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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.
| 51. |
The normality of `0.3 M` phosphorous acid `(H_(3) PO_(3))` isA. (a)0.1B. (b)0.9C. (c )0.3D. (d)0.6 |
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Answer» Correct Answer - D `H_(3)PO_(3)` is a dibasic acid and thus its molecular weight `=2xx "equivalent" wt`. `:. 1 M=2N` Hence, `0.3 M=2xx0.3=0.6 N` |
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| 52. |
Assertion: The weight percentage of a compound A in a solution is given by `% of A=("Mass A")/("Total mass of solution")xx100` Reason: The mole fraction of a component A is given by, Mole fraction of A `=("No. of moles of A")/("Total no. of moles of all components")`A. (a)If both assertion and reason are true and the reason is the correct 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 assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - B Assertion and reason both are two formula independent to each other. |
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| 53. |
The molarity of orthophosphoric acid having purity of 70% by weight and specific gravity 1.54 would beA. 11 MB. 22 MC. 33 MD. 44 M |
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Answer» Correct Answer - A 70% by weight 70 gm `H_(3)PO_(4) rarr 100` gm solution/sample `V=W/d=100/1.54" "M=(70xx1000)/(98xx100//1.54)=11M` |
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| 54. |
Sulphuric acid and orthophosphoric acid have the same molecular mass. Ratio of the masses of these acids needed to neutralise the same amount of an alakli if the sulphate and dihydrogen orthophosphate were formed, is:A. `1:2`B. `2:1`C. `1:3`D. `1:1` |
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Answer» Correct Answer - A `H_(2)SO_(4)+2OH^(-)toSO_(4)^(2-)+2H_(2)O` `H_(3)PO_(4)+OH^(-)toH_(2)PO_(4)^(-)+H_(2)O` For same amount of alkali, `H_(2)SO_(4)` and `H_(3)PO_(4)` required are in the ratio 1:2 (by number of moles or by weight) |
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| 55. |
Haemoglobin contains 0.33% of iron by weight. The molecular weight of heamoglobin is approximately 67200. The number of iron atoms (At. Wt. of Fe=56) present in one molecule of haemoglobin isA. (a)6B. (b)1C. (c )4D. (d)2 |
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Answer» Correct Answer - C `:. 100 g Hb` contains`=0.33 g Fe` `:. 67200 g Hb=(67200xx0.33)/100 g Fe` Gram atom of `Fe=(672xx0.33)/56=4`. |
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| 56. |
The haemoglobin from the red blood corpuscles of most mammals contains approximately `0.33%` of iron by weight. The molecular weight of haemoglobin as `67,200`. The number of iron atoms in each molecule of haemoglobin is (atomic weight of iron `=56`):A. `3`B. `4`C. `2`D. `6` |
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Answer» Correct Answer - B 100 g haemoglobin has = 0.33 g Fe 67200 g haemoglobin has Fe `= (0.33)/(100)xx67200=221.76g` 1 mole of `N_(0)` molecules of haemoglobin has `= (221.76)/(56)`g atom of Fe `= 3.96` g-atom of `Fe ~= 4.0 g` atoms |
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| 57. |
How many grams of a dibasic acid (Mol. Wt. =200) should be present in 100 ml of its aqueous solution to give decinormal strengthA. (a)`1 g`B. (b)`2 g`C. (c )`10 g`D. (d)`20 g` |
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Answer» Correct Answer - c For dibasic acid `E=M/2=2000/2=100` `N=(Wxx1000)/(ExxV("in mL"))` `1/10=(Wxx1000)/(100xx100)=W=1 g`. |
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| 58. |
How many grams of a dibasic acid (Mol. Wt. =200) should be present in 100 ml of its aqueous solution to give decinormal strengthA. (a)1 gB. (b)2 gC. (c )10 gD. (d)20 g |
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Answer» Correct Answer - A For Dibasic acid `E=M/2=200/2=100` `N=(Wxx1000)/(ExxV("in ml"))` `1/10=(Wxx1000)/(100xx100)=W =1 g` |
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| 59. |
