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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.
| 551. |
Which of the following represents 1 g-molecule of the substance?A. `6*02xx10^(24)` molecules of `NH_(3)`B. 4 g of heliumC. 40 g of CaOD. 127 g of iodine. |
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Answer» Correct Answer - B He is known as monoatomic molecule. |
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| 552. |
One atom of an element weighs `1*8xx10^(-22)` g its atomic mass isA. `29*9`B. `18`C. `108*36`D. `154` |
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Answer» Correct Answer - C Mass of 1 atom `= 1*8xx10^(-22) g` Mass of `6*02xx10^(23)` atoms `= 6*02xx10^(23)xx1*8xx10^(-22)g` `= 6*02xx1*8xx10g = 108*36g` `:.` Atomic mass of element `= 108*36`. |
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| 553. |
The number of neutrons in `1*8` g of water will approximately beA. `4*216xx10^(23)`B. `8*432x10^(23)`C. `4*816xx10^(23)`D. `4*216xx10^(24)` |
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Answer» Correct Answer - C `H_(2)O = 2_(1)^(1)H+._(8)^(16)O` `:.` neutrons, `._(0)^(1)n = 0+8` `.^(2)H_(2)O = (1*8)/(18) = 0*1` Hence `._(0)^(1)n = 8xx0*1xx6*02xx10^(23)` `=4*816xx10^(23)` |
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| 554. |
The mass of 1 mole of neutrons `(m_(n) = 1.675 xx 10^(-27)` kg) is:A. `1.800xx10^(-3)` kgB. `1.008xx10^(-4)` kgC. `1.080xx10^(-3)` kgD. `1.008xx10^(-3)` kg |
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Answer» Correct Answer - D Mass of 1 mole `=1.675xx10^(-27)xx6.02xx10^(23) kg=1.008xx10^(-3)` kg. |
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| 555. |
Which has maximum number of atomsA. 24 gms of `C_((12))`B. 56 gms of `Fe_((56))`C. 27 gms of `Al_((27))`D. 108 gms of `Ag_((108))` |
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Answer» Correct Answer - A 12 g. of carbon `=6.023xx10^(23)` 24 g. of Carbon `=(6.023xx10^(23))/12 xx24=12.046xx10^(23)` But 56 gm, 27 gm. And and 108 gm of Fe, Al the Ag respectively contains same no. (Avogardos number) of atoms. |
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| 556. |
Assertion (A): A reaction between `Fe` and `I_(2)` occurs, but a reaction between `Fe^(2+)` and `I^(ө)` does not occur. Reason (R ): `Fe` is a better reducing agent than `I^(ө)`.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 6B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 6C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A Both (1) and (2) are correct and (2) Is the correct explanatio for (1). `I_(2)+Fe rarr Fe^(2+)+2 I^(ɵ)` `I^(ɵ)+Fe^(2+) rarr` No reaction The oxidation potential of `Fe//Fe^(2+)` is greater than the oxidation potential of `2I//I_(2)^(ɵ)`. |
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| 557. |
No reaction occurs in which of the following equationsA. `I^(ɵ)+Fe^(2+) rarr`B. `F_(2)+2NaCl rarr`C. `Cl_(2) +2NaF rarr`D. `I_(2)+2NaBr rarr` |
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Answer» Correct Answer - A::C::D Reduction potential of `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. So `F_(2)` can displace `Cl^(Θ), Br^(Θ) and I^(Θ)` but not vice versa. Similarly, `Cl_(2)` can displace `Br^(Θ) and I^(Θ)` but not vice versa and `Br_(2)` can displace only `I^(Θ)` but not vice versa. In (a) `Fe(" not " Fe^(2+))` is a better reducing agent tahn `I^(Θ)`. |
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| 558. |
Which of the following is`//`are disproportionation redox changes?A. `(NH_(4))_(2) Cr_(2)O_(7) rarr N_(2)+Cr_(2)O_(3)+4 H_(2)O`B. `5H_(2)O_(2)+2 ClO_(2)+2 overset(ɵ)(OH) rarr 2 Cl +overset(ɵ)(%)O_(2)+6H_(2)O`C. `3 ClO^(ɵ) rarr ClO_(3)^(ɵ)+Cl^(ɵ)`D. `2HCuCl_(2) overset("Dilution")underset("with water")(rarr) Cu+Cu^(2+)+4Cl+.^(ɵ)2H^(o+)` |
