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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.
| 701. |
In reaction of hydrogen peroxide and sodium carbonate, `H_2O_2` acts as____.A. Oxidising agentB. Reducing agentC. Bleaching agentD. Both oxidising and bleaching agent |
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Answer» Correct Answer - B `H_(2)O_(2)` acts as a reducing agent in the reaction between `O_(3)` and `H_(2)O_(2)` |
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| 702. |
Which of the following sets of compounds correctly ltbr. Illustrate the law of reciprocal proportions?A. `P_(2)O_(3),PH_(3), H_(2)O`B. `P_(2)O_(5),PH_(3), H_(2)O`C. `N_(2) O_(5),NH_(3), H_(2)O`D. `N_(2) O,NH_(3), H_(2)O` |
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Answer» Correct Answer - A `P_(2) O_(3), PH_(3) and H_(2) O ` illustrates the law of reciprocal proportions, Ratio in the number of atoms of H and O combinig with one P is 3, i.e., `2 : 1`. |
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| 703. |
The percentage of hydrogen in water and hydrogen peroxide is 11 .1 and 5.9 respectively . These figures illustrateA. Law of multiple proportionsB. Law of conservation of massC. Law of constant proportionsD. Law of combining volumes |
| Answer» Correct Answer - A | |
| 704. |
An element forms two oxides containing respectively `53.33` and `36.36` percent of oxygen. These figures illustrate the law ofA. (a)Conservation of massB. (b)Constant proportionsC. (c )Reciprocal proportionsD. (d)Multiple proportions |
| Answer» Correct Answer - D | |
| 705. |
Number of hydrogen ions present in 10 millionth part of `1.33 cm^(3)` of pure water at `25^(@)C` isA. 6.023 millionB. 60 millionC. 8.01 millionD. 80.23 million |
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Answer» Correct Answer - C Now `[H^(+)]=10^(-7)` mole/litre Now 1000 ml contains `10^(-7)` mole `H^(+)` 1 ml contains `10^(-7)/1000` mole `H^(+)` `1.33xx10^(-7)` ml - contains `1.33xx10^(-17)` 10 million `=10^(-7)` so, `10 "million"^(th)` part of `1.33 cm^(3)=1.33xx10^(-7)` ml so, no. of `H^(+)` ions `=1.33xx10^(-17)xxN_(A)`. |
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| 706. |
The mass of `CaCO_(3)` produced when carbon dioxide is bubbled through 500 mL of `0*5 M` `Ca(OH)_(2)` will beA. 10 gB. 20 gC. 50 gD. 25 g |
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Answer» Correct Answer - D `Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O` `Ca(OH)_(2) = 0*5xx500xx10^(-3)` moles `=0*25` mol Now 1 mole `Ca(OH)_(2)` gives `CaCO_(3) = 100g` `0*25` mole `Ca(OH)_(2)` gives `CaCO_(3)` `=0*25xx100=25g` |
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| 707. |
If `CO_(2)` gas is passed through 500 ml of `0.5 (M) Ca(OH)_(2)`, the amount of `CaCO_(3)` produced isA. 10 gB. 20 gC. 50 gD. 25 g |
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Answer» Correct Answer - D By equating the equivalents `W/100xx2=(500xx0.5xx2)/1000` `W=25 g" "(W="mass of "CaCO_(3))` |
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| 708. |
If 500mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?A. 1.5 MB. 1.66 MC. 0.017 MD. 1.59 M |
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Answer» `M_1V_1 = M_2V_2` `5Mxx500mL = M_2xx1500mL` `therefore M_2=(5xx500)/(1500)=1.66M` |
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| 709. |
The empirical formula of an acid is `CH_(2)O_(2)`, the probable molecular formula of acid may beA. (a)`CH_(2)O`B. (b)`CH_(2)O_(2)`C. (c )`C_(2)H_(4)O_(2)`D. (d)`C_(3)H_(6)O_(4)` |
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Answer» Correct Answer - c Empirical formula of an acid is `CH_(2)O_(2)` `("Empirical formula")_(n)` = Molecular formula n= whole no. multiple i.e. 1, 2, 3, 4. If n=1 molecular formula `CH_(2)O_(2)`. |