How much water should be added to 200 mL of semi normal solution of NaOH to make it exactyly decinormal?A. 200 mLB. 800 mLC. 1000 mLD. 11200 mL |
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Answer» b) `N_(1)V_(1) = N_(2)V_(2)`, `1/2 xx 200 = 1/10 xx V_(2)` `V_(2) = 1/2 xx 200 xx 10 = 1000mL` `therefore` Volume of water added = 100-200 = 800 mL |
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| 60. |
1.5 litre of a solution of normality N and 2.5 litre of 2 M HCl are mixed together. The resutant solution had a normality 5. The value of N isA. (a)6B. (b)10C. (c )8D. (d)4 |
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Answer» Correct Answer - b Eq. of 1.5 litre solution +Eq. of 2.5 litres solution =Eq. of resultant of solution `implies 1.5xxN2.5xx2=4xx5` `:. N=(20-5)/1.5=15/1.5=10` |
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| 61. |
0.32 g of metal gave on treatment with an acid 112 mL of hydrogen at NTP. Calculate the equivalent weight of the metalA. 58B. 32C. 11.2D. 24 |
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Answer» Correct Answer - B Equivalent weight `=("Mass of metal "xx11200)/("Volume in mL of hydrogen")` Given, mass of metal `=0.32 g` volume of hydrogen at `NTP=112 mL` Equivalent weight `=(0.32xx11200)/112=32` |
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| 62. |
What volume at `STP` of ammonia gas will be required to be passed into `30 mL` of `1N H_(2) SO_(4)` solution to bring down the acid normality to `0.2 N`?A. (a)`556.5 mL`B. (b)`480.5 mL`C. (c )`537.6 mL`D. (d)`438.4 mL` |
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Answer» Correct Answer - c Meq. of `H_(2)SO_(4)` (original) `=30xx1=30` Meq. of `H_(2)SO_(4)` after passing `NH_(3)=30xx0.2=6` Meq. of `H_(2)SO_(4)` reacted =Meq. of `NH_(3)` `=30-6=24` `:. w_(NH_(3))/17xx1000=24, w_(NH_(3))=0.408 g` `:. V_(NH_(3))` at `STP=0.408/17xx22.4=0.5376 L` `=537.6 mL` |
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| 63. |
5 mL of N HCl, 20 mL of N/2 `H_(2)SO_(4)` and 30 mL of `N/3HNO_(3)` are mixed together and volume made to 1L. The normality of resulting solution isA. 0.45B. 0.025C. 0.9D. 0.05 |
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Answer» Normally equtaion is `N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(4)(V_(1)+V_(2)+V_(3))` or `1xx5+20xx(1)/(2)+30xx(1)/(3)=N_(4)(5+20+30)` `(HCI) (H_(2)SO_(4)) (HNO_(3))` `therefore` Resulting normality `(N_(4))=(25)/(55)=0.45N`. |
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| 64. |
1.82 g of a metal requires 32.5 ml of 1 N HCI to dissolve it . What is the equivalent weight of metal ?A. 46B. 65C. 56D. 42 |
| Answer» Correct Answer - C | |
| 65. |
The normality of solution obtained by mixing 100 ml of 0.2 M `H_(2) SO_(4)` with 100 ml of 0.2 M NaOH isA. 0.1B. 0.2C. 0.5D. 0.3 |
| Answer» Correct Answer - A | |
| 66. |
`1 g` of calcium was burnt in excess of `O_(2)` and the oxide was dissolved in water to make up `1 L` solution. Calculate the normality of alkaline soluiton.A. 0.04B. 0.4C. 0.05D. 0.5 |
| Answer» Correct Answer - C | |
| 67. |
Percent of solute: Calculate the mass of calcium suphate `(CaSO_(4))` contained in `200g` of a `6.00%` solution of `CaSO_(4)` Strategy: Percent infromation means the solution contains `6.00g` fo `CaSO_(4)` per `100g` of solution. To solve the problem, we can use either unitary method or Eq (1.16) or construct a unit factor. |
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Answer» (i) Unitary method: `100g` solution contains `6.00g CaSO_(4)` `1g` solution contains `(6.00)/(100)g CaSO_(4)` `200g` solution contains `(6.00)/(100) xx 200g CaSO_(4)` `= 12.0g CaSO_(4)` (ii) Using Eq (1.16) Percent solute `= ("Mass of solute")/("Mass of solution") xx100%` `:.` Mass of solute `= (("Percent solute") xx ("Mass of solution"))/(100%)` `= ((6.00%) (200g))/(100%) = 120g CaSO_(4)` (iii) Factor-label method `?g CaSO_(4) = 200g "soln" xx (6.00g CaSO_(4))/(100g "soln".) = 12.0g CaSO_(4)` |
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| 68. |
