31` `CN^(-) +H_(2)O rarr OCN^(-) +2H^(+) +2e^(-)` `:.` m. moles of `NaCN (31)/(2)` Weigth of `NaCN = (31)/(2) xx (49)/(1000) = 0.7595g` % of `NaCN = (0.7595)/(1) xx 100 = 75.95%` |
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| 3. |
A sample containing `0.496 gm` of `(NH_(4))_(2) C_(2)O_(4) (MW = 124)` and inert material was dissolved in water and made strongly alkaline with KOH which converts `NH_(4)^(+)` into `NH_(3)`. The liberated `NH_(3)` was distilled into exactly 50ml of `0.05M H_(2)SO_(4)`. The excess `H_(2)SO_(4)` was back titrated with 10ml of `0.1MNaOH`. The percentage of `(NH_(4))_(2) C_(2)O_(4)` with sample isA. `40%`B. `50%`C. `60%`D. `75%` |
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Answer» Correct Answer - B m. eqts of excess `H_(2)SO_(4) = 10 xx 0.1 = 1` m.eqts of `H_(2)SO_(4)` taken `= 50 xx 0.05 xx 2 = 5` m.eqts of `H_(2)SO_(4)` reacted with `NH_(4) = 4=` m.eqts of `(NH_(4))_(2) C_(2)O_(4)` wt of `(NH_(4))_(2)C_(2)O_(4) = 4 xx 62 xx 10^(-3)` % purity `=(4 xx 0.062)/(0.496) xx 100 = 50%` |
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| 4. |
1.575 g of oxalic acid `(CO OH)_(2).xH_(2)O` are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this soltuion requires 25 mL of `(N)/(15)NaOH` solution for complete neutralisation. Calculate x. |
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Answer» Correct Answer - 2 Meq of oxalic acid in `16.68 mL =` Meq of `NaOH` `= 25 xx (1)/(15)` `:.` Meq of oxalic acid in `250mL` `= 25 xx (1)/(15) xx (250)/(16.68) = 24.98` `:. (1.575)/((90 +18x)//2) xx 1000 = 24.98 :. x ~~2` |
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| 5. |
`A^(n+1)` is maximum oxidised by acidified `KMnO_(4)` solution into `AO_(3)^(-)`. If `2.68` m moles of `A^(+(n+1))` requires `32.16 mL` of a `0.05M` acidified `KMnO_(4)` solution for complete oxidation,value of n is |
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Answer» Correct Answer - 1 Change in oxidation no. of `A = 4n` Change in oxidation no. of `Mn = 5` `(4n)x 2.68 = 5 xx 32.16 xx 0.05 :. N = 1` |
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| 6. |
X litres of carbon monoxide is present at S.T.P. It is completely oxidised to `CO_(2)`. The volume of `CO_(2)` formed is 11.207 L at S.T.P. What is the value of X in litres?A. `22.414`B. `11.207`C. `5.6035`D. `44.828` |
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Answer» Correct Answer - B `CO +(1)/(2)O_(2) rarr CO_(2)` |
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| 7. |
Certain " mol of "HCN is oxidised completely by 25 " mL of " `KMnO_(4)`. The products are `CO_(2)` and `NO_(3)^(ɵ)` ion. When all `CO_(2)` is passed through lime water , 1 g of `CaCO_(3)` is obtained the molarity of the `KMnO_(4)` used isA. `1.44M`B. `0.72M`C. `0.36M`D. `0.8M` |
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Answer» Correct Answer - D `CN^(-) +5H_(2)O rarr CO_(2) +NO_(3)^(-) +10H^(+) +10e` `MnO_(4)^(-) +8H^(+) +5e^(-) rarr Mn^(2+) +4H_(2)O` `ulbar(CN^(-)+2MnO_(4)^(-)+6H^(+)rarr CO_(2)+NO_(3)^(-)+2Mn^(2+) +3H_(2)O)` `ca(OH)_(2) +CO_(2) rarr CaCO_(3)+H_(2)O` `100g` of `CaCO_(3) -= 44g` of `CO_(2)` `1g` of `CaCO_(3) -= 0.44g` of `CO_(2) = 0.01` moles of `CO_(2)` 1 mole of `CO_(2) -= 2` moles `MnO_(4)^(-)` `0.01` moles of `CO_(2) -= 0.02` moles of `MnO_(4)^(-)` Molarity of `KmnO_(4) = (0.02 xx 1000)/(25) = 0.8M` |
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| 8. |
Direct titration of `I_(2)` with a reducing agent is called iodimetry. If `I_(2)` is leberated by the oxidation of `I_(ɵ)` ion by a strong oxidising agent in neutral or acidic medium, the liberated `I_(2)` is then titrated with reducing agent. Iodometry is used to estimate the strngth of the oxidising agent. For example, in the estimation of `Cu^(2+)` with `S_(2)O_(3)^(2-)` `Cu^(2+)+I^(ɵ)toCuI_(2)+I_(2)` (iodometry) `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(ɵ)` (iodimetry) Strach is used as an indicator at the end point, which forms bluecoloured complex with `I_(3)^(ɵ)` Disappearance of blue colourindicates the end point whe free `I_(2)` in not present. Q. In the reaction `2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2)` The equivalent weight of `CuSO_(4)` is `(Mw=159.5g mol^(-1))`A. `10%`B. `20%`C. `5%`D. `30%` |