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| 710. |
The empirical formula of an acid is `CH_(2)O_(2)`, the probable molecular formula of acid may beA. `CH_(2)O`B. `CH_(2)O_(2)`C. `C_(2)H_(4)O_(2)`D. `C_(4)H_(6)O_(4)` |
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Answer» Correct Answer - B M. Formula = n(E. Formula) where n = 1,2,3, etc |
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| 711. |
How many meteres are equivalent to `5.00` in?A. `12.7 m`B. `127m`C. `0.127m`D. `1.27m` |
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Answer» Correct Answer - C Use the relations 1 in `= 2.54 cm` and `4 cm = 10^(-2) m`. Follow the sequence in `rarr cm rarr m` `(5.00 "in".) xx ((2.54 cm)/(1 "in".)) xx ((1xx10^(-2)m)/(1 cm))` |
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| 712. |
The density of a gas A is thrice that of a gas B at the same temperature. The molecular weight of gas B is twice that of A. What will be the ratio of the pressure acting on B and A?A. (a)`1/4`B. (b)`7/8`C. (c )`2/5`D. (d)`1/6` |
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Answer» Correct Answer - d `d=(PM)/(RT)` `implies d_(A)/d_(B)=(P_(A)M_(A))/(P_(B)M_(B))` T is same `d_(A)=3d_(B)` `M_(B)=2M_(A)` or `M_(A)=M_(B)/2` `implies (3d_(B))/d_(B)=P_(A)/P_(B)xxM_(B)/(2xxM_(B))` `P_(B)/P_(A)=1/6`. |
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| 713. |
Vapour density of a gas is 22. What is its molecular mass?A. `33`B. `22`C. `44`D. `11` |
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Answer» Correct Answer - C M. Mass `= 2xxV.D. = 2xx22 = 44` |
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| 714. |
The vapour density of gas A is four times that of B. If moelcular mass of B is M, then molecular mass of A isA. MB. 4 MC. `(M)/(4)`D. 2 M |
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Answer» Correct Answer - B Molecular mass of B = M V.D. of B = M/2 V.D. of A `= 4xxV.D.` of B `= 4xxM//2=2M` Molecular mass of `A = 2xx2M = 4M` |
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| 715. |
The vapour density of a gas A is twice that of a gas B. If the molecular weight of B is M, the molecular weight of A will be:A. (a)MB. (b)2MC. (c )3MD. (d)M/2 |
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Answer» Correct Answer - B `("Vapour density of A")/("Vapour density of B")=("molar mass of A")/("molar mass of B")` Molar mass of `A=2xxM/1` |
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| 716. |
How many grams are equivalent to `66 Ib` (pound) of sulphur?A. `20000`B. `30000`C. `40000`D. `25000` |
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Answer» Correct Answer - B Use the relations 1 pound `= 0.454 kg` and `1 kg = 1000g`. Follow the sequence `lb rarr kg rarr g` `(66 Ib) ((0.454 kg)/(1 lb)) ((1000 g)/(1 kg))` |
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| 717. |
The total charge (coulombs) required for complete electrolysis isA. 24125B. 48250C. 96500D. 193000 |
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Answer» `Na^(+) + e^(-) to Na` Moles of `Na^+` discharged at cathode =2 `therefore` The no. of electrons required for this purpuse = 2 moles `therefore` Total charge required = 2 faradays `=2xx96500` =193000 coulombs. |
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| 718. |
The charge in coulombs on 1 g ion of `N^(3-)` isA. (a)`5.2xx10^(6) C`B. (b)`2.894xx10^(5) C`C. (c )`6.6xx10^(6) C`D. (d)`8.2xx10^(6) C` |
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Answer» Correct Answer - B One ion carries `3xx1.6xx10^(-19)` coulombs Then `1 g` ion `N^(3-)(1 "mole")` carries `=3xx1.6xx10^(-19)xx6.2xx10^(23)=2.89xx10^(5)` coulombs |
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| 719. |
