Saved Bookmarks
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.
| 251. |
A hydrocarbon is burnt with just sufficient amount of oxygen. After combustion the volume contraction was 2.5 times of volume of hydrocarbon burnt. On passing this mixture through KOH, if volume contraction is twice the volume of hydrocarbon taken. Calculate no. of atoms in one molecule of the hydrocarbon. |
|
Answer» Correct Answer - 8 |
|
| 252. |
If 0.5 mole of `BaCl_(2)` mixed with 0.20 mole of `Na_(3) PO_(4)` the maximum number of moles of `Ba_(3)(PO)_(2)` then can be formed isA. 0.1B. 0.2C. 0.5D. 0.7 |
|
Answer» Correct Answer - A `underset(2 "mol")(3 BaCl_(2)) + underset(2 "mol")(2Na_(3)PO_(4)) rarr underset(6 "mol")(6NaCl) + underset(1 "mol")(Ba_(3) (PO_(4))_(2))` Given `implies 0.5 "mol of" BaCl_(2)` and `0.2 "mol of" Na_(3)PO_(4)` Find the limiting reagent: `2 "mol of" Na_(3)PO_(4) implies 2 "mol of" BaCl_(2)` `0.2 "mol of" Na_(3)PO_(4) implies 0.3 "mol of" BaCl_(2)` `:. Na_(3)PO_(4)` is the limiting reagent `:. "mol of" Na_(3)PO_(4) implies "mol of" Ba_(3) (PO_(4))_(2)` `0.2 "mol" of NaPO_(4) implies 0.1 "mol of" Ba_(3)(PO_(4))_(2)` |
|
| 253. |
How many gram water should be added in 200gm of `104.5%` labelled oleum sample to make the new labelling equal to `5225/52%?` |
|
Answer» Correct Answer - 8 |
|
| 254. |
A 50 gm sample which may contain either `(H_(2)S)_(4)` or `SO_(3)` or any combination of the two is mixed with 9 gm of water. The maximum oleum labelling possible of the sample formed can be:A. `118%`B. `109%`C. `105%`D. `103.8%` |
|
Answer» Correct Answer - D |
|
| 255. |
A sample of oleum is labelled as 112`%`. In 200gm of this sample, 18 gm water is added. The resulting solution will contain:A. 218 gm pure `H_(2)So_(4)`B. 218 gm `H_(2)SO_(4)` and 6 gm free `SO_(3)`C. 212 gm `H_(2)SO_(4)` and 6 gm free `SO_(3)`D. 191.33 gm `H_(2)SO_(4)` and 26.67 gm free `SO_(3)` |
|
Answer» Correct Answer - D |
|
| 256. |
100 gm oleum sample (labelled as 107.8%) is mixed with 7.8 gm water and requires, 1.1 L of x molar aq. Solution of `NaOH` for complete neutratization. The value of x is: |
|
Answer» Correct Answer - 2 |
|
| 257. |
A 10.0 g sample of a mixture of `CaCl_2` and `NaCl` is treated to precipitate all the calcium as calcium carbonate. Thus `CaCO_3` is heated to convert all the Ca to CaO and the final mass of CaO is 1.62 g. What is the percentage by mass of `CaCl_2` in the original mixture?A. 0.152B. 0.321C. 0.218D. 0.1107 |
|
Answer» Correct Answer - B |
|
| 258. |
100 gm oleum sample (labelled as X %) is mixed with excess water to make solution 4 litre. 1 L of this solution is neutralizes completely by `(1)/(3) M,1.8 L NaOH` solution. Calculate 10 X. |
|
Answer» Correct Answer - 1176 |
|
| 259. |
`10 g` of a sample of a mixture of `CaCl_(2)` and `NaCl` is treated to precipitate all the calcium as `CaCO_(3)`. This `CaCO_(3)` is heated to convert all the `Ca` to `CaO` and the final mass of `CaO` is 1.62 g.The percent by mass of `CaCl_(2)` in the origial mixture isA. 0.321B. 0.162C. 0.218D. 0.11 |
|
Answer» Correct Answer - A `CaCl_(2) + NaCl = 10 g` Let Weight of `CaCl_(2) = x g` `underset({:(1 "mol"),((x)/(111)"mol"):})(CaCl_(2)) rarr underset({:(1 "mol"),((x)/(111)"mol"):})(CaCO_(3)) rarr underset(1 "mol")(CaO)` Mole of `CaO = (1.62)/(56)` `:. (x)/(111) = (1.62)/(56)` `x = 3.21 g` % of `CaCl_(2) = (3.21)/(10) xx 100 = 32.1%` |