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Answer» Correct Answer - C::D (c) `2e^(-) + ClO^(Θ) rarr Cl^(Θ)` (Reduction) `x -2 = -1 " " x = -1` `x = +1` `ClO^(Θ) rarr ClO_(3)^(Θ) + 4e^(-)` (Oxidation) `x = 1 " " x = 5` (d) In `HCuCl_(2), Cu` is in +1 oxidation state which disproportionates to `Cu^(2+) and Cu^(0)` |
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| 559. |
Which of the following reactions does not involve oxidation-reduction ?A. `2 Rb+2 H_(2)O rarr 2 RbOH+H_(2)`B. `2 CuI_(2) rarr 2 CuI=I_(2)`C. `NH_(4)Cl +NaOH rarr NaCl +NH_(3)+H_(2)O`D. `4 KCN+Fe(CN)_(2) rarr K_(4)[Fe(CN)_(6)]` |
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Answer» Correct Answer - C::D In both (c) and (d), the oxidation numbers of the various elements in the molecule do not change. |
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| 560. |
In acidic medium dichromate oin osxidizes stannous ion as : `xSn^(2+)+yCr_(2)O_(7)^(2-)+zH^(+) to aSn^(4+)+bCr^(3+)cH_(2)O`A. The value of `x : y` is `1 : 3`B. The value of `x+y+z` is 18C. The value of `a : b` is `3 : 2`D. The value of `z-c` is 7 |
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Answer» Correct Answer - B::C::D `3 Sn^(2+) +14 H^(+) + Cr_(2)O_(7)^(2-) rarr 3 Sn^(3+) + 2Cr^(3+) +7H_(2)O`. |
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| 561. |
A hydrocarbon contain 86% carbon, 488 ml of the hydrocarbon weight 1.68 g at STP. Then the hydrocarbon is anA. (a)AlkaneB. (b)AlkeneC. (c )AlkyneD. (d)Arene |
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Answer» Correct Answer - B `{:("Element","At. wt".,"Mole","Ratio","Empirical formula"),(C=86%,12,7.1,1,CH_(2)),(H=14%,1,14,2,"Belongs to alkene" C_(n)H_(2n)):}` |
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| 562. |
Definig 1 amu as `1//12` of the mass of a `._(6)^(12)C` atom means that protons and neutrons each have a mass of almost exactly 1 amu, becauseA. atin as a whole is electrically neutralB. `C` atom contains equal number of protons and neutronsC. the mass of an electron is negligibel compared with the mass of proton and neutronD. neutrons are slightly heavier than protons |
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Answer» Correct Answer - C `m_(p) = 1.007276, m_(n) = 1.008665, m_(e) = 5.485799xx10^(-4)` |
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| 563. |
What is the % of `H_(2)O` in `Fe(CNS)_(3).3H_(2)O`?A. (a)45B. (b)30C. (c )19D. (d)25 |
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Answer» Correct Answer - C In Fe `(CNS)_(3).3H_(2)O` `%` of `H_(2)O=(3xx18)/284xx100=19%`. |
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| 564. |
The number of Na atom in 46 g Na (Atomic weight of Na = 23 ) isA. `6.023xx10^(23)`B. 2C. 1D. `12.046xx10^(23)` |
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Answer» Correct Answer - D `because ` 23 g of Na contains `= 6.023xx 10^(23)` atoms `therefore` 1 g of contains ` = (6.023 xx 10^(23))/23 `atoms `therefore` 46 g of Na contains `= (6.023 xx 10^(23)/23) xx 46` atoms `=12-046 xx 10^(23)` atoms |
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| 565. |
Complete combustion of 0.858 g of compound X gives 2.63 g of `CO_(2)` and 1.28 g of` H_(2)`O. The lowest molecular mass X can haveA. (a)43 gB. (b)86 gC. (c )129 gD. (d)172 g |
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Answer» Correct Answer - A `%C=12/44xxW_(CO_(2))/Wxx100=12/44xx2.63/0.858xx100=83.6%` `%H=2/18xxW_(H_(2)O)/Wxx100= =2/18xx1.28/.858xx100=16.4%` `{:(,%(a),At.wt.(b),a//b,Ratio),(C,83.6,12,6.96,1),(H,16.4,1,16.4,2.3):}` `C_(3)H_(7)=12xx3+7=43 g`. |
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| 566. |