A solution is prepared by dissolved 2 g of a solute in 36 g of water. Calculate the mass percent of the solute. |
| Answer» Correct Answer - `5.26%` | |
| 69. |
A molal solution is one that contains 1 mol of a solute dissolved inA. `22.4 L` of solutionB. `1 L` of solutionC. `1 L` of solventD. `1000 g` of solvent |
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Answer» Correct Answer - D Molarity of a solution is the moles of solute per kilogram `(1000g)` of solvent. |
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| 70. |
`HNO_(2)` acts both as reductant and as oxidant, while `HNO_(3)` acts only as oxidant. It is due to theirA. Solubility abilityB. Maximum oxidation numberC. Minimum oxidation numberD. Minimum number of valence electrons |
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Answer» Correct Answer - B In `Hoverset(**)(N)O_(2)` oxidation number of `N=+3` In `Hoverset(**)NO_(3)` oxidation number of `N=+5`. |
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| 71. |
In `C+H_(2)O rarr CO+H_(2)`, `H_(2)O` acts asA. Oxidising agentB. Reducing agentC. (a) and (b) bothD. None of these |
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Answer» Correct Answer - A In this reaction `H_(2)O` acts as oxidising agent. |
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| 72. |
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are `69.9` and `30.1`, respectively.A. FeOB. `Fe_3O_4`C. `Fe_2O_3`D. `FeO_2` |
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Answer» For element Fe, mole of atoms `=(69.9)/(56)=1.25` For element O, mole of atoms `=(30.1)/(16)=1.88` Mole ratio of Fe `=(1.25)/(1.25)=1`, Mole ratio of O `=(1.88)/(1.25)=1.5` Simplest whole number ratio of Fe and O= 2, 3 Empirical formula of compound `=Fe_2O_3` Molecular mass of `Fe_2O_3=160` `n=("Molecular mass")/("Empirical formula mass")=160/160=1` Molecular formula `=Fe_2O_3` |
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| 73. |
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are `69.9` and `30.1`, respectively.(molecular mass is 159.8). |
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Answer» Mass percent of iron (Fe) = 69.9 % (Given) Mass percent of oxygen (O) = 30.1% (Given ) number of moles of iron present in the oxide ` (69.90)/(55.85)` =1.25 Number ofmoles of oxygen present in the oxide `=(30.1)/(16.0)` =1.88 Ratio of iron to oxygen in the oxide , =1.25 :1.88 `=(1.25)/(1.25):(1.88)/(1.25)` =1:1.5 =2:3 ` therefore` The empirical formula of the oxide is `Fe_(2)O_(3)`. Empirical formula mass of `Fe_(20O_(3) =[2(55.85)+3(16.00)]g` Molar mass of `Fe_(2)O_(3) = 159.69 g` `therefore n=("molar mass")/("Emprical Formula Mass ") =(159.96 g)/(159.7 g)` =.0999 =1 ( approx) Molecular formula of a compound i9s obtined by mutiplying the empircal From with n thus , the empirical formula of the given oxide is `Fe_(2)O_(3)` and n is 1. Hence , the molecular formula of the oxide is `Fe_(2)O_(3)` |
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| 74. |
Mass of one atom of the element X is `1.66xx10^(-24) g`. Number of atoms in 1 g of the element is:A. `1.66xx10^(-24)`B. `1.66xx10^(24)`C. `1.66xx10^(-24)xxN_(A)`D. `6.02xx10^(23)` |
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Answer» Correct Answer - D `1 g = (1)/(1.66xx10^(-24)gatom^(-1)) = (10^(24))/(1.66) atom` `= 6.02xx10^(23) atoms` |
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| 75. |
Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc, Following reaction takes place `Zn+2HClrarrZnCl_(2)+H_(2)` Calculate the voluem of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP, atomic mass of Zn=65 .3uA. 10.03 LB. 11.35 LC. 11.57 LD. 9.53 L |
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Answer» `underset(65.3g)(Zn) + 2HCl to ZnCl_2 + underset (22.7L)(H_2)` According to the equation, 65.3g of Zn liberates = 22.7 L of `H_2` 32.65g of Zn will liberate `=(22.7)/(65.3)xx32.65=11.35L` |
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| 76. |