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Answer» Correct Answer - A eqt. of `CuSO_(4) -=` eqt. of Hypo `(w)/(159.5) = (100xx 1)/(1000), w = 15.95g` `:. %` purity `=(15.95)/(159.5) xx 100 = 10%` |
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| 9. |
`KMnO_(4)` is a strong oxidising agent in acidic medium. To provide acidic medium `H_(2)SO_(4)` is used instead of HCl. This is becauseA. `H_(2)SO_(4)` is a stronger acid than HClB. `HCl` is oxidised by `KMnO_(4)` to `Cl_(2)`C. `H_(2)SO_(4)` is a dibasic acidD. rate is faster in the presence of `H_(2)SO_(4)` |
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Answer» Correct Answer - B `5Cl_(2) +2MnCl_(2)+2KCl +8H_(2)O` |
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| 10. |
The vapour density of a tribasic acid is x. the equivalent mass of that acid isA. `(x)/(3)`B. `x-3`C. `(2x)/(3)`D. `2x -3` |
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Answer» Correct Answer - C Eq. mass `=(2 xx V.D)/("Basicity")` |
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| 11. |
When `Br_(2)` is mixed with phosphorous, first `PBr_(3)` is formed till all phosporous has reacted. If `Br_(2)` still remains, `PBr_(5)` is formed as far as possible. If equal masses of `Br_(2)` and P are mixed, which of the following will surely remain in the reaction mixtureA. PB. `PBr_(2)`C. `PBr_(5)`D. `Br_(2)` |
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Answer» Correct Answer - A::B `{:(2P+3Br_(2)rarr,2PBr_(3)),(2xx31,3xx 160),(62,480):}` If equal masses of phosphorous and bromine are mixed, phosphorous is excess reagent. |
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| 12. |
1 " mol of "`IO_(3)^(ɵ)` ions is heated with excess of `I^(ɵ)` ions in the presence of acidic conditions as per the following equation `IO_(3)^(ɵ)+I^(ɵ)toI_(2)` How many moles of acidified bypo solution will be required to react completely with `I_(2)` thus produced? |
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Answer» Correct Answer - 6 (i) `10e^(-) +2IO_(3)^(-) rarr I_(2), 2I^(-) rarr I_(2) +2e^(-) xx 5` `2IO_(3)^(-) +10I^(-) rarr 6I_(2)` 2 mol 10 mol 6 mol (ii) `{:(I_(2)+2S_(2)O_(3)^(2-),rarr S_(4)O_(6)^(2-) +2I^(-)),(1 mol ,1mol):}` 1 mol of `IO_(3)^(-) = 5` mol `5I^(-) -=3` mol of `I_(2) = 6` mol of `S_(2)O_(3)^(2-)` |
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| 13. |
The oxidation state of phosphorus is maximum inA. Phospine `(PH_(3))`B. Diphosphine `(P_(2)H_(4))`C. Metaphosphoric acid `(HPO_(3))`D. Phosphorus acid `(H_(3)PO_(3))` |
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Answer» Correct Answer - C `Hoverset(+5)(PO_(3)) gt H_(3)overset(+3)(PO_(3)) gt overset(-2)(P_(2))H_(4) gt overset(-3)(PH_(3))` |
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| 14. |
Oxidation state of phosphorus in pyrophosphosphate ion `(P_(2)O_(7)^(-4))` isA. `+7`B. `+3`C. `+8`D. `+5` |
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Answer» Correct Answer - D `P_(2)O_(7)^(-4)` `2x +7 (-2) = 4` `2x = +10 rArr x = +5` |
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| 15. |
Equivalent weight of Pyrophosphoric acid is `(H_(4)P_(2)O_(7))`A. `M.W//1`B. `M.W//2`C. `M.W//4`D. `M.W//3` |
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Answer» Correct Answer - C `H_(4)P_(2)O_(7), n`-factor =4 |
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| 16. |
When `Cl^(-)` ions are converted to `Cl_(2)`, the oxidation number of chlorine changes fromA. `-1`to 0B. `-1` to `+1`C. `-1` to `+2`D. `-2` to 0 |
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Answer» Correct Answer - A `Cl^(-) rarr (1)/(2)Cl_(2)^(@)` |