In an experiment it is shown that `10 mL` of `0.05 M` solution of chloride required `10 mL` of `0.1 M` solution of `AgNO_(3)`, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine):A. (a)`X_(2)Cl`B. (b)`X_(2)Cl_(2)`C. (c )`XCl_(2)`D. (d)`XCl_(4)` |
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Answer» Correct Answer - c Millimoles of solution of chloride`=0.05xx10=0.5` Millimoles of `AgVO_(3)` solution `=10xx0.1=1` So, the millimoles of `AgNO_(3)` are double than the chloride solution. `:. XCl_(2)+2AgNO_(3) rarr 2AgCl+X(NO_(3))_(2)` |
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| 567. |
If `2.73g` of oxide of vanadium contains `1.53g` fo the metal vanadium, the empirical formula of the oxide isA. `V_(2) O_(5)`B. `V_(2) O_(3)`C. `V_(3) O_(4)`D. `V O_(2)` |
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Answer» Correct Answer - A `n_(V) = (1.53g)/(51g mol^(-1)) = 0.03, n_(O) = (1.20g)/(16g mol^(-1)) = 0.075` `(n_(V))/(n_(O)) = (0.03)/(0.075) = (30)/(75) = (2xx15)/(5xx15) = 2:5` `:. EF` is `V_(2) O_(5)`. |
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| 568. |
`6.02xx10^(20)` molecules of urea are present in `100 mL` solution. The concentration of urea solution is:A. `0.1 M`B. `0.001 M`C. `0.1 M`D. `0.02 M` |
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Answer» Correct Answer - A Number of moles (urea) `= (6.02xx10^(20) "molecules")/(6.02xx10^(23) "molecules" mol^(-1))` `= 0.01 mol` Volume of soltuion in liters `= (100 mL)/(1000 mL L^(-1)) = 0.1L` Molarity `= ("Number of moles")/("Volume in liters") = (0.01 mol)/(0.1) = 0.01M` |
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| 569. |
`6.02xx10^(20)` molecules of urea are present in `100 mL` solution. The concentration of urea solution is:A. (a)`0.01 M`B. (b)`0.001 M`C. (c )`0.1 M`D. (d)`0.02 M` |
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Answer» Correct Answer - a `6.02xx10^(23)` molecules is 1 mole `6.02xx10^(20)` molecules is `10^(-3)` mole `100 mL` has `10^(-3)` mole `1000 mL` has `(10^(-3))/100xx1000=10^(-2)` |
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| 570. |
The empirical formula fo a compound is `CH_(2)O`. Its vapor density is `30`. It reacts with sodium metal. The compound isA. `HCOOCH_(3)`B. `CH_(2)O`C. `CH_(3)CH_(2) COOH`D. `CH_(3)COOH` |
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Answer» Correct Answer - D `MF = n xx EF` `n = ("Molecular mass")/(EFM) = (2 xx "Vapor density")/(EFM), n = 2` `:. MF = C_(2) H_(4) O_(2)` Since the compound contains acidic `H` atom, it is acetic acid,. `CH_(3) COOH + Na rarr (1)/(2) H_(2) + CH_(3) COONa` |
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| 571. |
`6.02xx10^(20)` molecules of urea are present in `100 mL` solution. The concentration of urea solution is:A. (a)0.1 MB. (b)0.01 MC. (c )0.02 MD. (d)0.001 M |
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Answer» Correct Answer - B `M=("Moles of urea")/("V in litre")` `=(6.02xx10^(20)xx1000)/(6.023xx10^(23)xx100)=10^(-2) M` |
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| 572. |
x g of Ag was dissolved in `HNO_(3)` and the solution was precipitated. The value of x isA. `1*08 g`B. `2*16 g`C. `2*70 g`D. `1*62 g` |
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Answer» Correct Answer - B `143.5 g` AgCl contains Ag = 108 g `2*87` g AgCl contains Ag `= (108xx2*87)/(143*5) = 2*16 g` |
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| 573. |
Standard molar volume of all gases isA. `22.4 L`B. `22.4 m^(3)`C. `22.4 ml`D. `22.4 cm^(3)` |
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Answer» Correct Answer - A Volume occupied by 1 mol of ideal gas under standard conditions is `22.4 L` |
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| 574. |
Which of the following has the minimum mass?A. `6.02xx10^(22)H_(2)` moleculesB. `1120 c c` of `CO_(2)` at `STP`C. `0.1` mol of `NH_(3)`D. `0.1g` atom of carbon |
