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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.
| 601. |
What volume of `NH_(3)` gas at STP would be needed to prepare 100 ml of 2.5 molal (2.5 m) ammonium hydroxide solution?A. (a)0.056 litreB. (b)0.56 litreC. (c )5.6 litresD. (d)11.2 litres |
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Answer» Correct Answer - C `2.5` molal `NH_(4)OH` means `2.5` moles of `NH_(3)` in `1000 gH_(2)O (1000 c c "of solution")` Hence, `100 c c` solution of `NH_(3)` requires`=0.25` mole `=0.25xx2.4 L=5.6 L`. |
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| 602. |
The number of moles of sodium oxide in 620 g of it isA. 1 molB. 10 molesC. 18 molesD. 100 moles |
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Answer» Correct Answer - B Sodium oxide `rarr Na_(2)O` Molecular weight `=46+16=62` 62 gm of `Na_(2)O=1` mole 620 gm of `Na_(2)O=10` mole. |
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| 603. |
Choose the correct match/matchesA. 18 ml of `H_(2)O` at `4^(@)C - N_(A)` molecule of `H_(2)O`B. 11.2 L of `CO_(2)` at - `(N_(A))/(2)` molecule of `CO_(2) 273^(@)C` and 1 atmC. 56 g of `Fe^(57) - N_(A)` atomD. 5.6 L of `CH_(4)` at STP - 4 g weight |
| Answer» Correct Answer - A::D | |
| 604. |
11.2 L of `CH_(4)` and 22.4 L of `C_(2)H_(6)` at STP are mixed. Then choose correct statement/ statementsA. Vapour density of the mixture is 12.67B. Average molecular wt. will be less than 16C. Average molecular wt. will be greater than 16 and less than 30D. Average molecular weight will be greater than 30 |
| Answer» Correct Answer - A::C | |
| 605. |
Choose the correct statement regarding equivalent weightA. Equivalent weight of a substance always remain sameB. Equivalent weight of substance depends on reactionC. Equivalent weight may be greater than atomic weightD. Equivalent weight may be less than atomic weight |
| Answer» Correct Answer - B::C::D | |
| 606. |
From the following statements regarding `H_(2)O_(2)`, choose the incorrect statements:A. It can act only as an oxidising agentB. It decomposed on exposure to lightC. It has to be stored in plastic or wax lined glass bottles in darkD. It has to be kept away from dust |
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Answer» Correct Answer - A `H_(2)O_(2)` acts as an oxidising as well as reducing agent, because oxidation number of oxygen in `H_(2)O_(20` is `-1`. So, it can be oxidised to oxidation state 0 or reduced to oxidation state `-2`. `H_(2)O_(2)` decomposes on exposure to light. So, it has to be stored in plastic or was lined glass bottle in dark for the prevention of exposure. It also has to be kept away from dust. |
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| 607. |
Which of the following is not true?A. `51.028xx1.31`, reported product is `66.8`B. `4.327xx2.8`, reported product is `1.3xx10^(1)`C. `1.235` is rounded off to `1.23`D. `1.225` is rounded off to `1.22` |
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Answer» Correct Answer - D The reported quotient should have the minimum number of significant figures. Thus, it should be 0.03 |
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| 608. |
In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` isA. (molecular weight)/2B. (molecular weight)/6C. (molecular weight)/3D. same as molecular weight |
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Answer» Correct Answer - B The following reaction occur between `S_(2)O_(3)^(2-)` and `Cr_(2)O_(7)^(2-)` `26H^(+) + 3S_(2)O_(3)^(2-) + 4Cr_(2)O_(7)^(2-) rarr 6SO_(4)^(2-) + 8Cr^(3+) +13 H_(2)O` Change in oxidation number of `Cr_(2)O_(7)^(2-)` per formula unit is 6 (it is always fixed for `Cr_(2)O_(7)^(2-)`). Hence, equivalent weight of `K_(2)Cr_(2)O_(7) = ("Molecular weight")/(6)` |
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| 609. |
In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` isA. (a)`"molecular weight"//2`B. (b)`"molecular weight"//6`C. (c )`"molecular weight"//3`D. (d)same as molecular weight |
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Answer» Correct Answer - B In this reaction `Cr_(2)O_(3)^(2-)` is changed into `Cr_(3)^(3+)` ion. So, equivalent weight of `K_(2)Cr_(2)O_(7)=("molecular wt".)/("decrease of oxidation number"xxn)` `("where n= No. of atoms of" Cr)` `=("molecular wt".)/(3xx2)=("molecular wt.")/6` |