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| 17. |
The answer of the calculation `(2.568 xx 5.8)/(4.168)` in significant figures will be :A. `3.579`B. `3.570`C. `3.57`D. `3.6` |
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Answer» Correct Answer - D `(2.568 xx 5.8)/(4.168) = 3.5735125 = 3.6` |
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| 18. |
`n g` of substance `X` reacts with `m g` of substance `Y` to from `p g` of substance `R` and p g of substance S. This reaction can be represented as, `X+Y=R+S`. The relation which can be established in the amounts of the reactants and the products will beA. `n - m = p-q`B. `n +m = p +q`C. `n=m`D. `p=q` |
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Answer» Correct Answer - B `n + m = p +q` |
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| 19. |
If 0.1 mole `H_(3)PO_(x)` is completely neutralised by 5.6g KOH then select the true statement.A. `x = 3` and given acid is dibasicB. `x = 4` and given acid has no P-H linkageC. `x = 2` and given acid does not form acid saltD. `x = 4` and given acid has P-H linkage |
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Answer» Correct Answer - C No. of equivalents of `H_(3)PO_(x) =` No. of equivalents of `KOH` `n xx 0.2 = (56)/(56) rArr n = 1` `:. H_(3)PO_(x)` is a monobasic acid. It is `H_(3)PO_(2)` `x = 2` and the acid does not form acidic salt |
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| 20. |
How many water molecules are there in one drop pf water (volume `0.0018ml)` at room temperature?A. `6.023 xx 10^(19)`B. `1.084 xx 10^(18)`C. `6.023 xx 10^(23)`D. `4.84 xx 10^(17)` |
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Answer» Correct Answer - A No. of molecules `=(W)/(MW) xx N`
Given volume in a drop of water=0.0018mL Density of Water is 1g/mL Therefore, In a drop of water there is 0.0018g of water. no. of moles of water in drop=0.0018/18=0.0001 no. of molecules=no. of moles × Avogadro's No. no. of molecules= 6.022×10¹⁹ |
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| 21. |
What is the percentage of deterium to heavy water ?A. `80%`B. `60%`C. `40%`D. `20%` |
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Answer» Correct Answer - D `%` of element `=(Wt)/(M.Wt) xx 100` |
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| 22. |
Equivalent weight of Ferrous ion is (At.wt.of Fe = 56)A. 56B. 28C. 14D. 32 |
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Answer» Correct Answer - B `GEW_(Fe^(+2)) = (GAW)/(2) = (56)/(2) = 28` |
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| 23. |
One mole of chlorine combines with certain weight of metal giving 111 gm of its chloride. The same amount of metal can displace. 2gm hydrogen from an acid. The equivalent weight of metal isA. 40B. 20C. 80D. 10 |
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Answer» Correct Answer - B `MW` of `MCl_(2) = 111g` `M +71 = 111` `M = 111 - 71 = 40` 2gm of `H_(2)` displaced by 40gm of meyal 1gm of `H_(2)` displaced by? |
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| 24. |
Ordinary water contain one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in drop of water of volume 0.01 ml is (Density of water is `1gm//ml)`A. `2.5 xx 10^(16)`B. `2.5 xx 10^(17)`C. `5 xx 10^(16)`D. `7.5 xx 10^(16)` |
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Answer» Correct Answer - C `6000 -1 0.01 -?` No. of molecules `(w)/(Mw) xx N` |
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| 25. |
One part of element A reacts with two parts of another element B. 6 parts of element C reacts with 4 parts of element B. If A and C combine together, the ratio of their weights be governed byA. Law of definite proportionB. Law of multiple proportionC. Law of reciprocal proportionD. Law of conservation of mass |
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Answer» Correct Answer - C Law of reciprocal proportion |
