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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.
| 401. |
A metal nitride `M_(3)N_(2)` contains 28 % of nitrogen. The atomic mass of metal M isA. `24`B. `54`C. `9`D. `87.62` |
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Answer» Correct Answer - A Moles of N `= (28)/(14) = 2` Moles of metal, M `= (100-28)/(a) = (72)/(a)` Mole ratio, `M:N = (72)/(a):2 = 3:2` `(72)/(a) = 3` `a = (72)/(a) = 24` |
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| 402. |
The number of significant figures in `3.040 dm` isA. infiniteB. threeC. twoD. four |
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Answer» Correct Answer - D Zeroes at the end of a number (greater than 1) that contains a decimal point are always significiant. |
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| 403. |
A compound was found to contain nitrogen and oxygen in the ratio `28 g` and `80 g` respectively. The formula of compound isA. (a)`NO`B. (b)`N_(2)O_(3)`C. (c )`N_(2)O_(5)`D. (d)`N_(2)O_(4)` |
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Answer» Correct Answer - C No. of nitrogen atoms=`("Mass in grams")/("Atomic wt.")=28/14=2` No. of oxygen atoms`=("Mass in grams")/("Atomic wt.")=80/16=5` `:.` Formula of compound is `N_(2)O_(5)`. |
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| 404. |
The number of significant figures in `60.0001` isA. (a)` 5`B. (b)` 6`C. (c )`3 `D. (d)`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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| 405. |
The number of significant figures in `40,501 kg` isA. threeB. twoC. fiveD. infinite |
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Answer» Correct Answer - C Zeroes between nonzero digits are always significant |
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| 406. |
Given `P=0.0030 m, Q=2.40 m, R=3000 m`, Significant figures in `P, Q` and `R` are respectivelyA. 2,2,1B. 2,3,4C. 4,2,1D. 4,2,3 |
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Answer» Correct Answer - B Given `P=0.0030m, Q=2.40m` & `R=3000m` In `P(0.0030)` initial zeros after the decimal point are not significant. Therefore, significant figures in `P(0.0030)` are 2. Similarly in `Q(2.40)` significant figures are 3 as in this case final zero is significant. In `R=(3000)` all the zeroes are significant hence, in R significant figures are 4 because they come from a measurement. |
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| 407. |
The significant figures in 3400 areA. 2B. 5C. 6D. 4 |
| Answer» Correct Answer - D | |
| 408. |
Given `P=0.0030 m, Q=2.40 m, R=3000 m`, Significant figures in `P, Q` and `R` are respectivelyA. (a)` 2, 2, 1`B. (b)` 1, 3, 4`C. (c )` 4, 2, 1`D. (d)` 4, 2, 3` |
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Answer» Correct Answer - B Given `P=0.0030 m, Q= 2.40 m` and `R=3000 m` In `P(0.0030)` initial zeros after the decimal point are not significant. Therefore, significant figures in `P(0.0030)` are 2. Similarly in `Q(2.40)` significant figures are 3 as in this case final zero is significant. In `R=(3000)` all the zeroes are significant hence, in R significant figures are 4. |
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| 409. |
The number of significant figures in `6.0023` areA. (a)` 5`B. (b)`4 `C. (c )`3 `D. (d)` 1` |
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Answer» Correct Answer - A Number of significant figures in 6.0023 are 5 because all the zeroes stand between two non-zero digit are counted towards significant figures. |
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| 410. |
Write the SI unit of thermodynamic temperature. |
| Answer» Correct Answer - Kelvin | |
| 411. |
The significant figures in 3400 areA. (a)` 2`B. (b)` 5`C. (c )`6 `D. (d)`4 ` |
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Answer» Correct Answer - A As we know that all non-zero unit are significant number. Therefore significant figure is 2. |
