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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.
| 501. |
The mass of `Mg_(3)N_(2)` produced if 48 gm metal is reacted with 34 gm `NH_(3)` gas is Mg + `NH_(3)` `to` `Mg_(3)N_(2)` + `H_(2)`A. 200/3 gmB. 100/3 gmC. 400/3 gmD. 150/3 gm |
| Answer» Correct Answer - A | |
| 502. |
6.0 gm of gaseous hydrogen produce 17.6 gm `CO_(2)` on complete combustion. The mass percent of carbon in the hydrogen is :A. `40%`B. `50%`C. `60%`D. `80%` |
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Answer» Correct Answer - D |
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| 503. |
The mass of `P_(4)O_(10)` produced if 440 gm `P_(4)S_(3)` is mixed with 384 gm of `O_(2)` isA. 568gmB. 426 gmC. 284 gmD. 396 gm |
| Answer» Correct Answer - B | |
| 504. |
The mass of `CO_(2)` produced from 620 gm mixture of `C_(2)H_(4)O_(2)` & `O_(2)` prepared to produce maximum energy is ( combustion reaction is exothermic )A. 413.33 gmB. 593.04 gmC. 440 gmD. 320 gm |
| Answer» Correct Answer - C | |
| 505. |
Volume of `O_(2)` obtained at 2 atm & 546K, by the complete decomposition of 8.5 g `NaNO_(3)` isA. 2.24 lt.B. 1.12 lt.C. 0.84 lt.D. 0.56 lt. |
| Answer» Correct Answer - B | |
| 506. |
Which of the following expressions is correct ( `n = no`. of moles of the gas, `N_(A) =` Avogadro constant `m = mass` of 1 molecule of the gas, N = no of molecules of the gas) ? .A. `n = mN_(A)`B. `m = N_(A)`C. `N = nN_(A)`D. `m = n//N_(A)` |
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Answer» Correct Answer - C No. of molecule = Mole ` xx N_(A)` `N = nN_(A)` . |
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| 507. |
Assertion : Both 32 g of `SO_(2)` and 8g of `CH_(4)` have same number of molecules Reason : Equal moles of substances have equal number of molecules.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 |
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Answer» Correct Answer - A Reason is the correcr explanation for assertion. Both `SO_(2)` and `CH_(4)` represent 0.5 g mole have same number of molecules. |
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| 508. |
A 12 gm sample of metallic element M reacts completely with 0.02 mole of `X_(3)` to form MX. Find atomic mass of M. (Fill your answer divinding it by 100) |
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Answer» Correct Answer - 2 |
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| 509. |
Suppose a chemist has chosen `10^(20)` as the number of particles in a mole. What would be the molecular mass of oxygen gas ? |
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Answer» Correct Answer - `5.31 xx 10^(-3) g` `6.022 xx 10^(23)` molecules of oxygen weigh = 32 g 1.0 molecule of oxygen weighs `= (32)/(6.022xx10^(23))=5.31xx10^(-23)g` `:.` Mass of `10^(20)` molecules of oxygen `= 5.31 xx 10^(-23) xx 10^(20) = 5.31 xx 10^(-3)g`. |
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| 510. |
Which of the following contains the greatest number of atoms?A. 1.0g of butane `(C_(4)H_(10))`B. 1.0 g of nitrogen `(N_(2))`C. `1.0g` of silver (Ag)D. `1.0g` of water `(H_(2)O)` |
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Answer» Correct Answer - A |
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| 511. |
One gram of an alloy of aluminium and magnesium when heated with excess of dil. `HCI` forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at `0^(0)C` has a volume of `1.2` litre at `0.92 atm` pressure. Calculate the composition of the alloy. |
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Answer» Correct Answer - `Al=0.546g,Mg=0.454g` `Al+3HClrarrAlCl_(3)+(3)/(2)H_(2)uarr` `1.5x+y=0.04925" "...(I)` `x` mole `1.5x " "x+27+yxx24=1` `Mg+2HClrarrMgCl_(2)+H_(2)uarr` `x xx9+yxx8=0.33" "...(II)` `y` mole `% Al =(x xx27)/(1)xx100=54.6%` `Mg=45.4%` |