The speed limit on a highway is 55 mile `h^(-1)`. Express this speed in `SI` base units.A. `25 ms^(-1)`B. `35 ms^(-1)`C. `20 ms^(-1)`D. `30 ms^(-1)` |
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Answer» Correct Answer - A Use the relations `1 ft = 12 In`. `1 "mile" = 5280 ft, 1 "in" = 2.54xx10^(-2) m`, and `1 h = 3600 s` `"mile" rarr ft rarr "in" rarr m` and `h rarr s` `(55 ("mile")/(h)) ((580 ft)/(1 "mile")) ((12 "in.")/(1 ft.)) ((2.54xx10^(-2) m)/(1 "in")) ((1h)/(3600 s))` `= 25 m s^(-1)` |
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| 720. |
The law of conservation of mass is not obeyed by aA. redox reactionB. double decomposition reactionC. nuclear reactionD. neutralization reaction |
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Answer» Correct Answer - C Because a nuclear reaction is not a chemical reaction. |
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| 721. |
The atomic weights of two alements A and B are 40 and 80 reapectively. If x g of A contains y atoms, how many atoms are present in 2x g of B?A. (a)`y/2`B. (b)`y/4`C. (c )yD. (d)2y |
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Answer» Correct Answer - C `{:(,,A,,B),("Atomic mass",,40,,80),("given weight",,x "gram",,2x "gram"),("No. of mole",,x/40,,(2x)/80),("No. of atom",,x/40xxN_(A),,x/40xxN_(A)),("But according to question",=,x/40xxN_(A)=y,,):}` |
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| 722. |
Formation of CO and `CO_(2)` illustrates the law ofA. Conservation of massB. multiple proportionC. reciprocal proportionD. constant proportion |
| Answer» b) Formation of CO and `CO_(2)` illustrates the law of multiple proportion. | |
| 723. |
A purified cytochrome protein was found to contain 0.376 % iron. What is the minimum molecular mass of the protein?A. `14,800 u`B. `1480 u`C. `148,000 u`D. `148 u` |
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Answer» Correct Answer - A Mass of iron (minor component) per gram of protien `=(0.376)/(100)=0.00376g` Atomic mass of iron = 56 u Minimum molecular mass `=("Atomic mass of Fe")/("Mass of Fe per gram of macromolecule")` `(56)/(0.00376)=14,800 u` |
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| 724. |
Oxidation number of oxygen in ozone `(Q_(3))` isA. `+3`B. `-3`C. `-2`D. 0 |
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Answer» Correct Answer - D Molecule and free atoms show zero oxidation state `O_(3)` is a molecule shows zero oxidation state. |
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| 725. |
Oxidation state of oxygen inhydrogen peroxide isA. `-1`B. `+1`C. `0`D. `-2` |
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Answer» Correct Answer - A In all peroxide shows -1 oxidation state. |
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| 726. |
Oxygen has an oxidation state of +2 inA. `H_(2)O_(2)`B. `CO_(2)`C. `H_(2)O`D. `OF_(2)` |
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Answer» Correct Answer - D Oxygen have +2 oxidation state in `OF_(2)`. |
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| 727. |
The sum of the oxidation numbers of all the carbons in `C_(6)H_(5)CHO` isA. `+2`B. 0C. `+4`D. `-4` |
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Answer» Correct Answer - D In benzaldehyde all carbon atoms show -4 oxidation state. |
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| 728. |
The number of mole of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:A. (a)`0.6` moleB. (b)`0.4` moleC. (c )`7.5` moleD. (d)`0.2` mole |
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Answer» Correct Answer - a In ferrous oxalate ferrous and oxalate ions are oxidized. `Fe^(2+) rarr Fe^(3+)+e^(-)` `C_(2)O_(4)^(2-) rarr 2CO_(2)+2e^(-)` `MnO_(4)^(2-) rarr Mn^(2+)` Thus eqn. will be `3MnO_(4)^(-)+5FeC_(2)O_(4) rarr` `implies` 1 mole of ferrous oxalate requires `3//5` moles of `KMnO_(4)`, i.e., 0.6 mole. |