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| 412. |
The prefix zepto stands forA. (a)`10^(9) `B. (b)` 10^(-12)`C. (c )` 10^(-15)`D. (d)` 10^(-21)` |
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Answer» Correct Answer - D `1 "zepto" =10^(-21)` |
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| 413. |
Write the SI unit of electric current. |
| Answer» Correct Answer - ampere | |
| 414. |
What temperature is `75^(@)` F on the Kelvin scale?A. 24 KB. 348 KC. 297 KD. 215 K |
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Answer» c) `75^(@)`F= 9/5^(@)`C + 32 = ((43 xx 5)/9)^(@)` C = `23.9^(@)`C = `(23.9 + 273)`K = 296.9 -~ 297 K |
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| 415. |
Express `20^(@)C` in kelvin scale. |
| Answer» Correct Answer - 293.15 K | |
| 416. |
The oxidation state of `M^(3+)` after removing three elelctrons isA. ZeroB. `+3`C. `+6`D. `-6` |
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Answer» Correct Answer - C After removing three electrons from `M^(3+)` the oxidation state is `M^(6+)`. |
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| 417. |
The mass of 60% HCl by mass required for the neutralisation of 10 L of `0*1` M NaOH isA. `60*8 g`B. `21*9 g`C. `100 g`D. `219 g` |
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Answer» Correct Answer - A `underset(36.5g)(HCl)+underset(40g)(NaOH)toNaCl+H_(2)O` Amount of NaOH to be neutralised `=0*1xx10xx40=40g` `:.` HCl required `= 36*5 g` 60% HCl required `= (36*5)/(60)xx100=60*8g` |
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| 418. |
Assertion : The oxidation numbers are artificial, they are useful as a book keeping device of elements in reactions Reason : The oxidation numbers do not usually represent real charge on atoms, they are simply conventions that indicate what the maximum charge could possibly be on an atom in a molecule.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 the assertion and reason both are false. |
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Answer» Correct Answer - A Both assertion and reason are true and reason is the correct explanation of assertion. |
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| 419. |
In the balanced chemical reaction `IO_(3)^(ө)+al^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)` `a, b,c`, and `d`, respectively, correspond toA. 5, 6, 3, 3B. 5, 3, 6, 3C. 3, 5, 3, 6D. 5, 6, 5, 5 |
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Answer» Correct Answer - A `IO_(3)^(-)+aI^(-)+bH^(+) rarr cH_(2)O+dI_(2)` Step 1 : `I^(-1) rarr I_(2)` (oxidation) `IO_(3)^(-) rarr I_(2)` (reduction) Step 2 : `2IO_(3)^(-)+12 H^(+) rarr I_(2)+6 H_(2)O` Step 3 : `2IO_(3)^(-)+12 H^(+)+10 e^(-) rarr I_(2)+6 H_(2)O` `2I^(-) rarr I_(2)+2e^(-)` Step 4 : `2IO_(3)^(-)+12 H^(+)+10 e^(-) rarr I_(2)+6 H_(2)O` `[2I^(-) rarr I_(2)+2e^(-)]5` Step 5 : `2IO_(3)^(-)+10 I^(-)+12 H^(+) rarr 6I_(2)+ 6H_(2)O` `IO_(3)^(-)+5I^(-)+6H^(+) rarr 3I_(2)+3 H_(2)O` On comparing, `a=5, b=6, c=3, d=3` |
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| 420. |
The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. One fifthB. FiveC. OneD. Two |
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Answer» Correct Answer - D In alkaline medium `2KMnO_(4)+KI+H_(2)O rarr 2MnO_(2)+2 KOH+KIO_(3)`. |
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| 421. |
Density of a metallic object is `5.6 g//cm^(3)`. Express it in `kg//m^(3)`. |
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Answer» We know, `1 g = 1xx10^(-3)kg` `1 cm = 1xx10^(-2) m` So, `5.6xx(10^(-3))/((10^(-2))^(3))=5.6xx10^(3)kg//m^(3)` `= 56000 kg m^(-3)` |
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| 422. |
Define metre. |
| Answer» The metre is the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second. | |
| 423. |
Which one of the following is not an element ?A. GraphiteB. WaterC. DiamondD. Nitrogen |
| Answer» Correct Answer - B | |