|
| 260. |
which of the following options temperature-dependent concentration term?A. ppmB. `%` w/wC. Volume strength of `H_(2)O_(2)`D. `%` labelling of oleum |
|
Answer» Correct Answer - A::C |
|
| 261. |
60 gm of oleum (labelled as 118`%`) is mixed with 11.8 gm of water. What will be the composition of final mixture?A. Only `H_(2)SO_(4)`, having mass 71.8 gmB. 118 gm of `H_(2)SO_(4)`C. 70.8 gm `H_(2)SO_(4)` and 1 gm waterD. 32 gm `SO_(3)` and 39.8 gm `H_(2)SO_(4)` |
|
Answer» Correct Answer - C |
|
| 262. |
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 1.12 g of CaO. Choose the correct statements. (Atomic weight :Ca=40=23, Cl =35.5)A. Mixture contains `50%` NaClB. Mixture contains 60% `CaCl_(2)`C. Mass of `CaCl_(2)` is 2.22 gD. Mass of `CaCl_(2)` 1.11 g |
|
Answer» Correct Answer - A::C |
|
| 263. |
0.6 mol of barium chloride in solution is mixed with 0.2 mol of sodium phosphate, the amount of barium phosphate produced is:A. 0.1molB. 0.3molC. 0.4molD. 0.5mol |
|
Answer» Correct Answer - A |
|
| 264. |
A piece of Al weighing 27 g is reacted with 200 ml of `H_(2)SO_(4)` (specific gravity=1.8 and 54.5% by weight). After the metal is completley dissolved, 73 gm `HCl` is added and solution is further is further diluted to 500 ml solution then find the concentration of `H^(+)` ion in `mol//litre.` |
|
Answer» Correct Answer - 6 |
|
| 265. |
A sample of a hydrate of barium chloride weighing 61g was heated until all the water of hydration is removed. The dried sample weighted 52g. The formular of the hydrated salt is: (Atomic mass, Ba = 137 amu, Cl = 35.5 amu)A. `BaCl_(2). 2H_(2)O`B. `BaCl_(2).4H_(2)O`C. `BaCl_(2).H_(2)O`D. `BaCl_(2).3H_(2)O` |
|
Answer» Correct Answer - A |
|
| 266. |
A solution with a mass of 1.263g containing an unkown amount of potassium ions was treated with excess sodium tetraphenylborate to precipitate 1.003g of KB `(C_(6)H_(5))_(4)` (M = 358.33). What is the mass percentage of potassium in the original solution?A. 0.0864B. 0.0916C. 0.109D. 0.138 |
|
Answer» Correct Answer - A |
|
| 267. |
Methyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation `C_(6)H_(5)COOH + CH_(3)OH rightarrow C_(6)H_(5)COOCH_(3) + H_(2)O` In an experiment 24.4gm of benzoic acid were reacted with 70.0mL of `CH_(3)OH`. The density of `CH_(3)OH` is 0.79 g `mL^(-1)`. The methyl benzoate produced had a mass of 21.6g. What is the percentage yeild of product?A. 0.917B. 0.794C. 0.715D. 0.217 |
|
Answer» Correct Answer - B |
|
| 268. |
In the reaction `2Al(s)+6HCl(aq)to6Cl^(-)(aq)+3H_2`A. 6 L HCl (aq) is consumed for every 3 L `H_2` producedB. 33.6 L `H_2(g)` is produced regardless temperature and pressure for every moles that reacts .C. 67.2 L `H_2(g)` at 1 atm , 273 K is produced for every mole Al that reacts .D. 11.2 L `H_2`(g) at 1 atm, 273 K is produced for every mole HCl(aq) consumed |
|
Answer» Correct Answer - D |
|
| 269. |
For the reaction : 2X + 3Y rightarrow 3Z, the combination of 2.00 moles of X with 2.00 moles of Y produces 1.75 moles of Z. What is the percent yeild of this reaction ?A. 0.438B. 0.583C. 0.667D. 0.875 |
|
Answer» Correct Answer - D |
|
| 270. |