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| 729. |
The number of moles of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:A. `3/5`B. `2/5`C. `4/5`D. `1` |
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Answer» Correct Answer - A `FeC_(2)O_(4) hArr Fe^(2+) + C_(2)P_(4)^(2-) : MnO_(4)^(-) " oxidises " Fe^(2+) and C_(2)O_(4)^(2-)` both `F^(2+) + C_(2)O_(4)^(2-) rarr Fe^(3+) + 2CO_(2) + 3e^(-) , MnO_(4)^(-) + 5e^(-) rarr Mn^(2+) , 5Fe C_(2)O_(4) -= 3MnO_(4)^(-) or 1 Fe C_(2)O_(4) -= (3)/(5) MnO_(4)^(-)` |
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| 730. |
If 8.5 g of hexane burns completely in oxygen, how many moles of `CO_(2)` is/are produced?A. 6B. 0.6C. 0.9D. 1.2 |
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Answer» `C_(6)H_(12)+9O_(2)to6CO_(2)+6H_(2)O` `6xx12+12=84` `because`84g`C_(6)H_(12)`gives=6moles of `CO_(2)` `therefore 8.4 g C_(6)H_(12)` will give `=(6xx8.4)/(84)`=0.6`mole of `CO_(2)` |
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| 731. |
19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered?A. 100B. `6.02 xx 10^(23)`C. `6.02 xx 10^(24)`D. `6.02 xx 10^(25)` |
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Answer» Atoms recovered `=(19700)/(197)xxN_(A)` `=100xx6.02xx10^(23)=6.02xx10^(23)` |
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| 732. |
1 mol of `CH_(4)` containsA. (a)`6.02xx10^(23)` atoms of `H`B. (b)`4 g` atom of HydrogenC. (c )`1.81xx10^(23)` molecules of `CH_(4)`D. (d)`3.0 g` of carbon |
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Answer» Correct Answer - B 1 mole of `CH_(4)` contains `4` mole of hydrogen atom, i.e., `4 g` atom of hydrogen. |
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| 733. |
The oxygen obtained form 72 kg of water isA. 72 kgB. 46 kgC. 50 kgD. 64 kg |
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Answer» Correct Answer - D The oxygen obtained from 72 kg of water is obtained by the following equation. `{:(2H_(2)O,rarr,2H_(2),+,O_(2)),(2(2+16)=36,,,,(2xx16)=32):}` Therefore, 36 kg of water produce 32 kg of oxygen. Hence, `72` kg of water produce oxygen `=(32xx72)/36=64` kg. |
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| 734. |
In order to prepare one litre normal solution of `KMnO_(4)`, how many grams of `KMnO_(4)` are required if the solution is used in acidic medium for oxidationA. 158 gB. 31.6 gC. 790 gD. 62 g |
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Answer» Correct Answer - B Acidic medium `E=M/5=158/5=31.6` gm. |
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| 735. |
In order to prepare one litre normal solution of `KMnO_(4)`, how many grams of `KMnO_(4)` are required if the solution is used in acidic medium for oxidationA. (a)`158 g`B. (b)`31.6 g`C. (c )`790 g`D. (d)`62 g` |
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Answer» Correct Answer - B Acidic medium `E=M/5=158/5=31.6 g`. |
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| 736. |
On reduction with hydrogen, `3.6 g` of an oxide of matel left `3.2 g` of metal. If the vapour density of metal is `32`, the simplest formula of the oxide would beA. `MO`B. `M_(2)O_(3)`C. `M_(2)O`D. `M_(2)O_(5)` |
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Answer» Correct Answer - C As we know that Equivalent weight `=("weight of metal")/("weight of oxygen")xx8` `=3.2/0.4xx8=64` Vapour density `=("mol. wt")/2` Mol. Wt`=2xxV.D=2xx32=64` As we know that `n=("mol. wt")/("eq. wt")=64/64=1` Suppose, the formula of metal oxide be `M_(2)O_(n)`. Hence the formula of metal oxide `=M_(2)O`. |
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| 737. |