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| 26. |
The number of parts by mass of the element or compound which combines or displaces directly or indirectly one part by mass of Hydrogen is calledA. Atomic massB. Molecular massC. Equivalent massD. Formula weight |
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Answer» Correct Answer - C Definition of equivalent mass |
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| 27. |
Equivalent weight of oxidizing or reducing agent isA. `("Formula weight")/("No.of" e^(-1) "gained(or)lossed for one formulae of compound")`B. `("Formula weight")/("Total change in oxidation state for one formula of compound")`C. Both 1 & 2D. `("Formula weight")/("Valency")` |
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Answer» Correct Answer - C Formula of Eq. Wt |
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| 28. |
`H_(2)SO_(4)+NaOH rarr NaHSO_(4)+H_(2)O` In this equation the equivalent weight of sulphuric acid is... |
| Answer» Correct Answer - `98//1 = 98` | |
| 29. |
18 " mL of " 1.0 M `Br_(2)` solution undergoes complete disproportionation in basic medium to `Br^(The hardness of water in terms of )` and `BrO_(3)^(ɵ)`. Then the resulting solution required 45 " mL of " `As^(3+)` solution to reduce `BrO_(3)^(ɵ)` to `Br^(ɵ)`. `As^(3+)` is oxidised to `As^(5+)` which statements are correct?A. `Ew (Br_(2)) = (M)/(10)`B. `Ew(Br_(2)) = (5M)/(3)`C. Molarity of `As^(+3) = 0.4M`D. Molarity of `As^(3+) = 0.2M` |
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Answer» Correct Answer - B::C `2e^(-) + Br_(2) rarr 2Br^(-) xx 5` `Br_(2) rarr 2BrO_(3)^(-) +10e^(-)` `6Br_(2) rarr 5Br^(-) +2BrO_(3)^(-)` eq of `Br_(2) = (M)/(2) + (10M)/(6) = (5M)/(3)` mmoles of `Br_(2) = 18 xx 1 = 18` So mmoles of `BrO_(3)^(-) = (18)/(6) xx 2 = 6` mEq of `underset((n=2))(As^(+3)) rarr` mEq. of `underset((n = 6))(BrO_(3)^(-))` `6e^(-) +BrO_(3)^(-) rarr Br^(-)` mEq of `As^(3+) -=` mEq of `BrO_(3)^(-)` `45 xx M xx 2` (n-factor) `-= 6 xx 6` `:. M_(As^(3+)) = (36)/(90) = 0.4M` |
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| 30. |
`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. 3B. 6C. 12D. 24 |
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Answer» Correct Answer - C `V_(a)N_(a) = V_(b)N_(b), 40 xx N_(a) = 28 xx (1)/(10) rArr N_(a) = 0.07` eqts. of metal =eqts. of acid reacted m. eqts of acid reacted with metal `= 100 - 70 = 30` `(0.36)/(E) = (30)/(1000) :. E w` of metal = 12 |
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| 31. |
Washing soda `(Na_(2)CO_(3).10H_(2)O)` is widely used in softening of hard waer. If 1 L of hard water requires 0.0286 g of washing soda, the hardness of `CaCO_(3)` in ppm isA. 10ppmB. 5 ppmC. 8ppmD. 6ppm |
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Answer» Correct Answer - A (a) Mw of washing soda `= 286`, `CaSO_(4) +Na_(2) CO_(3) rarr CaCO_(3)+Na_(2)SO_(4)` m. moles of `Na_(2)CO_(3).10H_(2)O =m`. Moles of `CaCO_(3) =` m. moles of salt causing hardness Hardness of `CaCO_(3) = (28.6)/(286) xx (100)/(1) = 10` ppm |
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| 32. |
40 ml of a hydrogen undergoes combustion in 260 ml of oxygen and gives 160 ml of carbondioxide. If all gases are measured under similar conditions of temperature and pressure the formula of hydrocarbon isA. `C_(3)H_(8)`B. `C_(4)H_(8)`C. `C_(6)H_(14)`D. `C_(3)H_(10)` |
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Answer» Correct Answer - D `C_(x)H_(y) +(x+(y)/(4)) O_(2) rarr xCO_(2) +(y)/(2)H_(2)O` |
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| 33. |
In the reaction, Hydrogen(g)+ Oxygen (g) `rarr` watervapour, the ratio of volumes is 2:1:2:. This illustrates the law ofA. Conservation of massB. Combining weightsC. Combining volumesD. All the above |
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Answer» Correct Answer - C Law of combining volumes |