Ammonia is produced in accordance with this equation . `N_(2)(g) + 3H_(2)(g) rightarrow 2NH_(3)(g)` In a particular experiment, 0.25mol of `NH_(3)` is formed when 0.5 mol of `N_(2)` is reacted with 0.5 mol of `H_(2)`. What is the percent yield?A. 0.75B. 0.5C. 0.33D. 0.25 |
|
Answer» Correct Answer - A |
|
| 271. |
When a mixture consisting of 10 moles of `SO_2` and 16 moles of `O_2` were passed over a catalyst , 8 mole of `SO_3` were formed at equilibrium.The number of moles of `SO_2` and `O_2` which did not enter into reation were :A. 2,12B. 12,2C. 3,10D. 10,3 |
|
Answer» Correct Answer - A |
|
| 272. |
Aluminium reacts with sulphur to form aluminium sulphide. If 31.9g of Al are reacted with 72.2 g of S, what is the theoretical yeild of aluminium sulphide in grams?A. 88.6gB. 69.7gC. 57.2gD. 113g |
|
Answer» Correct Answer - A |
|
| 273. |
The molar mass of an unkown organic liquid (M ~ 100) is determined by placing 5 mL of the liquid in a weighted 125mL conical flask with a piece of Al foil with a pin hole in it. The flask is heated in a Al foil with a pin hole in it. The flask is heated in a boiling water bath until the liquid evaporates to expel the air and fill the flsk with unknown vapour at atmospheric pressure. After cooling to vapour at atmospheric pressure. After cooling to room temperature the flask and its contents are room temperature the flask and its contents are reweighed. The uncertainty in which piece of apparatus causes the largest percentage error in the molar mass:A. balance ` (+-0.01g)`B. barometer ` (+-0.2mm Hg)`C. Flask` (+-1.0mL)`D. thermometer `(+-0.2^(@)C)` |
|
Answer» Correct Answer - A |
|
| 274. |
NX is produced by the following step of reactions `M + X_(2) rightarrow MX_(2)` `3MX_(2) + X_(2) rightarrow M_(3)X_(8)` `M_(3)X_(8) + N_(2)CO_(3) rightarrow NX + CO_(2) + M_(3)O_(4)` How much M (metal) is consumed to produce 206 gm of NX? (Take at. wt of M = 56, N=23, X = 80]A. 42gmB. 56gmC. 14/3 gmD. 7/4gm |
|
Answer» Correct Answer - A |
|
| 275. |
Ferric oxide can be obtained by oxidation of FeO: `4FeO + O_(2) rightarrow 2Fe_(2)O_(3)` The `O_(2)` gas required can be prepared by the following reaction. `2SO_(3) rightarrow 2SO_(2) + O_(2)(g)` What is the maximum amount of `Fe_(2)O_(3)` that can be produced by 144 g FeO and 160g of `SO_(3)`? [Atomic mass of Fe = 56]A. 320gB. 80gC. 120gD. 160g |
|
Answer» Correct Answer - D |
|
| 276. |
For a sequential reaction : A rightarrow B + C 2B rightarrow C + 2D If % yeild of (i) and (ii) reactions are 90% and 80% respectively, then the overall % yield is expected to be:A. 0.9B. 0.8C. 0.72D. 0.1 |
|
Answer» Correct Answer - C |
|
| 277. |
The number of alkoxy groups in an organic compound, `A(OR)_(x),` may be determined by the sequential reactions: `A(OR)_(x)+xHIrarrA(OH)_(x)+xRI` `RI+Ag^(+)+H_(2)OrarrROH+AgI(s)+H^(+)` When 4.8 gm of the organic compound, `A(OR)_(x),` (molar mass=240 gm//mol) is treated as above, 9.4 gm Agl is precipitated. The number of alkoxy groups in the compound is: |
|
Answer» Correct Answer - 2 |
|
| 278. |
For a sequential reaction. `NH_(3) rightarrow N_(2) + H_(2)` `H_(2) + O_(2) rightarrow H_(2)O` What will be the amount of water which will be obtained if 5 moles of `NH_(3)` is mixed with 3 moles of `O_(2)` and % yield of `1^(st)` and `2^(nd)` reaction is 50% and 80% respectively ?A. 3 molesB. 2.5 moleC. 2 moleD. 2.4 mole |