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Answer» Correct Answer - A (1) `n_(H_(2)) = (6.02xx10^(22))/(6.02xx10^(23)) = 0.1` mol `:. M_(H_(2)) = (0.1 mol) (2g mol^(-1)) = 0.2g` (2) `n_(CO_(2))=(1120 c c)/(22400c c"mol"^(-1))=0.05"mol"` `:. M_(CO_(2)) = (0.05 "mol") (44g "mol"^(-1)) = 2.2g` (3) `m_(NH_(3)) = (0.1 "mol") (17g "mol"^(-1)) = 1.7g` (4) `0.1g` atom of `C = 0.1` mol `= (0.1 "mol") (12 g "mol"^(-1)) = 1.25` |
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| 575. |
When 400 g of a `20%` solution by weight was cooled, 50 g of solute precipitated. What is the percentage by mass of solute in the remaining solution ?A. `8.57%`B. `15%`C. `12.25%`D. `9.5%` |
| Answer» Correct Answer - A | |
| 576. |
Normality of 2 M sulphuric acid isA. (a)2 NB. (b)4 NC. (c )`N/2`D. (d)`N/4` |
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Answer» Correct Answer - B `H_(2)PO_(4)` is dibasic `N=2 M=2xx2=4`. |
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| 577. |
The solution of sulphuric acid contains 80% by weight `H_(2)SO_(4)`. Specific gravity of this solution is 1.71. Its normality is aboutA. (a)18.0B. (b)27.9C. (c )1.0D. (d)10.0 |
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Answer» Correct Answer - B `N=(10xxsp. "Gr. Of the solution" xx" wt". % "of solute"xx"Mol. Wt".)/("Molecular wt. of solute"xx"Eq. Wt".)` `N=(10xx1.71xx80xx98)/(98xx49)=27.9` |
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| 578. |
Which of the following solution has normality equal to molarity ?A. `H_(2) SO_(4)` aqueous solutionB. `H_(3) PO_(4)` aqueous solutionC. `HNO_(3)` aqueous solutionD. `Mg (OH)_(2)` aqueous solution |
| Answer» Correct Answer - C | |
| 579. |
Normality of `2`M sulphuric acid isA. (a)`2 N`B. (b)`4N`C. (c )`N/2`D. (d)`N/4` |
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Answer» Correct Answer - a `H_(2)SO_(4)` is dibasic `N=2M=2xx2=4` |
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| 580. |
The set of numerical coefficients that balances the chemical equation `K_(2)CrO_(4)+HCltoK_(2)Cr_(2)O_(7)+KCl+H_(2)O`A. `1,1,2,2,1`B. `2,2,1,1,1`C. `2,1,1,2,1`D. `2,2,1,2,1` |
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Answer» Correct Answer - D The balanced equation is `2K_(2)CrO_(4)+2HCl to K_(2)Cr_(2)O_(7)+2KCl+H_(2)O` |
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| 581. |
The numerical value of N/n (where N is the number of molecules in a give sample of gas and n is the number of moles of gas) isA. `8.314`B. `6.02xx10^(23)`C. `0.0821`D. `1.66xx10^(-19)` |
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Answer» We know, `6.02xx10^(23)` molecules (N) = 1 mole (n) `:. (N)/(c) = (1)/(6.02xx10^(23))=1.62xx10^(-24)` |
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| 582. |
The mass of 11.2 L of ammonia gas at S.T.P. isA. `8.5 g`B. `85 g`C. `17 g`D. `1.7 g` |
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Answer» Correct Answer - A Molar mass of `NH_(3)=14+3xx1=17g mol^(-1)` `:.` Mass of 22.4 L of `NH_(3)` at S.T.P. = 17 g `:.` Mass of 11.2 L of `NH_(3)` at S.T.P. = 8.5 g |
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| 583. |
The same amount of a metal combines with 0.20 g of oxygen and with 3.17 g of a halogen. Hence equivalent mass of halogen isA. 127 gB. 80 gC. `36.5` gD. 9 g |
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Answer» Correct Answer - A `0.20` g oxygen `equiv 3.17` g halogen `therefore` Equivalent mass of halogen `equiv (3.17)/0.20 xx 8 = 126.8` g `~=127` g |
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| 584. |
The same amount of a metal combines with 0.20 g of oxygen and with 3.17 g of a halogen. Hence equivalent mass of halogen isA. `127 g`B. `80 g`C. `35.5 g`D. `9 g` |