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| 512. |
Calculate the amount of `KClO_(3)` needed to supply sufficient oxygen for burning 112 L of CO gas at N.T.P. |
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Answer» The chemical equations for the above reactions are : `{:(" "2KClO_(3)overset("Heat")(rarr)2KCl+3O_(2)),(" "2CO+O_(2)rarr2CO_(2)"]"xx3),(bar(undersetunderset(=245g)(2xx(39+35.5+48))(2KClO_(3))+undersetunderset(134.4L)(6xx22.4)(6CO)rarr3KCl+6CO_(2))):}` 134.4 L of CO require `KClO_(3)=245` g 1.0 L of CO require `KClO_(3) = (245)/(134.4)g` 112 L of CO require `KClO_(3)=(245)/(134.4)xx112 = 204.167` g. |
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| 513. |
A quantity of aluminium has a mass of `54.0 g`. What is the mass of the same number of magnesium atoms ?.A. `12.1 g`B. `23.3 g`C. `48 g`D. `97.2 g` |
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Answer» Correct Answer - C Mole of `Al = ("wt")/("At wt")=(54)/(27)=2"mol"` So wol of `Mg = ("wt")/(24)` `"wt" = 2 xx 24 = 48 g` . |
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| 514. |
Which of the following contains greatest number of oxygen atoms ? .A. 1g of OB. 1g of O_(2)C. g of `O_(3)`D. all have the same number of atoms |
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Answer» Correct Answer - D No of oxygen atom = mole `xx N_(A) xx "atomicity"` (A) `= (1)/(16) xx N_(A) xx 1 = (N_(A))/(16)` (B) `= (1)/(32) xx N_(A) xx 2 = (N_(A))/(16)` (C) `= (1)/(48) xx N_(A) xx 3 = (N_(A))/(16)` all are same . |
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| 515. |
What volume of hydrogen at N.T.P would be liberated by the action of 50 mL of dilute `H_(2)SO_(4)` of 40 % purity and having a specific gravity of `1.3 g mL^(-1)` on 65 g of zinc ? (Atomic mass of Zn = 65). |
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Answer» Step I. Calculation of mass of pure `H_(2)SO_(4)` Mass of 50 mL of dilute `H_(2)SO_(4)` = Volume `xx` Specific gravity `= 50 mL xx (1.3 "g mL"^(-1))=65` g 100 g of `H_(2)SO_(4)` solution contain pure acid = 40 g 65 g of `H_(2)SO_(4)` solution contain pure acid `= (40)/(100) xx 65 g = 26 g`. Step II. Calculate of volume of hydrogen evolved at N.T.P The reaction involved is : `{:("Zn"+,H_(2)SO_(4),rarr,ZnSO_(4)+,H_(2)),(65g,2xx1+32+4xx16,,,22.4 L),(,=98g,,,"at N.T.P."):}` 98 g of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P. = 22.4 L 26 g of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P `= (22.4)/(98) xx 26g = 5.94 L`. |
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| 516. |
A 1000 gram sample of `NaOH` contains 3 mole of `O` atoms, what is the `%` purity of `NaOH` :A. `14%`B. `100%`C. `12%`D. `24%` |
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Answer» Correct Answer - C `NaOH` contain 3 mole of `O` atoms so mole of `NaOH = 3 mol` wt of `NaOH = 3 xx 40 = 120g` `%` purity `= (120)/(1000) xx 100 = 12%` . |
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| 517. |
What volume of hydrogen at N.T.P would be liberated by the action of 50 mL of dilute `H_(2)SO_(4)` of 40 % purity and having a specific gravity of `1.3 g mL^(-1)` on 65 g of zinc ? (Atomic mass of Zn = 65) ? |
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Answer» Correct Answer - 5.94 L Mass of 50 mL of dilute `H_(2)SO_(4)` = volume `xx` density `=(50 mL )xx ("1.3 g mL"^(-1))=65g` Mass of pure `H_(2)SO_(4)` in the sample `= ((65g)xx40)/(100)=26g` The balanced chemical equation is : `Zn+underset(underset(98g)("1 mol"))(H_(2)SO_(4))rarrZnSO_(4)+underset(underset(22.4L)("1 mol"))(H_(2))` 98 g of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P = 22.4 L 26 g `H_(2)SO_(4)` evolve `H_(2)` at N.T.P `= ((22.4L))/((98.0g))xx(26.0g)=5.94L`. |
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| 518. |