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| 610. |
Which of the following equation does not obey the law of conservation of mass?A. (a)`4H+O_(2) rarr 2H_(2)O`B. (b)`H_(2)+O rarr 2H_(2)O`C. (c )`2H+O_(2) rarr 2H_(2)O`D. (d)`Both (b) and (c )` |
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Answer» Correct Answer - a According to law of conservation of mass, in all chemical changes the total mass of the system remains constant or in a chemical reaction, mass is neither be created nor be destroyed. This law was proposed by Lavosier and tested by Landolt. But in (b) and (c ), mass of reactants is not equal to the mass of products. |
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| 611. |
Which of the following cannot give iodometric titrations?A. (a)`Fe^(3+)`B. (b)`Cu^(2+)`C. (c )`Pb^(2+)`D. (d)`Ag^(+)` |
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Answer» Correct Answer - b Atom in highest oxidation state can oxidize iodide to liberate `I_(2)` which is volumetrically measured by iodometric titration using hypo. `2I^(-) rarr I_(2)` `Pb^(+2) rarr` Lowest oxidation state can not oxidise iodide to `I_(2)`. |
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| 612. |
Which of the following cannot give iodometric titrations?A. (a)`Fe^(3+)`B. (b)`Cu^(3+)`C. (c )`Pb^(2+)`D. (d)`Ag^(+)` |
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Answer» Correct Answer - C Atom in highest oxidation state can oxidize iodide to liberate `I_(2)` which is volumetrically measured by iodometric titration using hypo. `2I^(-) rarr I_(2)` `Pb^(+2)` Lowest oxidation state cannot oxidise iodide to `I_(2)`. |
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| 613. |
Which of the following is not a matter?A. FacebookB. Pen driveC. LightD. Mobile Phone |
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Answer» Correct Answer - C It is a from of energy |
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| 614. |
The specific heat of a metal os `0.16` its approximate atomic weight would beA. (a)`32`B. (b)`16`C. (c )`40`D. (d)`64` |
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Answer» Correct Answer - C Sp. `"Heat"xx "atomic wt".=6.4` `0.16xx "atomic wt".=6.4` Atomic wt.`=6.4/0.16=40` |
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| 615. |
The specific heat of a metal os `0.16` its approximate atomic weight would beA. 32B. 16C. 40D. 64 |
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Answer» Correct Answer - C Sp. Heat `xx`atomic wt. `=6.4` `0.16 xx` atomic wt. `=6.4` Atomic wt,. `=6.4/0.16=40`. |
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| 616. |
In `SI` basse units, the amount fo substance is expressed inA. kgB. `L`C. `mol L^(-1)`D. mol |
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Answer» Correct Answer - D Mole is the `SI` unit of amount of substance, it is abberviated as mol. |
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| 617. |
Chromium is added to steel to impart strength and shine. If the density of chromium is `7.19 g//cm^(3)`. Express in SI unit `(kg//m^(3))`. |
| Answer» `7.19 g//cm^(3) = (7.19xx10^(-3)kg)/(10^(-6)m^(3))=7190 kg//m^(3)` | |
| 618. |
Express the following in SI units. (i) `-20^(@)C` (ii) 60 mg |
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Answer» (i) `T = t^(@)C + 273 K =-20 + 273=253 K` (ii) `1 mg = 10^(-3)g = 10^(-6)kg` So, `60 mg = 60xx10^(-6)kg =6.0xx10^(-5)kg` |
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| 619. |
The largest number of molecules inA. (a)`36 g` of waterB. (b)`28 g` of carbon monoxideC. (c )`46 g` of ethyl alcoholD. (d)`54 g` of nitrogen pentoxide |
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Answer» Correct Answer - A (a)`18 g` of `H_(2)O=6.02xx10^(23)` molecules of `H_(2)O` `:. 36 g` of `H_(2)O=2xx6.02xx10^(23)` molecules of `H_(2)O` `=12.04xx10^(23)` molecules of `H_(2)O` (b) `28 g` of `CO=6.02xx10^(23)` molecules of `CO` (c ) `46 g` of `C_(2)H_(5)OH=6.02xx10^(23)`molecules of `C_(2)H_(5)OH` (d) `108 g` of `N_(2)O_(5)=6.02xx10^(23)` molecuiles of `N_(2)O_(5)` `:. 54 g` of `N_(2)O_(5)=1/2xx6.02xx10^(23)` molecules of `N_(2)O_(5)` `=3.01xx10^(23)` molecules of `N_(2)O_(5)` `:. 36 g` of water has highest number of molecules. |