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| 34. |
Iodine titrations: Compounds containing iodine are widely used in titrations, commonly known as iodine titration. It is of two kinds: (i) Iodometric titrations (ii) Iodimetric titrations. (i) Iodometric titrations: It is nothing but an indirect method of estimating the iodine. In this type of titration, an oxidising agent is made to react with excess of `KI`, in acidix medium or , basic medium in which `I^(-)` oxidises into `I_(2)`. Now the liberated `I_(2)` can be titrated with `Na_(2)S_(2)O_(3)` solution. `KI overset("Oxidising Agent")rarr I_(2) overset("Na_(2)S_(2)O_(3)//H^(+))rarr I^(-)+Na_(2)S_(4)O_(6)` Although solid `I_(2)` is black and insoluble in water, but it converts into soluble `I_(3)` ions `{:(I_(2)(s)+I^(-) hArr, I_(3)^(-)),("Black","dark brown"):}` Strach is used as indicator near the end point or equivalence point. Even small amount of `I_(2)` molecules, gives blue colour with strach. The completion of the reaction can be detected when blue colour disappears at the and point. In iodimetric titration, the strength of reducing agent is determined by reacting it with `I_(2)`. A solution containing `Cu^(+2)` and `C_(2)O_(4)^(-2)`ions M which on titration with `M//10 KMnO_(4)` requires `50mL`. The resulting solution is neutralized with `K_(2)CO_(3)`, then treated with excess of `KI`. M The liberated `I_(2)` required `25mL M//10 Na_(2)S_(2)O_(3)` in acidic solution, then what is the difference of the number of m mole of `Cu^(+2)` and `C_(2)O_(4)^(-2)` ions in the solution?A. 40B. 10C. 30D. 50 |
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Answer» Correct Answer - B m moles of `KMnO_(4)` used `=50 xx (1)/(10) = 5` m eq. of `KMnO_(4)` used (n f = 5) = 25 m eq. of `C_(2)O_(4)^(-2)` m moles of `C_(2)O_(4)^(-2) (n = 2) = (24)/(2) = 12.5` m eq of `Na_(2)S_(2)O_(3) = 2.5 =m` eq. of `Cu^(+2) (n = 1)` m eq. of `Cu^(+2) (n = 1) = 2.5` Differencein number of m moles of `Cu^(+2)` and `C_(2)O_(4)^(-2) = 12.5 - 2.5 = 10` |
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| 35. |
Sulphur forms the chlorides `S_(2)Cl_(2)` and `SCl_(2)`. The equivalent mass of sulphur in `SCl_(2)` isA. 8B. 16C. 64.8D. 32 |
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Answer» Correct Answer - B 71 g of `CI = 32g` of S `rarr 35.5 g` of `CI = ?` `rArr 16g` of `S = GEW` |
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| 36. |
Calculate the number of `Cl^(-)` and `Ca^(2+)` ions in `222 g` anhydrous `CaCl_(2)`.A. `4N,2N`B. `2N,4N`C. `1N,2N`D. `2N.1N` |
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Answer» Correct Answer - A 1 mole `CaCl_(2) = 1` mole `Ca^(+2) +2` moles `Cl^(-)` ions `rArr n_(CaCl_(2)) = (W_(CaCl_(2)))/(GMW)` |
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| 37. |
A sample of a mixture of `CaCl_(2)` and NaCl weighing 4.44 gm was treated to precipatate all the Ca as `CaCo_(3)` , which was then heated and quantitatively converted to 1.24 g of CaO. Choose the correct statements. (Atomic weight :Ca=40=23, Cl =35.5)A. Mixture contains `50% NaCI`B. Mixture contains `60% CaCI_(2)`C. Mass of `CaCI_(2)` is 2.22gD. Mass of `CaCI_(2) 1.11g` |
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Answer» Correct Answer - A::C `CaCl_(2) rarr CaCO_(3) rarr CaO (1.12)/(56) = 0.02` mole `CaO` `:.` Moles of `CaCl_(2) = 0.02` mole Mass of `CaCl_(2) = 0.02 xx 111 = 2.22g` `:. %` of `CaCl_(2) = (2.22)/(4.44) xx 100 = 50%` |
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| 38. |
A certain compound contains magnesium, carbona dn Nitrogen in the mass ratio `12: 12:14`. The formula of the compound isA. `MgCN`B. `Mg_(2)CN`C. `MgCN_(2)`D. `Mg(CN)_(2)` |
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Answer» Correct Answer - D `(Wt)/(At.wt) =` no. of gram atoms `Mg: C : N = (12)/(24): (12)/(12): (14)/(14)` |
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| 39. |