A polystryrene, having formula `Br_(3) C_(6) H_(2) (C_(8) H_(8))_(n)`, was perpared heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 1.46% bromine by weight, find the value of `n`. |
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Answer» Let the weigth of polystrene prepared be `100 g` Therefore, the number of moles of `Br` in `100 g` of polystyrene `(10.46)/(79.9) = 0.1309 mol` From the formula of polystyrene, we have, Number of mol `Br = 3 xx mol of Br_(3) C_(6) H_(2) (C_(8) H_(8))_(n)` `0.139 = 3 xx ("Weight")/("Molecular Weight")` `0.139 = (3 xx 100)/(314 + 104 n)` `n = 19` |
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| 519. |
Is law of conservation of mass always valid ? |
| Answer» No, its is not valid for nuclear reactions. In these reactions, a certain amount of mass gets converted into energy known as nuclear energy. Therefore, mass is not conserved and it does not remain constant. | |
| 520. |
1120 ml of ozonised oxygen `(O_2+O_3)` at 1 atm and passing the mixture through alkaline pyrogallol solution is :A. 896 mlB. 224 mlC. 448 mlD. 672 ml |
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Answer» Correct Answer - A |
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| 521. |
`H_(2)O_(2)` solution of 45.4 V at STP . In a hospital of Kota, patient under artifical respiration take s200ml `O_(2)` per min at 1 atm and 273 K and a cylinder last for 2.8 hours . After that it cannot be used for respiration though it still contains `H_(2)O_(2)` . [Assume volume of solution and rate of decomposition remain constant] Volume of oxygen used for respiration is :A. 11.2 LB. 22.4 LC. 33.6 LD. 5.6 L |
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Answer» Correct Answer - C |
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| 522. |
`H_(2)O_(2)` solution of 45.4 V at STP . In a hospital of Kota, patient under artifical respiration take s200ml `O_(2)` per min at 1 atm and 273 K and a cylinder last for 2.8 hours . After that it cannot be used for respiration though it still contains `H_(2)O_(2)` . [Assume volume of solution and rate of decomposition remain constant] Volume strenght of `H_(2)O_(2)` left in solution is:A. 11.35 VB. 22.7 VC. 5.67 VD. 34.05 V |
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Answer» Correct Answer - D |
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| 523. |
How many `g` atoms are there in one atoms? |
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Answer» `6.023 xx 10^(23) "atoms" = 1 g "atom" = 1` mole `1 "atom" (1)/(6.023 xx 10^(23))` `= 1.66 xx 10^(-24) g "atom or mol"` |
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| 524. |
When a mixture of 10 moles of `SO_(2)` and 15 moles of `O_(2)` was passed over catalyst, 8 moles of `SO_(3)` was formed. How many moles of `SO_(2)` and `O_(2)` did not enter into combination? |
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Answer» `underset("2 mol")(2SO_(2))+underset("1 mol")(O_(2))rarrunderset("2 mol")(2SO_(3))` 2 moles of `SO_(3)` are formed from `SO_(2)=2` mol 8 moles of `SO_(3)` are formed from `SO_(2)=8` mol `:.` Moles of `SO_(2)` left unreacted `= (10-8)=2 mol` 2 moles of `SO_(3)` are formed from `O_(2)=1` mol 8 moles of `SO_(3)` are formed from `O_(2)=4` mol `:.` Moles of `O_(2)` left unreacted `= 15 - 4 = 11 mol`. |
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| 525. |
Which combustion prouduct is produced THE LEAST by gasoline-powered vehicles ?A. `CO_2`B. `H_2O`C. `NO_2`D. `SO_2` |
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Answer» Correct Answer - D |
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| 526. |
A compound has the molecular formula `X_(4)O_(6)`. If `10 g "of" X_(4)O_(6)` has `5.72 g X`, atomic mass of `X` is:A. 32amuB. 37 amuC. 42 amuD. 98amu |