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Answer» Correct Answer - A 0.20 g of oxygen = 3.17 g of halogen 8 g of oxygen `= (3.17)/(0.20)xx8`g of halogen `= 126.8g` `:.` Eq. mass of halogen `=126.8~~127` |
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| 585. |
0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metal isA. 12B. 24C. 18D. 36 |
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Answer» a) 1 mole of `O_(2)` = 4 equivalent of oxygen 22400 mL of `O_(2)` = 4 equivalent of oxygen 46.6 mL of `O_(2)` = (4/22400) xx 46.6 = 0.00832 eq. Equivalent of metal = Equivalent of oxygen (Weight)/(Equivalent) = 0.00832 0.1/E = 0.00832 `rARr` E= (0.1)/(0.00832) = 12.0 |
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| 586. |
The sulphate of a metal `M` contains `9.87%` of `M`, This sulphate is isomorphous with `ZnSO_(4).7H_(2)O`. The atomic weight of `M` isA. `40.3`B. `36.3`C. `24.3`D. `11.3` |
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Answer» Correct Answer - C As the given sulphate is isomorphous with `ZnSO_(4).7H_(2)O` its formula would be `MSO_(4).7H_(2)O`m is the atomic weight of M, molecular weight of `mSO_(4).7H_(2)O = m + 32 + 64 + 126 m + 222` Hence % of M `= (m)/(m + 222) xx 100 = 9.87` (given) or 100m `= 9.87 m + 222 xx 9.87 or 90.13m = 222 xx 9.87` or `m = (222 xx 9.87)/(90.13) = 24.3` |
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| 587. |
The sulphate of a metal A contains 20% of M. This sulphate is isomorphous with `ZnSO_(4).7H_(2)O`. The atomic mass of M isA. `12`B. `24`C. `36`D. `48` |
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Answer» Correct Answer - B `("Mass of metal sulphate")/("Mass of metal")` `("Eq. mass of M + Eq. mass of" `SO_(4)^(2-)`)/("Eq. mass of M")` `(100)/(20)=(E+48)/(E)` `5E =E+48` `4E=48` or `E=12` As metal sulphate is isomosphous with `ZnSO_(4).7H_(2)O` Valency of M = Valency of Zn =2 Hence At. mass of M = Eq. mass `xx` valency `=12xx2=24` |
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| 588. |
Express `-20^(@)C` in kelvin scale. |
| Answer» Correct Answer - 253.15 K | |
| 589. |
Express 900.00 in scientific notation. |
| Answer» Correct Answer - `9xx10^(2)` | |
| 590. |
On heating `100g` of `Na_(2)SO_(4) 10 H_(2) O` will loose ..... % of water.A. `44.1`B. `65.3`C. `55.9`D. `34.7` |
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Answer» Correct Answer - C It is equal to the mass precent of wate. Percent water `= ("Mass of" H_(2) O "in 1 mol of compound")/("Molar mass of compound") xx 100%` `= (180)/(320) xx 100% = 59.9%` |
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| 591. |
At `100^(@)C` and `1 atm`, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.0006 g cm^(-3)`, then the volume occupied by water molecules in `1 L` of steam at this temperature isA. (a)`6 cm^(3)`B. (b)`60 cm^(3)`C. (c )`0.6 cm^(3)`D. (d)`0.06 cm^(3)` |
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Answer» Correct Answer - D For water vapour, `P=0.0006 g c c^(-1)` `:. 0.0006="Mass"/"Volume"="Mass"/1000` Mass`=1000xx0.0006=0.6` g`:. 0.0006="Mass"/"Volume"="Mass"/1000` Mass`=1000xx0.0006=0.6` g the density of liquid water is `1 g c c^(-1)` So, the volume occupied by water is `"Mass"/"Density"=0.6/1=0.6 c c` |
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| 592. |
The modern atomic weight scale is based onA. `C^(12)`B. `O^(16)`C. `H^(1)`D. `C^(13)` |
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Answer» Correct Answer - A The modern atomic weight scale is based on `C^(12)` |
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| 593. |
Express the following in the scientific notation: a. `0.0048` b. `234000` c. `8008` d. `500.0` e. `6.0012` |