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| 620. |
One gram of oxygen is equal toA. `10^(5) mu g`B. `10^(6) mu g`C. `10^(4) mu g`D. `10^(3) mu g` |
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Answer» Correct Answer - B Micro `(mu)` is `10^(-6)`, thus, `10^(6) (10^(-6)) g = 1g`. |
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| 621. |
`20 g` of an acid furnishes `0.5 "mole of" H_(3)O^(+)` ions in its aqueous solution. The value of `1` equivalent of the acid will be:A. (a)`40 g`B. (b)`20 g`C. (c )`10 g`D. (d)`100 g` |
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Answer» Correct Answer - A `0.5 mol H_(3)O^(+)-=20 g` acid `1 mol H_(3)O^(+)-=40 g` acid |
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| 622. |
The number of atoms in `4.25 g NH_(3)` is approximately:A. `1.505 xx 10^(23)`B. `6.02 xx 10^(23)`C. `3.01 xx 10^(23)`D. None of these |
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Answer» a) `therefore` 17 g of `NH_(3)` contains = `6.023 xx 10^(23)` molecules of `NH_(3)` `therefore` 4.25 g of `NH_(3)` will contain = `(6.023 xx 10^(23)/(17) xx 425` `=1.505 xx 10^(23)` molecules of `NH_(3)` |
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| 623. |
The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1xx10^(23)`B. `1.5xx10^(23)`C. `2xx10^(23)`D. `6xx10^(23)` |
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Answer» Correct Answer - D Total no. of atoms in 4.25g of `NH_(3)` `= (4xx6.02xx10^(23))/(17)xx4.25 = 6xx10^(23)` |
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| 624. |
The number of atoms in 20 g of `SO_(3)` is approximatelyA. `1 xx 10^(23)`B. `1.5 xx 10^(23)`C. `2 xx 10^(23)`D. `6 xx 10^(23)` |
| Answer» Correct Answer - D | |
| 625. |
The number of oxygen atoms in 4.4 g of `CO_(2)` is approxA. `1.2xx10^(23)`B. `6xx10^(22)`C. `6xx10^(23)`D. `12xx10^(23)` |
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Answer» Correct Answer - A 44 g of `CO_(2)` has `2xx6xx10^(23)` atoms of oxygen 4.4 g of `CO_(2)` has `=(12xx10^(23))/44 xx4.4=1.2xx10^(23)` atoms. |
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| 626. |
Vapour density of a gas is 22. What is its molecular mass?A. (a)`33`B. (b)`22`C. (c )`44`D. (d)`11` |
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Answer» Correct Answer - c `MW=2xxV.D. =2xx22=44` |
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| 627. |
Assertion: Analytical molarity of `1 M HCl` is zero. Reason: Equilibrium molarity of `1M HCl` is zero.A. (a)If both assertion and reason are true and the reason is the true explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If the assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - d After dissociation `underset(0)underset(1)(HCl) rarr underset(1)underset(0)(H^(+))+underset(1)underset(0)(Cl^(-))` |
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| 628. |
When potassium permanganate is titrated against ferrous ammonoum sulphate, the equivalent weight of potassium permanganent isA. Molecular weight /10B. Molecular weight /5C. Molecular weight /2D. Molecular weight |
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Answer» Correct Answer - B `E=M/5` |
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| 629. |
When potassium permanganate is titrated against ferrous ammonoum sulphate, the equivalent weight of potassium permanganent isA. (a)`"Molecular weight"//10`B. (b)`"Molecular weight"//5`C. (c )`"Molecular weight"//2`D. (d)Molecular weight |
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Answer» Correct Answer - B `E=M/5` |
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| 630. |
Assertion: Molecular weight of oxygen is 16. Reason: Atomic weight of oxygen is 16.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D We known that molecular weight of substance is calculated by adding the atomic weight of atoms present in one molecules. We also know that molecular weight of oxygen `(O_(2))=2x` (Atomic weight of oxygen ) `=2xx16=32` a.m.u. Atoic weight of oxygen is 16, because it is 16 times heavier than `1//12^(th)` of carbon atom. Therefore assertion is false but reason is true. |