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Answer» Correct Answer - A |
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| 527. |
A compound with the formula `X_(2)O_(5)` contains `34.8%` oxygen by mass. Identify element X.A. ArsenicB. CarbonC. PhosphorousD. Samarium |
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Answer» Correct Answer - A |
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| 528. |
A 1.50g sample of an ore containing silver was dissolved, and all of the `Ag^(+)` was converted to 0.124 g of `Ag_(2)S`. What was the percentage of silver in the ore?A. 0.0641B. 0.072C. 0.0827D. 0.108 |
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Answer» Correct Answer - B |
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| 529. |
84 gm of iron (Fe) is required with sufficient amount of steam of produce 45.4 L, `H_(2) ` gas at S.T.P. according the following reaction, `aFe + bH_(2)O rArr c Fe_(3)O_(4) + d H_(2)`.The stoichimetric coefficients of the reaction is (At. Wt., Fe = 56, O=16, H=1):A. 4,3,1,4B. 3,4,1,4C. 1,4,2,3D. none of these |
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Answer» Correct Answer - B |
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| 530. |
The number of moles of oxygen obtained by the electrolytic decomposition of 90g water is :A. 2.5B. 5C. 7.5D. 10 |
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Answer» Correct Answer - A |
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| 531. |
What volume of `0.108` M `H_(2)SO_(4)` is required to neutralize `25.0` mL of `0.145` M KOH?A. `16.8` mLB. `33.6` mLC. `37.2` mLD. `67.1` mL |
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Answer» Correct Answer - A |
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| 532. |
What is mass of `O_(2)` required to produce `960gm` of `O_(3)` if `%` yield of reaction `30_(2)rarr 2O_(3)` is `50%` ? . |
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Answer» `3O_(2)rarr 2O_(3)` `(960)/(48) = 20mol` = Actual yield Let `X` mol `O_(2)` be required Theoretical yield of `O_(3)((2)/(3)X)rArr50((2)/(2/(3)X))(100)` `X = 60 mol = 60 xx 32 = 1920 g O_(2)` required . |
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| 533. |
What is mass of `C` obtained on reacting `20` moles of `A` with excess `B` by reaction `{:(2A,+,B,rarr,3C):}` If `%` yield of reaction is `80%` ? . |
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Answer» `{:(2A,+,B,rarr,3C,),(20mol,,,,30mol="Theoreticalyield",),(,,,,"Actualyield=Xmol"=?,):}` `rArr (X)/(30) xx 100 = 80 rArr X = 24"moles"` . |
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| 534. |
A `0.534 g` of a sample of haemoglobin on analysis was found to contain `0.34% Fe`. If each haemoglobin molecule has four `Fe^(2+)` ions, what is the molecular mass of haemoglobin ? `(Fe = 56 "amu")` |
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Answer» `100 g` of haemoglobin contains `= 0.34 g of Fe` 1 mol of haemoglobin contains `= Fe^(2+)` ions `= 4 xx 56 g = 224 g` `:. 0.34 g of Fe` is present in `100 g` of Haemoglobin `224 g of Fe` is present in `= (100 xx 224)/(0.34) = 65882.4 g` `Mw` of haemoglobin `= 65882.4 g` |
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| 535. |
`0.492g` sample of haemoglobin has `0.34%` by mass of Fe. If each molecule of haemoglobin has `4` Fe atoms find its molecular mass ? . |
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Answer» Let `M` = Molecular mass of haemoglobin `rArr (0.34)/(100) xx M = 4 xx 56` `M = 65882.3` amu . |
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| 536. |
A certain metal M forms an insoluble oxalate complex `M_(4)O_(3)(C_(2)O_(4))_(3).12 H_(2)O`. If `2.38` gm of the complex are formed from 1 gm of oxalic acid `(H_(2)C_(2)O_(4))`, what is the atomic weight of M ? [Write nearest integral valve]. |