| 424. |
Dimensions of pressure are same as that ofA. forceB. energy per unit volumeC. force per unit volumeD. energy |
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Answer» Correct Answer - B `("Energy")/("Volume") = ("Force" xx "Length")/("Area" xx "Length") = ("Force")/("Area") = "Pressure"` |
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| 425. |
Dimensions of pressure are same as that ofA. (a)EnergyB. (b)ForceC. (c )Energy per unit volumeD. (d)Force per unit volume |
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Answer» Correct Answer - C `Pressure= (Force)/(Area)=([MLT^(-2)])/([L^(3)])=[ML^(-1)T^(-2)]` Energy per unit volume `=([ML^(2)T^(-2)])/([L^(3)])=[ML^(-1)T^(-2)]` |
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| 426. |
The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` is |
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Answer» Correct Answer - `4.14g` Molar mass of `CuSO_(4).5H_(2)O` `= 63.5 + 32+4 xx 16 + 5 xx 18 = 249.5 g` Also, molar mass represents mass of Avogadro number of molecules in gram unit, therefore `because 6.023 xx 10^(23)` molecules of `CuSO_(4).5H_(2)O` weigh `249.5 g` `:. 10^(22)` molecules will weigh `(249.5)/(6.023 xx 10^(23)) xx 10^(22) = 4.14 g` |
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| 427. |
Three grams of salt of molecular weight 30 is dissolved in `250 g` of water. The molality of the solution is…. |
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Answer» Correct Answer - `(0.4)` `"Molarity" = ("Number of moles of solute")/("Volume of solution in litre")` `= ("Weight of solute")/("Molar mass") xx (1000)/("Volume in mL") = (3)/(30) xx (1000)/(250) = 0.4 M` |
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| 428. |
In the conversion fo ` Br_2` to `BrO_3^(-)` , the oxidation state of `Br` changes from.A. `-1` to `-1`B. 0 to -1C. `0` to `+5`D. `0` to `-5` |
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Answer» Correct Answer - C `overset(0)(B)r_(2) rarr overset(+5)(Br)O_(3)^(-)`, in this reaction oxidation state change from 0 to +5. |
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| 429. |
Chlorine is in +1 oxidation state inA. `HCl`B. `HClO_(4)`C. `ICl`D. `Cl_(2)O` |
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Answer» Correct Answer - D In case of `Cl_(2)O` chlorine shows +1 oxidation state. |
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| 430. |
Phosphorus has the oxidation state `+3` inA. Orthophosphoric acidB. Phosphorus acidC. Metaphosphoric acidD. Pyrophosphoric acid |
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Answer» Correct Answer - B `H_(3) overset(**)(P)O_(3)` `3+x-2xx3=0, x=6-3=+3`. |
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| 431. |
The oxidation state of nitrogen in `N_(3)H` isA. `+1/3`B. `+3`C. `-1`D. `- 1/3` |
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Answer» Correct Answer - D In hydrazoic acid `(N_(3)H)` nitrogen shows `-1/3` oxidation state. `overset(**)(N)_(3)H` `3x+1=0, 3x=-1, x=-1/3`. |
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| 432. |
The oxidation number of N in `NH_(4)Cl` isA. `+5`B. `+3`C. `-5`D. `-3` |
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Answer» Correct Answer - D `NH_(4)Cl" "NH_(4)^(+)+Cl^(-)` `overset(**)(N)H_(4)^(+)` `x+4=+1, x=1-=-3`. |
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| 433. |
The oxidation number and the electronic configuration of sulphur in `H_(2)SO_(4)` isA. `+4, 1s^(2)2s^(2)2p^(6)3s^(2)`B. `+2, 1s^(2)2s^(2)2p^(6) 3s^(2)3p^(2)`C. `+3, 1s^(2) 2s^(2)2p^(6)3s^(2)3p^(1)`D. `+6, 1s^(2) 2s^(2) 2p^(6)` |
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Answer» Correct Answer - D `H_(2) overset(**)(S)O_(4)` `2xx(+1)+x+4xx(-2)=0` `+2+x-8=0, x=8-2=+6` Electronic configuration of sulphur in `H_(2)SO_(4)` is `1s^(2), 2s^(2), 2p^(6)` |
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| 434. |