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Answer» Correct Answer - `(i) 4.8 xx 10^(-3) ` `(ii) 2.34 xx10^(5)` (iii) `8.008xx 10^(3)` (iv) `5.000xx10^(2)` `(v) 6.0012` `(i) 0.0048=4.8 xx10^(-3)` `(ii) 234,000=2.34xx10^(5)` `(iii) 8008=8.008 xx10^(3)` (iv) `500.0=5.000xx10^(2)` (v) `6.0012=6.0012` |
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| 594. |
A resucing agent is a substance which canA. Accept electronB. Donate electronsC. Accept protonsD. Donate protons |
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Answer» Correct Answer - B A substance which is capable of reducing other substances and is capable of donating electrons during reduction is called agent or reductant. |
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| 595. |
Which one is oxidising substance?A. `C_(2)H_(2)O_(2)`B. `CO`C. `H_(2)S`D. `CO_(2)` |
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Answer» Correct Answer - D `CO_(2)` is an oxidizing agent. |
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| 596. |
A divalent metal has 12 equivalent weight. The molecular weight of its oxide isA. 16B. 32C. 40D. 52 |
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Answer» Correct Answer - C Equivalent weight of metal = 12 Equivalent weight of oxygen = 8 Equivalent weight of MO (oxide) with divalent metal = 20 Thus, molecular weight of metal oxide = 40 |
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| 597. |
The equivalent mass of chlorine is 35.5, and the molar mass of copper is 63.5. The equivalent mass of copper chloride is 99.0. Hence, formula of copper chloride isA. `CuCl`B. `Cu_(2)Cl`C. `CuCl_(2)`D. None of these |
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Answer» Correct Answer - A Eq. mass of copper chloride = 99 Eq. mass of chlorine = 35.5 `:.` Eq. mass of copper = 99-35.5 = 63.5 `:.` Valency of copper `= ("At mass of copper")/("Eq. mass of copper") = 1` `:.` Formula of copper chloride is `CuCl` |
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| 598. |
In the combustion of 5.00 g of a metal, 9.44 g of metal oxide are formed. Hence, equivalent mass of the metal isA. `4.44` gB. `9.00` gC. `5.00` gD. `2.22` g |
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Answer» Correct Answer - B Matel `= 5. 00` g Metal oxide ` = 9. 44` g Oxygen combined `= 4. 44` g `4.44` g of oxyen combined with `= 5. 00` g metal `therefore 8 ` g oxygen combined with `= 5.00/4.44 xx 8` g metal `=9.00` g |
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| 599. |
Arsenic forms two oxides, one of which contains 65.2 % and the other 75.7% of the element. Hence, equivalent masses of arsenic are in the ratioA. `1:2`B. `3:5`C. `13:15`D. `2:1` |
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Answer» Correct Answer - B In first oxide Mass of arsenic = 65.2 Mass of oxygen = 34.8 `:.` Eq. mass of arsenic `= (65.2)/(34.8)xx8 = 14.99` In second oxide, Mass of arsenic = 75.7 g Mass of oxygen = 24.3 g `:.` Eq. mass of arsenic `= (75.7)/(24.3)xx8` `=24.92` `underset(("oxide I"))("Eq. mass of arsenic")" : "underset(("oxide II"))("Eq. mass of arsenic")` `14.99:24.92` or 3:5 |
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| 600. |
Red colored compound , hemoglobin present in blood contains `0.355% Fe (AM = 56u)` If four atoms of `Fe` are present per molucule of hemoglobin, its molecular mass would beA. `63098 u`B. `78654 u`C. `54786 u`D. `98036 u` |
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Answer» Correct Answer - A As there are 4 atoms of `Fe` per molecule, we have `% Fe = (4 xx AM of Fe)/("MM of hemoglobin") xx 100%` (`AM` is atomic mass, `MM` is molar mass) `0.335% = (4 (56u))/(MM) xx 100%` `:. MM = (4(56u))/(0.335%) xx 100% = 63098 u` |
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