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Answer» Correct Answer - 96 |
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| 537. |
`Pb(NO_(3))_(2)` and `KI` reacts in aqueous solution to form an yellow precipitate of `PbI_(2)`. In one series of experiments, the masses of two reactants varied, but the total mass of the two was held constant at `5.0g`. What maximum mass of `PbI_(2)` can be produced in the above experiment : |
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Answer» Correct Answer - `3.464g` In order to obtain maximum yield from a reaction, the reactants must be supplied in stoichiometric amount so that no reactant should be left unreacted. The balanced chemical reaction is, `Pb(NO_(3))_(2)+2KIrarrPbI_(2)+2KNO_(3)` Let `x g` of `KI` is taken implies moles of `KI=(x)/(166) `impliesmoles of `Pb(NO_(3))_(2)` present `=(x)/(2xx166)` `implies (x)/(2xx166)=(5-x)/(330)implies x=2.5 g` implies mass of `PbI_(2)` `=(x)/(332)xx460=3.464 g` |
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| 538. |
The hydrated salt `Na_(2)CO_(3)x H_(2)O` undergoes 63 % loss in mass on heating and becomes anhydrous. What is the value of x ? |
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Answer» Correct Answer - 10 Molecular mass of `Na_(2)CO_(2)X.H_(2)O=106+18x` Loss in mass due to dehydration = 63 % `(18x xx 100)/(106+18x)=63or(18x)/((106+18x))=0.63` `18x=0.63(100+18x)=66.78+11.34x` `(18 x - 11.34 x)=6.66 x = 66.78` `x=(66.78)/(6.66)~~10`. |
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| 539. |
What is the percentage of water of crystallisation in a pure sample of washing soda `(Na_(2)CO_(3).10 H_(2)O)` ? |
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Answer» Correct Answer - 0.6294 Molecular mass of `Na_(2)CO_(3).10H_(2)O=46+12+48+180=286.0u` Percentage of water of crystallisation `= ((180u))/((286u))xx100=62.94%`. |
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| 540. |
In an organic compound of molar mass greater than `100` containing only `C`, `H` and `N`, the percentage of `C` is `6` times the percentage of `H` while the sun of the percentage of `C` and H is `1.5` times the percentage of `N`. What is the least molar mass :A. `175`B. `140`C. `105`D. `210` |
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Answer» Correct Answer - B Let assume `%` of `H` is `x` `%` of `H = x` `%` of `C = 6x` `%` of `N=(7x)/(1.5)` `{:(,"Element",%,"Ratio of mol","Simplest"),(,H,x,x//1=1,6),(,C,6x,6x//12=1//2,3),(,N,(7x)/(1.5),(7x)/(1.5xx14)=(1)/(3),2),(,because,F.F=C_(3)H_(6)N_(2),,):}` atomic mass `=70` molar mass `=140` |
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| 541. |
A reacts by following two parallel reactions to give B and C. If half of A goes into reaction I and other half goes to reaction-II, then select the correct statement(s) `A + N overset(I)rarr B+L` `A+Noverset(II)rarr (1)/(2)B+(1)/(2)(C) +L`A. B will be always greater than CB. If 2 moles of C are formed then total 2 moles of B are also formedC. If 2 moles of C are formed then total 4 moles of B are also formedD. If 2 moles of C are formed then total 6 moles of B are also formed |
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Answer» Correct Answer - A::D |
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| 542. |
The mass of `560 cm^3` of a gas at `0^@C` and 1 atm is 1.60 g. Which gas could it be ?A. `O_2`B. `CO_2`C. `SO_2`D. `Cl_2` |
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Answer» Correct Answer - C |
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| 543. |
A certain organic substance used as a solvent in many reactions contains carbon, hydrogen, oxygen and sulphur. Weight % of hydrogen in the compound is 7.7. The weight ratio `C : O : S = 3:2:4`. What is the least possible molar mas of the compound?A. 86B. 63C. 94D. 78 |