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| 631. |
A sample of `AIF_(3)` contains `3.0 xx 10^(24)` `F^(-)` ions. The number of formula units of the sample areA. `9.0 xx 10^(24)`B. `3.0 xx 10^(24)`C. `0.75 xx 10^(24)`D. `1.0 xx 10^(24)` |
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Answer» d) `therefore` `3F^(-)` = 1 Formula units of AI`F_(3)` `therefore` `3.0 xx 10^(24)F^(-)` = `1 xx 10^(24)` formula units of `AIF_(3)` |
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| 632. |
Which of the following is correct?A. When we are dealing with objects which can be counted, we always get an exact answer.B. Though the height of a person is an exact quantity, it is not possible to measure it exactly.C. Chairs are measured by a continous variable, height is measured by a discrete variableD. The measurement of a continous variable can only be as precise as the choice of the measuring apparatus but no matter what we do, some uncertainty always remains. |
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Answer» Correct Answer - C Chairs are measured by a discrete varible but height is measured by a continuous varibale. |
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| 633. |
A substance which participates readily in both acid-base and oxidation-reduction reactions is:A. (a)`Na_(2) CO_(3)`B. (b)`KOH`C. (c )`KMnO_(4)`D. (d)`H_(2) C_(2) O_(4)` |
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Answer» Correct Answer - D `H_(2)C_(2)O_(4)` readily participates in oxidation-reduction and acid-base reactions both (due to 2 acidic H and C in +3 oxidation state). |
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| 634. |
Which of the following is wrong?A. `1.234` is rounded off to `1.23`B. `1.236` is rounded off to `1.24`C. `1.235` is rounded off to `1.23`D. `1.225` is rounded off to `1.22` |
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Answer» Correct Answer - C When the numnber to be dropped is equal to 5, the preceding number to be dropped is equal to 5, the preceding number is increased by 1 if it is an odd number, thus, it should round off to 1.24. |
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| 635. |
Which of the following is wrong?A. `22.2 + 2.22 + 0.222`, reported sum is `24.6`B. `4.53 + 2.3 + 6.24`, reported sum is `13.063`C. `7.21 + 12.141 + 0.0028`, reported sum is `19.35`D. `3.74 - 0.0016`, reported difference is `3.74` |
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Answer» Correct Answer - B The reported sum should have least number of significant figures to the demical point. |
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| 636. |
Choose the wrong statementA. Molar mass is the mass of one moleculeB. Molar mass is the masss of one mole of substanceC. 1 mole means `6.023 xx 10^(23)` particlesD. Molar mass is the molecular mass (g) |
| Answer» a) molar mass is the mass of 1 mole or `6.022 xx 10^(23)` molecules of the substance | |
| 637. |
Forr 14 g of CO, the wrong statement isA. it occupies `2.24` L at NTPB. It correspends to `1//2` mole of COC. It corresponds ot same mole of CO and nitrogen gasD. it corresponds to `3.01xx 10 ^(23)` molecules of CO |
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Answer» Correct Answer - A 28 g of CO occuupies `= 22.4` L of CO at NTP `therefore` 14 g of CO occupies `=22.4/28 xx 14 ` L of CO at NTP `= 11, 2 ` L of CO at NTP Thus, 14 g of CO occupies `11.2` L volume at NTP. Thus, given statement is wrong. |
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| 638. |
The incorrect statement for 14 g of `CO` isA. It occupies 2.24 litre at NTPB. It corresponds to `1/2` mole of `CO`C. It corresponds to same mole of `CO` and `N_(2)`D. It corresponds to `3.01xx10^(23)` molecules of `CO` |
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Answer» Correct Answer - A 14 g of `CO=14/28=0.5` mole 0.5 mole `=3.01xx10^(23)` molecules The molecular weight of `CO=(15+16)=28` The molecular weight of `N_(2)=14+14=28` Therefore, the incorrect statement is that it occupies 2.24 litre at NTP. |
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| 639. |