The highest oxidation state of `Mn` is shown byA. `KMnO_(4)`B. `K_(2)MnO_(4)`C. `Mn_(2)O_(3)`D. `MnO_(2)` |
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Answer» Correct Answer - A `{:("Compound","O.S"),(KMnO_(4),+7),(K_(2)MnO_(4),+6),(Mn_(2)O_(3),+3),(MnO_(2),+4):}` |
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| 435. |
The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is `:`A. `+4`B. `+6`C. `+2`D. `+3` |
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Answer» Correct Answer - D `K_(2)Cr_(2)O_(7)+6KI+7 H_(2)SO_(4) rarr 4K_(2)SO_(4)+Cr_(2) (SO_(4))_(3)+ 7H_(2)O+3I_(2)` `overset(**)(C)r_(2) (SO_(4))_(3) rarr 2 overset+3)(C)r+3 SO_(4)^(2-)` |
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| 436. |
Maximum oxidation state of Cr isA. 3B. 4C. 6D. 7 |
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Answer» Correct Answer - C Maximum oxidation state for `Cr` is +6. |
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| 437. |
In the Haber process, `30 L` of dhyrgen and `30L` of dintrogen were taken for reaction which yielded only `50%` of the expectedf product. What will be the xomposition of the gaseous mixturre under the aforesaid condition in the end?A. 20 L of ammonia, 25 L of nitrogen, 15 L of hydrogenB. 20 L of ammonia, 20L of nitrogen, 20 L hydrogenC. 10L of ammonia, 25 L of nitrogen, 15 L of hydrogenD. 20L of ammonia, 10 L of nitrogen, 30 L of hydrogen |
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Answer» Correct Answer - C `underset("1vol")(N_(2))+underset("3vol")(3H_(2))hArr underset("2vol")(2NH_(3))` `:.` 30 L of `H_(2)` will react with 10 L of `N_(2)` to produce 20 L of `NH_(3)` (Here `H_(2)` is the limiting reactant). As the yield is only 50%, 15 L of `H_(2)` will react with 5 L of `N_(2)` to produce 10 L of `NH_(3)` `:.` Reaction mixture will contain, 15 L of `H_(2)` (30L-15L), and 10 L of `NH_(3)` |
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| 438. |
Oxidation number of sulphur in `Na_(2)S_(2)O_(3)` isA. `+1`B. `+2`C. `+3`D. `-3` |
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Answer» Correct Answer - B Suppose the oxidation number of sulphur in `Na_(2)S_(2)O_(3)` is x. `:. 2(+1)+x+3(-2)=0` `2+2x-6=0` `2x-4=0` `:. X=4/2=+2` |
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| 439. |
The oxidation number of `S` in `Na_(2)S_(4)O_(6)` isA. `-2`B. `+2`C. `-6`D. `+6` |
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Answer» Correct Answer - D `Na_(2)SO_(4)` `2+x-2xx4=0` `x=+6`. |
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| 440. |
One mole of calciium phosphide on reaction with excess water givesA. 1 mole of phosphineB. 2 moles of phosphoric acidC. 2 moles of phosphineD. 1 mole of phosphorus pentaoxide |
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Answer» Correct Answer - C `Ca_(3)P_(2) + 6H_(2)O rarr 2PH_(3) + 3Ca(OH)_(2)` |
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| 441. |
Oxidation number of `N` in `(NH_(4))_(2)SO_(4)` isA. `-1//3`B. `-1`C. `+1`D. `-3` |
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Answer» Correct Answer - D `(NH_(4))_(2)SO_(4) hArr 2NH_(4)^(+)+SO_(4)^(--)` `overset(**)(N)H_(4)^(+)` `x+4=+1," "x=1-4=-3`. |
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| 442. |
Yellowish-green gas chlorine `(Cl_(2))` can be prepared in the laboratory by heating hydrochloric acid `(HCl, aq)` with pyrolusinte (manganese dioxide, `MnO_(2)`): `4Cl(aq.) + MnO_(2)(s) rarr Cl_(2)(g) + 2H_(2)O(l) + MnCl_(2)(aq.)`. How many grams of `HCl` reacts with `5.00g` of manganses dioxide?A. `4.80g HCl`B. `8.40g HCl`C. `6.95g HCl`D. `5.69g HCl` |
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Answer» Correct Answer - B `g_(MnO_(2)) rarr n_(MnO_(2)) rarr n_(HCl) rarr g_(HCl)` `5.00 g_(MnO_(2)) xx (1 mol MnO_(2))/(86.9g MnO_(2)) xx (4 mol HCl)/(1 mol MnO_(2))` `xx (36.5g HCl)/(1 mol HCl)` `= 8.40g HCl` |
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| 443. |