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Answer» Correct Answer - D |
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| 544. |
If equal mass of following substance are taken the which will have maximum number of molecles.A. `C_(6)H_(12)O_(6)`B. `C_(12)H_(22)O_(11)`C. `C_(2)H_(6)`D. `CO_(2)` |
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Answer» Correct Answer - C |
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| 545. |
An antifreeze mixture consists of `40%` ethylene glycol `(C_(2)H_(6)O_(2))` by weight in aqueous solution. If the density of this solution is `1.05 g//"mol"`, what is the molar concentration :A. `6.77 M`B. `6.45 M`C. `0.017 M`D. `16.9 M` |
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Answer» Correct Answer - A Molarity `=(40xx1.05xx10)/(62)=6.77M` |
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| 546. |
(a) A solution is prepared by dissolving `3.65 g` of `HCl` in 500 mL of the solution. Calculate the normality of the solution (b) Calculate the volume of this solution required to prepare 250 mL of 0.05 N solution. |
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Answer» Step I. Calculation of normality of solution Mass of `HCl = 3.65 g` Molar mass `HCl = 1+35.5 = 36.5 u = 36.5 "g mol"^(-1)` Basicity of `HCl` from the formula =1 `:.` Equivalent mass `HCl=("Molar mass")/("Bascity")=((36.5"g mol"^(-1)))/(1)=(("3.65 g"))/(("36.5 g equiv"^(-1)))=0.1 "equiv"` Volume of solution in litres `= 500 mL = (500)/(1000)=0.5 L` Normality of solution `(N)=("No. of equivalent of HCl")/("Volume of solution in litres")=("0.1 equiv.")/(("0.5 L"))="0.2 equiv L"^(-1)=0.2N` Step II. Calculation of volume of concentrated `HCl` solution required Normality of conc. `HCl = (N_(1))=0.2 N` Let the volume of conc. HCl needed `= V_(1)` Normality of dilute `HCl (N_(2))=0.05 N` Volume of dilute `HCl (V_(2))=250 mL` Applying normality equation : `N_(1)V_(1) -= N_(2)V_(2)` `(0.2 N)xxV_(1) = (0.05 N) xx ("250 mL")` `V_(1)=((0.05N)xx(250 mL))/((0.2 N))=62.5mL`. |
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| 547. |
When `2.86 g` of a mixture of `1-`butene, `C_(4)H_(8)` and butane `C_(4)H_(10)` was burned in excess of oxygen `8.80 g` of `CO_(2)` and `4:14 g` of `H_(2)O` were obtained. What is percentage by mass of butane in the mixture |
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Answer» Correct Answer - `60.8%` `{:(C_(4)H_(8)+6O_(2)rarr4CO_(2)+4H_(2)O),(x" "4x" "5x),(C_(4)H_(10)+(13)/(2)O_(2)rarr4CO_(2)+5H_(2)O),(y" "4y" "5y),((4x+5y)xx44=8.8),(x+y=0.05" "....(I)),((4x+5y)xx18=4.14),(4x+5y=0.23" "....(II)),(y=0.03):}` `%` by mass of `C_(6)H_(10)` `=(0.03xx58)/(2.86)xx100=60.8%` |
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| 548. |
For the complete combustion of 4 litre ethane, how much oxygen is required ?A. 14 litreB. 4 litreC. 8 litreD. 12 litre |
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Answer» Correct Answer - A |
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| 549. |
What volume of liquid `A_(2)O_(3)` has same number of atoms as there are atom in `BO_(2)`(I) having volume 20 Ml? [Given: Density of `A_(2)O_(3)=1.5 gm//ml` and density of `BO_(2)=0.7` gm/ml, Atomic mass of `A=50,` Atomic mass of `B=60` and O represents oxygen] |
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Answer» Correct Answer - 9 |
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| 550. |
Which of the following solutions will have maximum amount of NaOH?A. 4 L of `0.1` M NaOH solutionB. 2 L 0f `5%w//v` NaOH solutionC. 540 gm of 2m NaOH solutionD. 300 gm of `20%w//w` NaOH solution |
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Answer» Correct Answer - B |
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