`18.4 gms` of a mixture of calcium carbonate and magnesium carbonate on heating gives `4.0` gms of magnesium oxide. The volume of `CO_(2)` produced at STP in this process isA. `1.12` litB. `4.48` litC. `2.24` litD. `3.36`lit |
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Answer» Correct Answer - C `MgCO_(3) +CaCO_(3) rarr MgO +CaO +2CO_(2)` |
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| 40. |
How many moles of magnesium phosphate, `Mg_(3)(PO_(4)_(2)` will contain 0.25 mole of oxygen atoms?A. `3.125 xx 10^(-2)`B. `1.25 xx 10^(-2)`C. `2.5 xx 10^(-2)`D. `0.02` |
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Answer» Correct Answer - A 1 mole of magnesium phosphate contains 8 moles of oxygen atoms. |
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| 41. |
The number of atoms in 0.004g of magnesium will beA. `4 xx 10^(20)`B. `8 xx 10^(20)`C. `10^(20)`D. `6.02 xx 10^(20)` |
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Answer» Correct Answer - C No. of atoms =No. of moles `xx N` |
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| 42. |
Which one of the following gases contain the least number of molecukes ? (At Wt of `P = 32, Cl = 35.5, N = 14, O = 16, C = 12, H = 1)`.A. 4.0g of laughing gasB. 3.0g of PhosphineC. 2.0 g of marsh gasD. 10.0 g of Phosgene |
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Answer» Correct Answer - B No. of molecules = No. of moles `xxN` |
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| 43. |
`27 g` of `Al` will react completely with…… `g` of `O_(2)`A. 8gmB. 16gmC. 32mD. 24gm |
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Answer» Correct Answer - D `4Al +3O_(2) rarr 2Al_(2)O_(3)` |
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| 44. |
Number of gram equivalents of solute in 100ml of `5NHCl` solution isA. 50B. 500C. 5D. 0.5 |
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Answer» Correct Answer - D No. gram equivalents `= (N xx V("in ml"))/(1000)` |
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| 45. |
A solution is prepared by adding 2g of substance A to 18g of water. The mass percent of the solute.A. 10B. 20C. 40D. 25 |
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Answer» Correct Answer - A `w = (M xx GMW xx V("in ml"))/(1000)` |
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| 46. |
"The total mass of reactants is always equal to the total mass of products in a chemical reaction." This statement is known asA. Law of conservation of massB. Law of definite proportionsC. Law of equivalent weightsD. Law of combining masses |
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Answer» Correct Answer - A Law of conservation of mass |
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| 47. |
Which is larger quantity.A. MegaB. FemtoC. MilliD. Giga |
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Answer» Correct Answer - D Giga `= 10^(9)`, Mega `= 10^(6)`, milli `= 10^(-3)`, femto `= 10^(-15)` |
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| 48. |
Dimensions of pressure are same as that ofA. EnergyB. ForceC. Energy per unit volumeD. Force per unit volume |
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Answer» Correct Answer - C Pressure `= ML^(-1)T^(-2)` & `(E)/(V) = ML^(-1)T^(-2)` |
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| 49. |
The prefix `10^(18)` isA. GigaB. NanoC. MegaD. Exa |
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Answer» Correct Answer - D Gida `= 10^(-9)`, Mega `= 10^(6)`, nano `= 10^(-9)`, Exa `= 10^(18)` |
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| 50. |
An unbalanced chemical equation is against the law ofA. The law of gaseous volumesB. The law of constant proportionsC. The law of mass actionD. The law of conservation of mass |
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Answer» Correct Answer - D Unbalanced chemical equation does not follows law of conservation of mass |
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