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InterviewSolution
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. |
A compound `H_(2)X` with molar mass of 80 g is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, what is the molality of 3.2 M solution of the compound ? |
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Answer» Let the olume of 3.2 M solution be 1000 mL Since there is no change in volume on dissolution Volume of the solvent = 1000 mL Mass of the solvent = Volume `xx` density `(1000 mL xx 0.4 "g mL"^(-1)) = 400 g` No. of moles of the solute present in the solution = 3.2 mol `:.` Molality of the sodium `(m)=("No. of moles of the solute")/("Mass of the solvent in Kg")` `=((3.2 "mol"))/(400//1000kg)=8 "mol kg"^(-1)=8m` |
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| 2. |
Fill in the blanks. a. The equivalent weight of `NaHCO_(3)` is ……..and of `SO_(2)` is ……… b. 2 mol of 50% pure `Ca(HCO_(3))_(2)` on heating forms 1 mol of `CO_(2)`. The % yield of `CO_(2)` is …….. c. `5 g` of `K_(2) SO_(4)` was dissolved in `250 mL` of solution. The volume of this solution that should be used so that `1.2 g` of `BaSO_(4)` be precipitated fromk `BaCl_(2)` is ....... (molecular mass of `K_(2) SO_(4) = 174` and `BaSO_(4) = 233`) d. The residue obtained on strongly heating `2.76 g Ag_(2) CO_(3)` is ........ `[Ag_(2)CO_(3) overset(Delta)rarr Ag + CO_(2) + O_(2)]` Atomic mass of `Ag = 108` |
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Answer» Correct Answer - A::B::C::D a. 84,32 b. 100% c. `44.8 mL` d. `2.16 g` a. `Ew of NaHCO_(3) = (84)/(1)` (valnecy factor = 2) `Ew of SO_(2) = (64)/(2) = 32` (Valency factor = 2) b. `underset(2 mol)(2Ca(HCO_(3))_(2)) rarr CaCO_(3) + underset(1 mol)(CO_(2)) + H_(2) O` The percent yield of `CO_(2)` is 100% whether `Ca(HCO_(3))_(2)` is 100% pure or 50% pure. c. `underset(1 mol)(K_(2) SO_(4)) + BaCl_(2) rarr underset(1 mol)(BaSO_(4)) + 2 KCl` `M_(K_(2)SO_(4)) = (5 xx 1000)/(147 xx 250) = (20)/(174)` `M xx V_(L) (K_(2) SO_(4)) =` moles of `BaSO_(4)` `(20)/(174) xx V_(L) = (1.2)/(233)` `V_(L) = 0.0488 L = 44.8 mL` d. `underset((Mw = 276 g))(Ag_(2) CO_(3)) rarr (2 xx 108 g)(2Ag ) + CO_(2) uarr + O_(2) uarr` The residue is due to `Ag`. `276 g of Ag_(2) CO_(3) = 2 xx 108 g Ag` `2.76 g of Ag_(2) CO_(3) = (2 xx 108 xx 2.76)/(276) = 2.16 g` |
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| 3. |
`5.6 g` of a metal forms `12.7 g` of metal chloride. Hence equivalent weight of the metal isA. 127B. 254C. 56D. 25 |
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Answer» Correct Answer - D Weight of `Cl` reacted `= 12.7 - 5.6 - 7.1 g` `7.1 g Cl -= 5.6` of metal `35.5 g Cl = 28 g` of metal |
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| 4. |
Assertion: The average mass of one Mg atom is `24.305 amu`, which is not actual mass of one Mg atom. Reason: Three isotopes, `24 Mg, 25 Mg and 26 Mg`, of Mg are found in nature.A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is Ture, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - A |
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| 5. |
` 1 g` of impure `Na_(2)CO_(3)` is dissolved in water and the solution is made upto `250 mL`. To `50 mL` of this solution, `50 mL` of `0.1 N HCl` is added and the mixture after shaking well required `10 mL` of `0.16 N NaOH` solution for complete neutralization. Calculation percent purity of the sample of `Na_(2) CO_(3)`. |
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Answer» `1.0 g` of (impure) `Na_(2) CO_(3) + H_(2)O rarr 250 mL` `(50 mL of Na_(2)CO_(3) + 50 mL of 0.1 N NaOH)` `= 10 mL of 0.16 N NaOH` In this question, `HCl` is in excess. `implies` Excess mEq of `HCl = mEq of NaOH = 0.16 xx 10 = 1.6` mEq of `HCl` added to `Na_(2) CO_(3) = 0.1 xx 50 = 5` `implies` mEq of `HCl` used to neutralised `Na_(2) CO_(3) = 5 - 1.6 = 3.4` So, mEq of `Na_(2)CO_(3)` (pure) in `50 mL = 3.4` `implies` mEq of pure `Na_(2) CO_(3)` is `250 mL` `3.4 xx (250)/(50) = 17` or `("Weight")/(Ew) xx 100 = 17` `implies "Weight = (17 xx (106 // 2))/(1000) = 0.901 g` `(Na_(2) CO_(3)` is diacidic base `implies Ew = 106//2)` So mass of pure `Na_(2)CO_(3) = (0.901)/(1) xx 100 = 90.1%` |