Which of the following reactions is not correct according to the law of conservationof mas?A. 2Mg(s) + `O_(2)`(s) `to` 2MgO(s)`B. `C_(3)H_(8)(g) + O_(2) to CO_(2)(g) + H_(2)O(g)`C. `P_(4)(s) + 5O_(2)(g) to P_(4)O_(10)(s)`D. `CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(g)` |
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Answer» b) `C_(3)H_(8)(g) + O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(g)` is not correct. Correct equation is `C_(3)H_(8)(g) + 5O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(g)` |
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| 640. |
The number of significant figure in 60.001 areA. 5B. 6C. 3D. 2 |
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Answer» Correct Answer - B All the zeroes between two non zero digit are significant. Hence in 60.0001 significant figures is 6. |
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| 641. |
In which of the following numbers all zeroes are significant?A. 30000B. 0.7C. 0.0005D. 0.003 |
| Answer» a) 30,000 has all significant zeroes. | |
| 642. |
The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4g of `AgNO_3` isA. 28gB. 9.25gC. 2.870gD. 58g |
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Answer» `AgNO_3+NaCl to AgCl + NaNO_3` No. of moles of `AgNO_3=(3.4)/(170)=0.02` No. of moles of NaCl `=(5.85)/(58.5)=0.1` limiting reagent = `AgNO_3` 1 mole of `AgNO_3` produces 1 mole of AgCl 0.02 mole of `AgCO_3` produce 0.02 mole of AgCl Weight of AgCl produced = `0.02xx143.5=2.870 g` |
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| 643. |
The oxidation number of carbon in ` CH_2O` is.A. `-2`B. `+2`C. 0D. `+4` |
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Answer» Correct Answer - C `overset(**)(CH_(2))O` `x+2-2=0` `x=0`. |
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| 644. |
The ultimate products oxidation of most of hydrogen and carbon in food stuffs areA. `H_(2)O` aloneB. `CO_(2)` aloneC. `H_(2)O` and `CO_(2)`D. None of these |
| Answer» Correct Answer - C | |
| 645. |
the oxidation number of C in `CO_(2)` isA. `-2`B. `+2`C. `-4`D. `+4` |
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Answer» Correct Answer - D `overset(**)(C)O_(2)` `x+2(-2)=0, x-4=0, x=+4`. |
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| 646. |
Oxidation number of S in `SO_(4)^(2-)`A. `+6`B. `+3`C. `+2`D. `-2` |
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Answer» Correct Answer - A `x=8-2=+6` |
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| 647. |
Which one of the following statements is correct ?A. All elements are neterogenous in natureB. Compounds made up of a number of elements are heterogenousC. A mixture is always heterogenousD. Air is a homogenous mixture |
| Answer» Correct Answer - D | |
| 648. |
Which of the following is not correct regarding 14 g gram of carbon nonoxide?A. It corresponds to `0*5` mole of COB. It occupies `2*24` litres at S.T.P.C. It corresponds to `3*01xx10^(23)` molecules of COD. It corresponds to same number of moles of `CO_(2)` and nitrogen (I) oxide gases |
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Answer» Correct Answer - B (A) Mol mass of CO = 28 `:.` 14 g CO = 1/2 mol of CO (B) 28 g of CO at S.T.P. `= 22*4 L` 14 g of CO at S.T.P. `=11*2 L` (C) 14 g of CO `= (14)/(28) = 0*5` mol `= 11*2L` at S.T.P. `=3*01xx10^(23)` molecules. (D) Mol. mass of `CO_(2)` = Mol mass of `N_(2)O = 44` |
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| 649. |
`4*48` litre of methane at S.T.P. correspods toA. `1*2xx10^(22)` molecules of methaneB. `0*5` mole of methaneC. `3*2` g of methaneD. `0*1` mole of methane |
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Answer» Correct Answer - C `4*48` litre `CH_(4)=(4*48)/(22*4) = (1)/(5) = 0.2` mol `=16xx0*2=3*2gCH_(4)` `=0*2xx6*02xx10^(22)` molecules. |
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| 650. |
How many picometers `(pm)` are equivalent to `0.48 Å`?A. `84`B. `8.4`C. `8.40`D. `8400` |
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Answer» Correct Answer - A Use the relations `1 Å = 10^(-10)m` and `1 "pm" = 10^(-12) m`. Follow the sequence `Å rarr m rarr "pm"` `(0.84 Å) xx ((10^(-10) m)/(1 Å)) xx ((1 "pm")/(10^(-12) m)) = 84 pm` |
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