The most abundant elements by mas in the body of a healthy human adult are Oxygen `(61.4%)`, Carbon `(22.9%)`. Hydrogen `(10.0)%)`, and Nitrogen `(2.6%)`. The weight which a `75 kg` person would gain if all `.^(1)H` atoms are replaced by `.^(2)H` atoms isA. 15 kgB. 37.5 kgC. 7.5 kgD. 10 kg |
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Answer» Correct Answer - C Given, abundance of element by mass oxygen `= 61.4 %` carbon `= 22.9%` hydrogen `= 10%` and nitrogen `= 2.6%` Total weight of person `= 75 kg` Mass due to `.^(1)H = (75 xx 10)/(100) = 7.5 kg` `.^(1)H` atoms are replaced by `.^(2)H` atoms, Mass due to `.^(2)H = (7.5 xx 2) kg` `:.` Mass gain by person `= 7.5 kg` |
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| 738. |
The most abundant elements by mas in the body of a healthy human adult are Oxygen `(61.4%)`, Carbon `(22.9%)`. Hydrogen `(10.0)%)`, and Nitrogen `(2.6%)`. The weight which a `75 kg` person would gain if all `.^(1)H` atoms are replaced by `.^(2)H` atoms isA. 37.5 kgB. 7.5 kgC. 10 kgD. 15 kg |
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Answer» Correct Answer - B 75 kg person contain 10% hydrogen i.e. 7.5 kg Hydrogen. If all H atom are replaced by `.^(2)H`, the weight of Hydrogen become twice i.e., it increased by 7.5 kg |
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| 739. |
What volume of oxygen gas `(O_(2))` measured at `0^(@)C` and 1 atm is needed to burn completely `1 L` of propane gas `(C_(3)H_(8))` measured under the same condition?A. `5L`B. `10L`C. `7L`D. `6L` |
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Answer» Correct Answer - A `{:(C_(3)H_(8),+,5O_(2),to,3CO_(2),+,4H_(2)O),("Propane",,,,,,),(1vol,,5vol,,3vol,,4vol):}` According to the balanced equation , one volume of 1 L propane needs 5 volumes or 5L of `O_(2)` for complete combustion |
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| 740. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will beA. `3mol`B. `4mol`C. `1mol`D. `2mol` |
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Answer» Correct Answer - B `10g H_(2) = (10)/(2)=5` mol, 64 g of `O_(2) = (64)/(32) = 2` mol `{:(H_(2),+,(1)/(2)O_(2),to,H_(2)O),(1mol,,0.5mol,,1mol),(5mol,,2.5mol,,5mol),(,,2mol,,?):}` In 5 mol of oxygen needed = 2.5 mol which not present `:. O_(2)` is the limiting reactant. Hence `H_(2)O` formed from 2 mol is 4 mol. |
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| 741. |
If we consider that `1/6` in place of `1/12`, mass of carbon atom is taken to be the relative atomic mass unit, the mas of one mole of a substance willA. decrease twiceB. increase two foldC. remain unchangedD. be a function of the molecular mass of substance |
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Answer» Correct Answer - C 1 mole = No. of atoms in 12.0000 g of `C^(12) = 6.022xx10^(23)` Mass of 1 mole of a substance = Actual mass of 1 atom (or molecule) of the substanc `xx6.022xx10^(23)` As mass of 1 mole of substance does not depend upon the atomic mass unit scale, the mass of one mole of substance will not change |
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| 742. |
The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by Hall process isA. `180 kg`B. `270 kg`C. `540 kg`D. `90 kg` |
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Answer» Correct Answer - D `2Al_(2)O_(3)+underset(3xx12g)(3C)to3CO_(2)+underset(4xx27g)(4Al)` `4xx27g` of Al are obtained when carbon anode consumed `= 12xx3 g` |
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| 743. |
If 30 mL of `H_(2)` and 20 mL of `O_(2)` reacts to form water, what is left at the end of the reaction?A. 10 mL of `H_(2)`B. 5 mL of `H_(2)`C. 10 mL of `O_(2)`D. 5 mL of `O_(2)` |