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| 6. |
`0.7 g` of iron reacts directly with `0.4 g` of sulphur to form ferrous sulphide. If `2.8 g` of iron is dissolved in dilute `HCl` and excess of sodium sulphide solution is added, `4.4 g` of iron sulphide is precipitated. Prove the law of constant composition. |
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Answer» The ratio of the weight of iron and sulphur in the first sample of the compound is `Fe : S : : 0.7 : 0.4` or `7 : 4`. According to the second experiment, `2.8 g` or iron gives `4.4 g` ferrous sulphide, or `2.8 g Fe` combines with `S = 4.4 - 2.8 = 1.6 g` Therefore, the ratio of the weights of `Fe : S : : 2.8 : 1.6` or `7 : 4`. Since the ratio of the weights of the two elements is same both the cases, the law of constant composition is true. |
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| 7. |
`4.0 g` of a mixture of `Nacl` and `Na_(2) CO_(3)` was dissolved in water and volume made up to `250 mL`. `25 mL` of this solution required `50 mL` of `N//10 HCl` for complete neutralisation. Calculate the percentage composition of the original mixture. |
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Answer» Correct Answer - B `NaCl` does not react with `HCl` `NaCO_(3) + 2HCl rarr 2NaCl + H_(2)O + CO_(2)` mEq of `Na_(2)CO_(3) -= mEq "of" HCl` `-= 50 xx (1)/(10) = 5` mEq per `25 mL` `-= 50 mEq` per `250 mL` Weight of `Na_(2)CO_(3) = 50 xx 10^(-3) xx (106)/(2) = 2.65 g` % of `Na_(2)CO_(3) = (2.65)/(4.0) xx 100 = 66.25%` |
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| 8. |
Molarity and Molality of a solution of a liquid (molecular weight = 50) in aqueous solution is 9 and 18 respectively. What is the density of solution?A. 1 g/ccB. `0.95` g/ccC. `1.05` g/ccD. `0.66` g/cc |
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Answer» Correct Answer - B |
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| 9. |
Objective question (single correct answer). i. `H_(3) PO_(4)` is a tribasic acid and one of its salt is `NaH_(2) PO_(4)`. What volume of `1M NaOH` solution should be added to `12 g` of `NaH_(2) PO_(4)` to convert in into `Na_(3) PO_(4)` ? a. `100 mL` b. 2 mol of `Ca (OH)_(2)` c. Both d. None iii. The normality of a mixture obtained mixing `100 mL` of `0.2 m H_(2) SO_(4)` with `100 mL` of `0.2 M NaOH` is: a. `0.05 N` b. `0.1 N` c. `0.15 N` d. `0.2 N` iv `100 mL` solution of `0.1 N HCl` was titrated with `0.2 N` `NaOH` solutions. The titration was discontinued after adding `30 mL` of `NaOH` solution. The reamining titration was completed by adding `0.25 N KOH` solution. The volume of `KOH` required from completing the titration is: a. `70 mL` b. `35 mL` c. `32 mL` d. `16 mL` |
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Answer» Correct Answer - B::D a. b. `Mw "of" NaH_(2) PO_(4) = 120 g` (it is dibasic) (valency factor = 2) Eq of `NaOH = Eq "of" NaH_(2)PO_(4)` `(n = 1)` `(n = 2)` `1 M xx 1 xx V_(L) = (12)/(120) xx 2`. `V_(L) = 0.26 L = 200 mL` ii.b. `SO_(2) Cl_(2) + 2H_(2) O rarr H_(2) SO_(4) + 2HCl` `{:(1"mol" H_(2)SO_(4) -= 2NaOH),(2"mol"HCl -= 2NaOH):}] = 4 "mol" NaOH` `{:(1"mol" H_(2)SO_(4) -= 1"mol" Ba(OH)_(2)),(2"mol"HCl -= 1"mol"Ba(OH)_(2)):}] implies {:(2"mol"),(Ba(OH)_(2)):}` iii. b. `mEq "of" H_(2)SO_(4) = 100 xx 0.2 xx 1 = 20` `mEq NaOH = 100 xx 0.2 xx 1 = 20` `mEq "of" H_(2)SO_(4) "left" = 40 - 20 = 20` Total volume `= 100 + 100 = 20` `N` solution `= (20 mEq)/(200 mL) = 0.1 N` iv. d. Total mEq of `HCl = 100 xx 0.1 = 10` Total mEq of `NaOH = 300 xx 0.2 = 6` mEq of `HCl` left `= 10 - 6 = 4` mEq of `HCl = mEq "of" KOH` `4 = 0.25 xx V, V implies 16 mL` |
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| 10. |