Which of the following statement is correctA. Hydrogen has oxidation number -1 and +1B. Hydrogen has same electronegativity as halogensC. Hydrogen will not be liberated at anodeD. Hydrogen has same ionization potential as alkali metals |
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Answer» Correct Answer - A Hydrogen have oxidation no. +1 and -1. |
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| 444. |
10 mL of 1 NHCl is mixed with 20 mL of 1`MH_(2)SO_(4)` and 30 mL of 1M NaOH. The resultant solution has:A. (a)20 meq of `H^(+)` ionsB. (b)20 meq of `OH^(-)`C. (c )0 meq of `H^(+)` or `OH^(-)`D. (d)30 millimoles of `H^(+)` |
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Answer» Correct Answer - A Exlpanation: No. of meq of `H^(+)=10xx1+20xx2=50` `[Q H_(2)SO_(4), N=2M]` No. of meq of `OH- =30xx1=30` No. of meq of `H^(+)` left uncreacted`=50-30=20 `meq Hence, (a) is correct, (b), (c ) and (d) are ruled out. |
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| 445. |
What mass of calcium chloride in grams would be enough to produce `14*35` g of AgCl? (At. mass: Ca=40, Ag=108)A. `5*55 g`B. `8*295 g`C. `16*59 g`D. `11*19g` |
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Answer» Correct Answer - A `underset(111g)(CaCl_(2)) + 2AgNO_(3) to Ca(NO_(3))_(2)+underset(2xx143.5 g)(2AgCl)` `CaCl_(2)` required to produced `2xx143*5g` of AgCl `= 111g` `CaCl_(2)` required to produce `14*35g` of AgCl `=(111xx14.35)/(2xx143.5)=5*55g` |
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| 446. |
How many grams of oxygen `(O_(2))` gas reacts completely with `1.0g` of calcium?A. `0.6 g`B. `1.6 g`C. `4.0 g`D. `0.4 g` |
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Answer» Correct Answer - D `Ca + O_(2) rarr CaO` According to formula, Moles of `C` = Moles of `O` `("Mass"_(Ca))/("Molar mass"_(Ca)) = ("Mass"_(O))/("Molar Mass"_(O))` `(1g Ca)/(40g Ca) = (m_(O))/(16g O)` `:. M_(O) = 0.4g` We can use this approach because the enite `Ca` is changed into `CaO`. |
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| 447. |
Oxidation number of fluorine in `F_(2)O` is:A. `-1`B. `+1`C. `+2`D. `-2` |
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Answer» Correct Answer - A Fluorine always shows -1 oxidation state in oxides. |
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| 448. |
Assume that the nucleus of the fluorine atom is a sphere of radius `5xx10^(-3)`cm. What is the density of matter in the nucleus?A. `6.02xx10^(23)g//mL`B. `6.02xx10^(13)g//mL`C. `12.02xx10^(23)g//mL`D. `12.02xx10^(13)g//mL` |
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Answer» Correct Answer - B Since almost, whole the mass of an atom is in the nucleus, Mass of F nucleus = Mass of F atom `=(19)/(6.02xx10^(23))g` `=3.15xx10^(-23)g` (At. Mass of F = 19) Volume of nucleus `=(4)/(3)pir^(3)` `=(4)/(3) xx(22)/(7)xx(5xx10^(-13)^(3)cm^(3)` `=5.24xx10^(-37)cm^(3)` Density `= ("Mass")/("Volume")=(3.15xx10^(-23))/(5.24xx10^(-37)cm^(3))` `=6.02xx10^(23)gcm^(-3)=6.02xx10^(13)g//mL` |
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| 449. |
The oxidation number of oxygen in `KO_(3), Na_(2)O_(2)` isA. `3, 2`B. `1,0`C. `0, 1`D. `-0.33, -1` |
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Answer» Correct Answer - D Let oxidation number of oxygen `=x` O.N. of oxygen in `KO_(3)=1+3x=0` `:. X=-1//3=-0.33` O.N. of oxygen in `Na_(2)O_(2)=2xx1+2x=0" ":. X=-1`. |
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| 450. |
Common salt obtained from sea water contains `95% NaCl` by mass. The appoximate number of molecules present in `10.0g` of the salt isA. `10^(24)`B. `10^(23)`C. `10^(22)`D. `10^(21)` |
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Answer» Correct Answer - B `m_(NaCl) = 95%` of `10g = 9.5g` ? Fomula units `NaCl = 9.5g NaCl xx (1 mol NaCl)/(58.5g NaCl)` `xx (6.022xx10^(23) "formula units")/(1 mol NaCl)` `= (57.209)/(58.5) xx10^(23)` formula units `= 10^(23)` formula unit |
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