|
Answer» Correct Answer - A |
|
| 279. |
Three substances A,B and C can react to form D and E as shown : `2A + 3B + C rArr 4D + 2E` If molar masses of A,B, C and D are 40,30,20 and 15 respectively and 285 gm of mixture of A, B and C is reacted then maximum mass of E which can be obtained will be:A. 285 gmB. 200gmC. 195gmD. 100gm |
|
Answer» Correct Answer - C |
|
| 280. |
For the sequential reactions: `2A overset(80%)to 3B + C` `2B overset(50%)to 5D + E` Moles of A needed for the formation of 1.5 moles of D is:A. 0.6B. 0.4C. 1D. 1.16 |
|
Answer» Correct Answer - C |
|
| 281. |
Statement-1 : For reaction `2A(g)+3B(g)to4C(g)+D(g)` vapour density remains constant throughout the course of reaction. Statement-2 : In all gaseous chemical reactions vapour density reamain constant.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. |
|
Answer» Correct Answer - C |
|
| 282. |
The number of moles of C and D produced on mixing 5 moles of A and 7 moles of B are respectively: `(3A + 5B rArr 7C + 9D)`A. 9 moles and 11 molesB. 11.66 moles and 15 molesC. 9.8 moles and 12.6 molesD. 1 mole and 13 moles |
|
Answer» Correct Answer - C |
|
| 283. |
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio. (a) Is this statement true? (b) It yes, according to which law? (c) Give one example related to this law. |
|
Answer» (a) Yes, the statement is true (b) If is according to law of multiple proportions |
|
| 284. |
Calculate the mass per cent of calcium, phosphorus and oxygen in calcium phosphate `Ca_(3)(PO_(4))_(2)`. |
|
Answer» Molecular mass of calcium phosphate `Ca_(3)(PO_(4))_(2)` is : `3 xx` Atomic mass of `Ca + 2 xx` Atomic mass of `P + 8 xx` Atomic mass of O. `= (3xx40 u)+2(2xx31u)+(8xx16u)=310u` Mass percent of calcium `(Ca)=((12u))/((310u))xx100=38.71%` Mass percent of phosphorus `(P)=((62u))/((310u))xx100=20.0%` Mass percent of oxygen `(O)=((128u))/((310u))xx100=41.29%`. |
|
| 285. |
The charge on 1 gram ions of `AI^(3+)` is .(a). `(1)/(27) N_(A)e` coulomb(b). `(1)/(3) N_(A)e` coulomb(c). `(1)/(9) N_(A)e` coulomb(d). `3 xx N_(A)e` coulombA. `1/27N_(A^(e))` coulombB. `1/3N_(A^(e))` coulombC. `1/9N_(A^(e))` coulombD. `13 xx N_(A^(e)` coulomb |
|
Answer» Correct Answer - D |
|
| 286. |
Assertion : Relative atomic mass of boron is 10.8 Reason : Boron has two isotopes B-10 and B-11 with percentage abundance of 19.6 % and 80.4 % respectively.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
|
Answer» Correct Answer - A Reason is the correcr explanation for assertion `= (10xx19.6+11xx80.4)/(19.6+80.4)=10.80`. |
|
| 287. |
A gaseous mixture contains `CO_(2)`(g) and `N_(2)O(g)` in `2:5` ratio by mass. The ratio of the number of molecules of `CO_(2)(g)` and `N_(2)O(g)` is:A. `5:2`B. `2:5`C. `1:2`D. `5:4` |
|
Answer» Correct Answer - B |
|
| 288. |
Statement-1 : Boron has relative atomic mass 10.81. Statement-2 : Boron has two isotopes `._5^10B` and `._5^11B` and their relative abundance is `19%` and `81%`.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. |
|
Answer» Correct Answer - A |
|
| 289. |
Oxygen exists as three isotopes `._(8)^(16)O,_(8)^(17)O,_(18)^(18)O` with relative abundance `90%,7%` and `3%` repectively What is average atomic mass of oxygen? . |