`H_(2) SO_(2)` solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified is just dilute `H_(2) SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium. simulataneously forming a dark brown precipitate of hydrated `MnO_(2)`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2) SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2) O_(2)` solution. |
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Answer» Take the balanced equations as follows: `2MnO_(4)^(ɵ) + 6 H^(o+) + 5 H_(2) O_(2) rarr 2 Mn^(2+) + 8 H_(2) O + 5 O_(2)` `2 MnO_(4)^(ɵ) + 2 Mn^(2+) rarr 5 MnO_(2)` `MnO_(2) + C_(2) O_(4)^(2-) + 4 H^(o+) rarr Mn^(2+) + 2 H_(2) O + 2CO_(2)` To find molarity of `H_(2) O_(2)` Given: `20 mL H_(2) O_(2) -= 20 mL KMnO_(4)//H^(o+)` `20 mL KMnO_(4) -= mL MnSO_(4) rarr MnO_(2)` `MnO_(2) -= 10 mL (0.2 M) Na_(2) C_(2) O_(4)` From stoichiometry of above reactions: 1 mmol of `Na_(2) C_(2) O_(4) -= 1 mmol MnO_(2)` `implies 0.2 xx 10 mmol of Na_(2) C_(2) O_(4) -= 2 mmol of MnO_(2)` Also, 5 mmol of `MnO_(2) -= 3 mmol of Mn^(2+)` `implies 2 mmol of MnO_(2) -= (3)/(5) xx 2` `-= (6)/(5) mmol of Mn^(2+) (MnSO_(4))` Also, 3 mmol of `MnSO_(4) -= 2 mmol KMnO_(4)` `implies (6)/(5) mmol of MnSO_(4) -= (2)/(3) xx (6)/(5) = mmol KMnO_(4)` Now same amount of `KMnO_(4)` reacts completely with `H_(2) O_(2)`. `2 mmol of H_(2) SO_(4) -= 5 mmol of H_(2) O_(2)` `implies 2 mmol of KMnO_(4) -= (5)/(2) xx (4)/(5) = 2 mmol of H_(2) O_(2)` Now mmol of `H_(2) O_(2) = MV_(mL)`. `implies 2 = M xx 20 implies M = 0.1` |
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| 11. |
`0.607g` of silver salt of tribasic organic acid was quantitatively reduced to `0.37g` of pure `Ag`. What is the mol. Wt. of the acid ?A. 207B. 210C. 531D. 324 |
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Answer» Correct Answer - 2 Moles of `Ag_(3)A=("Moles of Ag")/(3)=(0.607)/(M)=(0.37)/(108)xx(1)/(3)` `rArr " "M=531` `:. " " mol. wt. `of `H_(2)A=mol. wt. ` of `Ag_(3)A-3xxAt. wt. ` of `Ag+3xxAt. wt. `of `H=210` |
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| 12. |
The mineral rutile is an oxide of titanium containing 39.35% oxygen and is isomorphous with cassiterite `(SnO_(2))`. The atomic weight of titanium isA. 68.1B. 58.1C. 48.1D. 38.1 |
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Answer» Correct Answer - C `[{:("Rutile" = TiO_(2)),("casiterite" = SnO_(2)):}]` Both are isomorphous `Mw "of" TiO_(2) = x + 32` `(x + 32) g "of" TiO_(2) implies (32)/((32 + x)) xx 100` `:. (32 xx 100)/((32 + x)) = 39.95` `:. X = 48.10` |
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| 13. |
Cholrine gas can be produced by the reaction of `HCl (aq)` with `MnO_(2) (s)`. Only `MnCl_(2)` and `H_(2)O(l)` are the by products. What volume of `Cl_(2) (g) ("in litre")` of density `2.84 g//L` will be product from the reaction of `400 mL` of `0.1 M HCl(aq)` with an excess of `MnO_(2)` ? |
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Answer» Correct Answer - `0.25L` ` M " "+ " "Cl_(2) " "rarr " "MCl_(2)` `MCl_(2)" "rarr" "M(OH)_(2)` `2.375 gm` `M(OH)_(2) " "+ " "2HCl " "rarr " "4MCl_(2)+2H_(2)O` `25 m` mole `50 m` mole `(2.375)/((x+71))=(25)/(1000)` `implies x=24` |
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| 14. |
200 ml of an aqueous solution of glucose `(C_(6)H_(12)O_(6))` has molarity of `0.01M`. Which of the following operations can be done to this solution so as to increase molarity to `0.015` M?A. Evaporate 50 ml water from this solutionB. Add `0.18` g glucose to solution without changing its volumeC. Add 50 ml water to this solutionD. None of the above |
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Answer» Correct Answer - B |
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| 15. |