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Answer» Correct Answer - D `underset(2vol)(2H_(2))+underset(1vol)(O_(2))to2H_(2)O` 2 mole of `H_(2)` reacts with 1 vol of `O_(2)` to give water. Therefore, 30 mL of `H_(2)` will reacts with 15 mL of `O_(2)` to give water. Here, `H_(2)` is the limiting reactant. Volume of `O_(2)` left unreacted = 20-15=5 mL |
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| 744. |
How will you separate a solution (miscible) of benzene + `CHCl_(3)`?A. SublimationB. FiltrationC. DistillationD. Crystallisation |
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Answer» Correct Answer - C Boiling points of benzene and chloroform are `80^(@)C` and `61.5^(@)C` respectively. As such they can be separated by fractional distillation |
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| 745. |
Two elements X and Y combine in gaseous state to form XY in the ratio `1:35*5` by mass. The mass of Y that will be required to reat with 2 g of X isA. `7*1 g`B. `3*55 g`C. `71 g`D. `35*5 g` |
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Answer» Correct Answer - C 1 g of X combines with Y `= 35*5 g` `:.` 2 g of X combines with Y `= 2xx35*5 = 71 g` |
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| 746. |
Two compounds A and B have some percentage composition. The compouds A and BA. are identicalB. are isomersC. are neither identical nor isomersD. All the three are correct |
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Answer» Correct Answer - D Converse of law of definite proportion is not true. For example, butane and isobutane have same percentage composition (Choice B), acetylene, `C_(2)H_(2)` and benzene, `C_(6)H_(6)` have same percentage composition. As such choice (D) is correct |
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| 747. |
Which of the following has the smallest number of molecules?A. 0.1 mol of `CO_(2)` gasB. 11.2 L of `CO_(2)` gas at N.T.P.C. 22 g of `CO_(2)` gasD. `22.4xx10^(3)mL` of `CO_(2)` gas |
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Answer» Correct Answer - A 0.1 mol `CO_(2), (11.2)/(22.4) = 0.5 mol CO_(2),` `(22)/(44) = 0.5 mol CO_(2), (22400)/(22400) = 1 mol CO_(2)` |
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| 748. |
Among the species given below which have same number of molecules?A. 3.2 g of `O_(2)`B. 0.1 mol of `NH_(3)`C. 4.0 g of `He`D. 11.2 L of `Cl_(2)` at S.T.P. |
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Answer» Correct Answer - A::B::D Same number of molecules are present in 3.2 g of `O_(2)` (0.1 mol) and 0.1 mol of `NH_(3)`. |
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| 749. |
A solution is prepared by dissolving 5.3 g of `Na_(2)CO_(3)` in 250 `cm^(3)` of solution. The solution can be described asA. Decinormal solutionB. Decimolar solutionC. 0.4 N solutionD. 0.2 M solution |
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Answer» Correct Answer - C::D Mol. Mass of `Na_(2)CO_(3)=106` Eq. mass of `Na_(2)CO_(3)=106//2=53` Molarity `=(5.3xx1000)/(106xx250)=0.2M` Normality `=(5.3xx1000)/(53xx250)=0.4N` |
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| 750. |
What is the mass of the precipitate formed when 50 mL of 16.9% solution of `AgNO_(3)` is mixed with 50 mL of 5.8% NaCl solution?A. (a)`7 g`B. (b)`14 g`C. (c )`28 g`D. (d)`3.5 g` |
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Answer» Correct Answer - a `16.9 g AgNO_(3)` is present in `100 mL` solution. `:. 8.45 g AgNO_(3)` is present in `50 mL` solution. ` 5.8 g NaOl` is present in `100 mL` solution. `:. 2.9 g NaCl` is present in `50 mL` solution. `AgNO_(3)+NaCl rarr AgCl+ NaNO_(3)` `8.45/170 mol" " 2.9/58.5` `=0.049 mol=0.049 mol rarr 0" " 0` `After "0 " 0 rarr 0.049 mol 0.049` reaction" " mol Mass of `AgCl` precipitated `=0.049xx143.5 g =7 g AgCl`. |
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