| Answer» `M_(avg) = sum x_(i) M_(i) = 0.9 xx 16 + 0.07 xx 17 + 0.03 xx 18 = 16.13` amu or `16.13 g//mol` . | |
| 290. |
From 392 mg of `H_(2)SO_(4)`, `1.204 xx 10^(21)` molecules of `H_(2)SO_(4)` are removed. How many moles of `H_(2)SO_(4)` are left?A. `2 xx 10^(-3)`B. `1.2 xx 10^(-3)`C. `4 xx 10^(-3)`D. `1.5 xx 10^(-3)` |
|
Answer» Correct Answer - A |
|
| 291. |
`10^(21)` molecules are removed from 200 mg of `CO_(2)`. The moles of `CO_(2)` left are:A. `2.88 xx 10^(-3)`B. `28.2 xx 10^(-23)`gC. `1.5 xx 10^(-23)`gD. `2.5 xx 10^(-24)`g |
|
Answer» Correct Answer - A |
|
| 292. |
If `1//6`, in place of `1//12`, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one one of a substance will:A. be a function of the molecular mass of thesubstanceB. remain unchangedC. increase two foldD. decrease twice |
| Answer» Correct Answer - A | |
| 293. |
In the reaction: `2Al_((s))+6HCl_((aq.))rarr2Al_((aq.))^(3+)+6Cl_((aq.))^(-)+3H_(2(g))`A. `6LHCl_(aq)` is consumed for every `3LH_(2)(g)` producedB. `33.6 LH_(2)(g)` is produced regardless of temperature and pressure for every mole Al that reactsC. `67.2 H_(2)(g)` at `STP` is produced for every mole Al that reactsD. `11.2 L H_(2)(g)` at `STP` is produced for every mole `HCl_(aq)` consumed |
| Answer» Correct Answer - D | |
| 294. |
A sample of aluminium has a mass of `54.0g`. What is the mass of the same number of magnesium atoms? `(At. Wt. Al=27, Mg=24)`A. 12gB. 24gC. 48gD. 96g |
|
Answer» Correct Answer - C |
|
| 295. |
How many mole of magnesium phosphate `Mg_(3)(PO_(4))_(2)` will contain `0.25 "mole"` of oxygen atoms?A. `3.125xx10^(-2)`B. `1.25xx10^(-2)`C. `2.5xx10^(-2)`D. `0.02` |
| Answer» Correct Answer - A | |
| 296. |
How many mole of magnesium phosphate `Mg_(3)(PO_(4))_(2)` will contain `0.25 "mole"` of oxygen atoms?A. 0.02B. `3.125 xx 10^(-2)`C. `1.25 xx 10^(-3)`D. `2.5 xx 10^(-2)` |
|
Answer» Correct Answer - B |
|
| 297. |
How many moles of magnesium phosphate, `Mg_(3)(PO_(4))_(2)` will contain 0.25 mole of oxygen atoms ?A. 0.02B. `3.125 xx 10^(-2)`C. `1.25 xx 10^(-2)`D. `2.5 xx 10^(-2)` |
|
Answer» Correct Answer - B `underset("1 mole")(Mg_(3)(PO_(4))_(2))-=underset("8 mole")(8(O))` 8 mole of oxygen atoms are present in `Mg_(3)(PO_(4))_(2)=1` mole 0.25 mole of oxygen atoms are present in `Mg_(3)(PO_(4))_(2)` `=(1xx0.25)/(8)=3.125xx10^(-2)` mole. |
|
| 298. |
On dividing `0*25` by `22*1176`, the actual answer is `0*011303`. The correctly reported answer will beA. 0.011B. 0.01C. 0.0113D. 0.013 |
|
Answer» Correct Answer - A `0.25 div 22.1176=0.011303`. As the least precise term involved has two significant figure (vizm 0.25) therefore reported answer is 0.011. |
|
| 299. |
For an infinitely dilute aqueous solution molality will be equal to :A. formalityB. molarityC. mole fractionD. ppm |
| Answer» Correct Answer - B | |
| 300. |
In a gaseous reaction of the type `aA+bBrarrcC+dD`, which is wrong:A. a litre of A combines with b litre of B to give `C & D`B. a mole of `A` combines with `b` mole of `B` to give `C % D`C. `a g` of `A` combines with `b g` litre of `B` to give `C & D`D. a molecules of A combines with b molecule of B to give C & D |
| Answer» Correct Answer - C | |