`11.2 g` of carbon reacts with 21.1 litres of oxygen at `18^(@)C` and `750 mm` of `Hg`. The cooled gases are passed through 2 litre of `2.5 N NaOH`. Determine the concetration of `NaOH` remaining in solution which is not converted to `Na_(2) CO_(3)`. Assume that `CO` does not react with `NaOH`: a. Whatis the mole fraction of `CO` in the gases? b. What is the concetration of `NaOH` which is not converted to `Na_(2) CO_(3)` in the remaining solution? |
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Answer» Mole of `C = (11.2)/(12) = 0.933 mol` `Pv = nRT` `(750)/(760) xx 21.2 = n xx 0.82 xx 291` `n = (750 xx 21.2)/(760 xx 0.082 xx 291) = 0.876 mol of O_(2)` a. `C + O_(2) rarr CO_(2)` `x` `x` `x` b. `C + (1)/(2) O_(2) rarr CO` `(0.933 - x) (0.933 - x)/(2) (0.933 - x)` Total moles of `O_(2) = x + (0.933 - x)/(2) = 0.876` `:. x = 0.821 mol` `CO_(2) = 0.821 mol` `CO = 0.933 - 0.821= 0.112 mol` a. `:. x_(CO) = (n_(CO))/(n_(CO) + n_(CO_(2))) = (0.122)/(0.93) = 0.12` `underset({:(1 mol),(0.821 mol):})(CO_(2)) + underset({:(2 mol),(2 xx 0.821 mol):})(2NaOH) rarr underset(1 mol)(Na_(2)CO_(3)) + H_(2)O` Total moles of `NaOH = 2 xx 821` Moles of `NaOH` left `= 5 - 2 xx 0.821` `= 3.558 mol` Concentration of `NaOH = ("Moles")/("Volume") = (3.385)/(2)` `= 1.679 mol L^(-1)` |
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| 16. |
What is the density of propane, `C_(3)H_(8)`, at `25^(@)C` and 740 mmHg?A. `0.509 gL^(-1)`B. `0.570gL^(-1)`C. `1.75 gL^(-1)`D. `1.96gL^(-1)` |
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Answer» Correct Answer - C |
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| 17. |
The density of liquid mercury is `13.6 g//cm^(3).` How many moles of mercury are there in 1 litre of the metal? (Atomic mass of `Hg=200).` |
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Answer» Correct Answer - 68 |
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| 18. |
A 200 mL sample of a gaseous hydrocarbons has a density of `2.52gL^(-1)` at `55^(@)C` and 720mm Hg. What Is its formula?A. `C_(2)H_(6)`B. `C_(4)H_(10)`C. `C_(5)H_(12)`D. `C_(6)H_(6)` |
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Answer» Correct Answer - C |
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| 19. |
For the given series of reaction, `4NH_(3)(g) + 5O_(2)(g) rarr 4NO(g) + 6H_(2)O(l)` `2NO(g) + O_(2)(g) rarr 2NO(g) +O_(2)(g) rarr 2NO_(2)(g)` To obtain maximum mass of `NO_(2)` from a given mass of a mixture of `NH_(3)` and `O_(2)` the ratio of mass of `NH_(3)` ot `O_(2)` should be:A. `(4)/(7)`B. `(17)/(56)`C. `(17)/(40)`D. none of these |
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Answer» Correct Answer - B |
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| 20. |
For the given series of reaction, `4NH_(3)(g) + 5O_(2)(g) rarr 4NO(g) + 6H_(2)O(l)` `2NO(g) + O_(2)(g) rarr 2NO(g) +O_(2)(g) rarr 2NO_(2)(g)` If 20 ml of `NH_(3)` is mixed with 100 ml of `O_(2)` .Volume contraction at the completion of above reaction is:A. 20 mlB. 85 mlC. 35 mlD. 100 ml |
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Answer» Correct Answer - C |
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| 21. |
Nitric acid can be produced from ammonia in three step process. `{:(,4NH_(3)(g)+5O_(2)(g)rarr4NO_(g)+6H_(2)O(g),...(1)),(,2NO(g)+O_(2)(g)rarr2NO_(2)(g),...(2)),(,3NO_(2)(g)+H_(2)O(l)rarr2HNO_(3)(aq.)+NO(g),...(3)):}` Calculate weight of `NH_(3)(g)` required to produce `1260kg` of `HNO_(3)`. When `%` teild of `1^(st), 2^(nd)` and `3^(rd)` reaction are respectively `69%,60%` and `68%` respectively. |
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Answer» Correct Answer - `1811.5kg` `(1260xx1000)/(63) = (W_(NH_(3)))/(7) xx (4)/(4) xx (2)/(2) xx (2)/(3)` `W_(NH_(3))=1811.5 kg` |
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| 22. |
When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.A. `1 :3`B. `3:1`C. `2:3`D. `3:2` |
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Answer» Correct Answer - A Moles `prop` Pressure `(2)/(Mw_(A)) prop 1 "atm"` Pressure of `B = 1.5 - 1 = 0.5 "atm"` `(3)/(Mw_(B)) prop 0.5 "atm"` `(3)/(Mw_(B)) xx (Mw_(A))/(2) = (0.5)/(1)` `(MW_(A))/(Mw_(B)) = 0.5 xx (2)/(3) = (1)/(3)` `(Mw_(A) : Mw_(B) = 1:3` |
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| 23. |
How many moles of ferric alum `(NH_(4))_(2) SO_(4) Fe_(2) (SO_(4))_(3). 24 H_(2)O` can be made from the sample of `Fe` containing `0.0056 g` of it?A. `10^(-4) mol`B. `0.5 xx 10^(-4) mol`C. `0.33 xx 10^(-4) mol`D. `2 xx 10^(-4) mol` |
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Answer» Correct Answer - B Moles of `Fe = (0.0056)/(56) = 10^(-4) "mol"` 1 mol of alum = 2 mol of `Fe` 2 mol of `Fe = 1` mol of alum `10^(-4) "mol of" Fe (1)/(2) xx 10^(-4) "mol"` `= 0.5 xx 10^(-4) "mol"` |
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| 24. |
5 moles of VO and 6 moles of `Fe_(2)O_(3)` are allowed to react completely according to reaction `VO + Fe_(2)O_(3) rArr FeO + V_(2)O_(5)` The number of moles of `V_(2)O_(5)` formed is:A. 6B. 2C. 3D. 5 |
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Answer» Correct Answer - B |
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| 25. |
`XeF_(6)` fluorinates `I_(2)` to `IF_(7)` and liberates Xenon(g). 210 mmol of `XeF_(6)` can yield a maximum of…. Mmol of `IF_(7)`A. 420B. 180C. 1D. 0.15 |
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Answer» Correct Answer - B |
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| 26. |
In aviation gasoline of 100 octane number, `1.0 mL` of tetraethy lead `(TEL), (C_(2)H_(5))_(4) Pb`, of density `1.615 g mL^(-1)`, per litre is added to the product. `TEL` is prepared as follows: `4C_(2) H_(5) Cl + 4Na(Pb) rarr (C_(2)H_(5))_(4) Pb + 4NaCl + 3 Pb` Calculate the amount of `C_(2) H_(5) Cl` required to make enough `TEL` for `1.0 L` of gasoline.A. `0.645 g`B. ` 1.29 g`C. `1.935 g`D. `2.58 g` |
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Answer» Correct Answer - B The weight of `1.0 mL Tel = 1.0 mL xx 1.615 = 1.615 g` The weight of `TEL` needed per litre of gasoline `= 1.615 g` `[Mw "of" TEL [C_(2) H_(5))_(4) Pb] = 4 xx 29 + 207 = 323 g mol^(-1))` moles of `TEL = (1.615)/(323 g mol^(-1)) 0.005 "mol"` 1 mol of `TEL` resquires `= 4 xx 0.005` `= 0.02 "mo of" C_(2)H_(5) Cl` Weight of `C_(2)H_(5) Cl = 0.02 xx (64.5 g//"mol") = 1.29 g` |
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| 27. |
Calcium carbonate reacts with aqueous HCl to give `CaCl_(2)` and `CO_(2)` according to the reaction given below `CaCO_(3)(s)+2HCl(aq)rarrCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)` What mass of `CaCl_(2)` will be formed when 250mL of 0.76 M HCl reac ts with 1000 g of `CaCO_(3)`? Name the limiting reagent. Calculate the number of moles of `CaCl_(2)` formed in the reaction. |
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Answer» Calculation of mass of HCl Molarity of solution `(M)=("Mass of HCl/Molar mass of HCl")/("Volume of solution in litres")` `0.76 "mol L"^(-1)=(("Mass of HCl/36.5 g mol"))/((250//1000L))` Mass of `HCl=(0.76"mol L"^(-1)xx(0.25L)(36.5"g mol"^(-1))=6.935g` Step-II Identification of limiting reactant `underset((100 g))(CaCO_(3)(s))+underset((2xx36.5=73g))(2HCl(aq))rarrunderset((111g))(CaCl_(2))(aq)+CO_(2)(g)+H_(2)O(l)` 100g of `CaCO_(3)` need HCl = 73 g 1000 g of `CaCO_(3)` need `HCl=((73g))/((100g))xx(1000g)=730g` But the amount HCl actually available = 6.935 g `:.` HCl is the limiting reactant Step-III Calculation of mass/moles of `CaCl_(2)` in the reaction 73 g of HCl form `CaCl_(2) = 111g` 6.935 g of HCl form `CaCl_(2)=((111g))/((73g))xx(6.935g)=10.54g` Moles of `CaCl_(2)` formed `= ("Mass of "CaCl_(2))/("Molar mass of CaCl"_(2))=((10.5g))/((111"g mol"^(-1)))=0.095` mol. |
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| 28. |
Copper gives two oxides. On heating `1g` of each in hydrogen, we get `0.888 g and 0.798 g` of the metal respectively. Show that these results are in agreement with the Law of Multiple Proportions. |
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Answer» Let us fix `1g` of copper as the fixed weight in the two oxides In the first oxide Weight of copper `= 0.888 g` Weight of oxygen `= 1-0.888 = 0.112g` Now, `0.888g` of copper combine with oxygen `= 0.112 g` `:. 1.0 g` of copper combine with oxygen `= (0.112)/(0.888)g=0.126 g` In the second oxide Weight of copper `= 0.798 g` Weight of oxygen `= 1-0.798 = 0.202 g` `0.798 g` of copper combine with oxygen `=0.202 g` `1.0 g` of copper combine with oxygen `= (0.202)/(0.798)g=0.253 g` Ratio by weight of oxygen which combine with `1g` of copper in the two oxides is `0.126 : 0.253 or 1:2` As the ratio is simple whole number in nature, the law of Multiple Proportions is proved. |
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| 29. |
Statement-1 : Both 10 g of sodium carbonate and 12 g of carbon have same number of carbon atoms Statement-2 : Both contain 1g atoms of carbon which contains `6.022 xx 10^(23)` carbon atoms.A. Statement 1 is true, statement 2 is also true , statement 2 is the correct explanation of statement 1.B. Statement 1 is true, statement 2 is also true , statement 2 is not the correct explanation of statement 5C. Statement 1 is true, statement 2 is falseD. Statement 1 is fale, statement 2 is true |
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Answer» Correct Answer - A Statement-2 is the correct explanation for statement-1 |
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| 30. |
Calculate the molality of a solution containing 20.7 g of potassium carbonate in 500 mL of solution (assume density of solution `= "1g mL"^(-1)`) |
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Answer» Correct Answer - 0.313 m Mass of 500 mL of the solution `= 500 xx 1 = 500 g` Mass of solvent (water) `= 500 - 20.7 = 479.3 g = 0.4693 kg` Molality of solution `(m)=((20.7g))/(("138 g mol"^(-1))xx("0.4793 kg"))=0.313 "mol kg"^(-1)=0.313` m. |
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| 31. |
An alloy of iron and carbon was treated with suphuric acid, in which only iron reacts `2Fe(s) + 3H_(2)SO_(4)(aq) rarr Fe_(2)(SO_(4))_(3)(aq) + 3H_(2)(g)` If a sample of alloy weighing 140 g gave 6 g hydrogen, What is the percentage of iron in the alloy? |
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Answer» Correct Answer - 80 |
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| 32. |
In what volume ratio can 0.1 M `Fe_(2)(SO_(3))_(4)` and 0.1 M `Al_(2)(SO_(4))_(3)` be mixed so that the ratio of total numebr the solution is 2:3? Consider `Fe^(3+), Al^(3+) and (SO^(2-))_(4)` as only ions in the solution. Salts are completely dissociated and do not undergo any hydrolysis.A. `1:1`B. `4:7`C. `3:16`D. Any of these values |
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Answer» Correct Answer - D |
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| 33. |
Air sample from an industrial town, heavily polluted by `CO_(2)` was collected and analyzed . In one anaylsis , 56L of air measured at 1 atm and 273 K was passed through a 250 mL of 0.025 M NaOH solution , where `CO_(2)(g)` was absorbed completely . 25 mL of the above solution was then treated with excess of `BaCl_(2)` solution where all the carbonate was precipitated as `BaCo_(3)(s)` . The solution was filtered off and the filtrate required 25 mL of a 0.005 MHCl solution for neutralization. Weight (in milligrams ) of precipitate `BaCO_(3)(s)` obtained from the 25 ml of test solution was: [Atomic weight :Ba =137, C=12, O=16]A. 27.58B. 275.8C. 492.5D. 49.25 |
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Answer» Correct Answer - D |
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| 34. |
Air sample from an industrial town, heavily polluted by `CO_(2)` was collected and analyzed . In one anaylsis , 56L of air measured at 1 atm and 273 K was passed through a 250 mL of 0.025 M NaOH solution , where `CO_(2)(g)` was absorbed completely . 25 mL of the above solution was then treated with excess of `BaCl_(2)` solution where all the carbonate was precipitated as `BaCo_(3)(s)` . The solution was filtered off and the filtrate required 25 mL of a 0.005 MHCl solution for neutralization. ppm strenght of `CO_(2)(g)` , volume by volume i.e. , mL of `CO_(2)` per `10^(6)` mL of air was :A. 560B. 5600C. 100D. 1000 |
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Answer» Correct Answer - D |
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| 35. |
Air sample from an industrial town, heavily polluted by `CO_(2)` was collected and analyzed . In one anaylsis , 56L of air measured at 1 atm and 273 K was passed through a 250 mL of 0.025 M NaOH solution , where `CO_(2)(g)` was absorbed completely . 25 mL of the above solution was then treated with excess of `BaCl_(2)` solution where all the carbonate was precipitated as `BaCo_(3)(s)` . The solution was filtered off and the filtrate required 25 mL of a 0.005 MHCl solution for neutralization. Fraction of original NaOH (by mole) that reached with `CO_(2)` was:A. 0.2B. 0.4C. 0.6D. 0.8 |
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Answer» Correct Answer - D |
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| 36. |
4.9 g of `KClO_(3)` when heated produced 1.92 g of oxygen and the residue of KCl left was found to weigh 2.96 g. Show that the data illustrates the law of conservation of mass. |
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Answer» `2KClO_(3)rarr2KCl+3O_(2)` Mass of reactants =4.9 g , Mass of products `= (1.92 + 2.96) = 4.88 g` As the mass of the products is nearly same as that of the reactants, this illustrates the Law of conservation of mass. |
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| 37. |
`KClO_(3)` on heating decomposes to give `KCl and O_(2)`. What is the volume of `O_(2)` at N.T.P liberated by 0.1 mole of `KClO_(3)` ? |
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Answer» The chemical equation for the decomposition of `KClO_(3)` is `{:(2KClO_(3),overset("Heat")(rarr),2KCl+3O_(2)),("2 mol",,"3 mol"),(,,(3xx22.4L=67.2L)):}` 2 moles of `KClO_(3)` evolve `O_(2)` at N.T.P = 67.2 L 1 mole of `KClO_(3)` evolve `O_(2)` at N.T.P `= (67.2)/(2)L` 0.1 mole of `KClO_(3)` evolve `O_(1)` at N.T.P `= (67.2)/(2)xx0.1 L =3.36 L`. |
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| 38. |
A given initial mass of `KClO_(3)` on 50% decompostion produces 67.2 litre oxygen gas at `0^(@)C` and 1 atm. The other product of decompostion is KCl. The initial mass of `KClO_(3)` (in gm) taken is:A. 245B. 122.5C. 490D. none of these |
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Answer» Correct Answer - C |
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| 39. |
`18.625g KCl` is formed due to decomposition of `KClO_(4)` in reaction `KClO_(4)(s)rarr KCl(s)+2O_(2)(g)` Find volume of `O_(2)` obtained at `STP`[Atomic mass `K = 39 Cl = 35.5]` |
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Answer» `{:(KClO_(4)(s),rarr,KCl(s),+,2O_(2)(g)),(,,(18.625)/(74.5)mol,,(2xx18.626)/(74.5)),(,,=0.25,,=0.5mol):}` Vol of `O_(2)` at `STP = 0.5 xx 22.4 = 11.2L` . |
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| 40. |
Polyethene can be prepard by `CaC_(2)` by the following sequence of reactions . `CaC_(2)` + `H_(2)O to CaO + C_(2)H_(2)` `C_(2)H_(2) + H_(2) to C_(2)H_(4)` `nC_(2)H_(4)to (C_(2)H_(4))_(n)` `" (Polythene)"` The mass in kg of polythene that can be prepared by 20 kg `CaC_(2)` .A. 4.1 kgB. 8.75 kgC. 3.78 kgD. 10 kg |
| Answer» Correct Answer - B | |
| 41. |
Statement-1: Limiting reagent is the reactant that gets completed (or consumed) in a chemical reaction. Statement-2: Limiting reagent always has either least mass or the least moles among all the reactant available for a chemical reaction.A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is Ture, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - C |
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| 42. |
Silver metal reacts with nitric acid according to the equation : `3Ag(s)+4HNO_(3)(aq)rarr3AgNO(aq)+NO(g)+2h_(2)O(l)` What volume of `1.15"M HNO"_(3)(aq)` is required to react with `0.784g` of silver?A. `4.74` mLB. `6.32` mLC. `8.43` mLD. `25.3` mL |
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Answer» Correct Answer - C |
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| 43. |
The density of 2.45 M aqueous methanol `(CH_(3)OH)` is `0.976 g/mL`. What is the molatiy of the solution `(CH_(3)OH=32)` ?A. `27.3 m`B. `0.273 m`C. `7.23 m`D. `2.73 m` |
| Answer» Correct Answer - D | |
| 44. |
Statement-I : `A` reactant that is entirely consumed when a reaction goes to completion is known as limiting reactant. Because Statement-II : The amount of reactant limits the amount of product formed.A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true , Statement-II is `NOT` a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - A | |
| 45. |
Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation `2H_(2)O_(2)(aq)rarr2H_(2)O(l)+O_(2)(g)` under conditions where 1 mole of gas occupies 24 `dm^(3)`. `100cm^(3)` of XM solution of `H_(2)O_(2)` produces 3 `dm^(3)` of `O_(2)`. Thus, X is :A. 2.5B. 1C. 0.5D. 0.25 |
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Answer» Correct Answer - A |
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| 46. |
Two gases `A` and `B` which react according to the equation `aA_((g)) +bB_((g)) rarr cC_((g)) +dD_((g))` to give two gases `C` and `D` are taken (amount not known) in an Eudiometer tube (operating at a constant Pressur and temperature) to cause the above. If one causing the reaction there is no volume change observed then which of the following statement is/are correct.A. `(a+b)=(c+d)`B. average molecular mass may increase or decrease if either of `A` or `B` is present in limited amount.C. Vapour Density of the mixture will remain same throughout the course of reaction.D. Total moles of all the component of mixture will change. |
| Answer» Correct Answer - A::C | |
| 47. |
Statement-I : `16g` each `O_(2)` and `O_(3)` contains `(N_(A))/(2)` and `(N_(A))/(3)` atoms repectively Because Statement-II : `16g` of `O_(2)`, and `O_(3)` contains same no. of atoms.A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true , Statement-II is `NOT` a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - D | |
| 48. |
`{:(,"Column-I",,"Column-II",),(,,,("mass of product"),),((A),2H_(2)+O_(2)rarr2H_(2)O,(p),1.028 g,),(,lg" "lg,,,),((B),3H_(2) +N_(2) rarr2NH_(3),(q),1.333 g,),(,lg " " lg,,,),((C),H_(2) + CI_(2) rarr 2HCI ,(r),1.125 g,),(,lg " "lg ,,,),((D),2H_(2) + C rarrCH_(4),(s),1.214 g,),(,lg " "lg,,,):}` |
| Answer» Correct Answer - A::B::C::D | |
| 49. |
Hydrazine reacts with `KIO_(3)` in presence of `HCl` as `:` `N_(2)H_(4)+IO_(3)^(-)+2H^(+)+Cl^(-) rarrICI+N_(2)+3H_(2)O` The equivalent masses of `N_(2)H_(4)` and `KIO_(3)` respectively are `:`A. 8 and `53.5`B. 16 and `53.5`C. 8 and `35.6`D. 8 and 87 |
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Answer» Correct Answer - 1 Oxidation number of `N` in `N_(2)H_(4)` is `-2` which change `s` to `0` in `N_(2)`. Hence , equivalent mass of `N_(2)H_(4)=("Molar mass")/(2xx2)` Oxidation number of iodine changes from `+5` to `+1` in `ICI` Hence, equivalent mass of `IO_(3)=("Molar mass")/(4)` |
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| 50. |
How many of the prefixes are correctly matched with their multiples `:` `(i)`pico `(p) -10^(-12)" "(ii)` tera`(T)-10^(12)" "(iii)`giga `(G)-10^(9)` `(iv)` nano `(n) -10^(-9)" "(v)` mega`(M)-10^(6)" "(vi)`micro`(mu)-10^(-6)` `(vii)` centi`(cm)-10^(-1)" " (vii)` deci`(D)-10" "(ix)` milli`(m)-10^(-3)`A. 5B. 6C. 7D. 8 |
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Answer» Correct Answer - 3 `(vi)` centi`(cm)-10^(-2)" "(vii)`deci`(D)-10^(-1)` |
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