The maximum amount of `BaSO_(4)` precipitated on mixing equal volumes of `BaCl_(2)` (0.5 M) with `H_(2)SO_(4)` (1 M) will correspond toA. 0.5 MB. 1.0 MC. 18 molesD. 100 moles |
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Answer» Correct Answer - A `BaCl_(2)+H_(2)SO_(4) rarr BaSO_(4)+2HCl` One mole of `BaCl_(2)` reacts with one mole of `H_(2)SO_(4)`. Hence 0.5 mole will react with 0.5 mole of `H_(2)SO_(4)` i.e. `BaCl_(2)` is the limiting reagent. |
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| 77. |
An element X has the following isotopic composition: `.^(200)X:90% .^(199)X:8.0% .^(202)X:2.0%` The weight average atomic mass of the naturally occurring element X is closest toA. (a)`201` amuB. (b)202 amuC. (c )199 amuD. (d)200 amu |
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Answer» Correct Answer - d Average isotopic mass of `X=(200xx90+199xx8+202xx2)/(90+8+2)` `=(18000+1592+404)/100=19996/100` `=199.96` amu |
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| 78. |
An element X has the following isotopic composition `.^(200)X : 90%, .^(199)X: 8.0%, .^(202)X:2%` . The Weighed average atomic mass of naturally occurring element X is closet toA. 199 a.m.u.B. 200 a.m.u.C. 201 a.m.u.D. 202 a.m.u. |
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Answer» Correct Answer - B Average atomic mass, `overset(-)(A)=Sigmaf_(i)A_(i)` (`f_(i)`=fractional abundance, `A_(i)`=atomic mass) Atomic mass `=0.90xx200+0.08xx199+0.02xx202` `=180+15.92+4.04~~200` |
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| 79. |
An element `X` has the following istopic compositon: `.^(200)X(90%), .^(199)X(8.0%), .^(202)X(2.0%)` The weighted average atomic mass of the naturally occuring element `X` is closent toA. `202 am u`B. `200 am u`C. `199 am u`D. `201 am u` |
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Answer» Correct Answer - B Average atomic mass = sum total of (Fractional abundance `xx` Isotopic mass) `= (0.9)(200) + (0.08)(199) + (0.02) (202)` `= 180 + 15.92 + 4.04` `= 199.96 ~~ 200` |
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| 80. |
The ionic stregth of `Na^(+)` on mixing `100 mL 0.1 NaCl` and `100 mL 0.1 N Na_(2)SO_(4)` is:A. (a)`0.2`B. (b)`0.1`C. (c )`0.3`D. (d)`0.075` |
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Answer» Correct Answer - d `M_(1)V_(1)+M_(2)V_(2)=M_(3)V_(3)` `0.1xx100+2xx0.1xx100=x+200` Conc. Of `Na^(+)=(100xx0.1)/200+(100xx0.1xx2)/200=0.15 M` :. Ionic strength`=1/2 CZ^(2)=1/2xx[0.15xx1^(2)]=0.075` |
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| 81. |
The oxidation number of Ba in barium peroxide isA. `+6`B. `+2`C. `1`D. `+4` |
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Answer» Correct Answer - B `+2` it is a second group element. |
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| 82. |
One mole of `KClO_(3)` is heated in presence of `MnO_(2)`. The produced oxygen is used in burning of Al. Then oxide of Al that will be formed.A. 2 moleB. 1 moleC. 4 moleD. 3 mole |
| Answer» Correct Answer - B | |
| 83. |
The total number of ions present in 1 ml of 0.1 M barium nitrate solution isA. `6.02 xx 10^(8)`B. `6.02 xx 10^(10)`C. `3.0 xx 6.03 xx 10^(19)`D. `3.0 xx 6.02 xx 10^(8)` |
| Answer» Correct Answer - C | |
| 84. |
31.3 gram mixture NaBr and NaCl treated with `H_(2)SO_(4).28.4` gram of `Na_(2)SO_(4)` is produced. Then calculate the amount of NaCl and NaBr in the mixture. |
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Answer» Correct Answer - NaCl = 13.01 g NaBr = 18.28 g |
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| 85. |
What is the molarity of NaOH solution if 250 mL of it contains 1 mg of NaOH ?A. `10^(-1) M`B. `10^(-2) M`C. `10^(-4) M`D. `10^(-3) M` |
| Answer» Correct Answer - C | |
| 86. |
Assertion : Components of a homogeneous mixture cannot be separated by using physical methods Reason : Composition of homogeneous mixture is uniform throughtout as the components react to form a single compound.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» In homogeneous mixture, the components completely mix with each other and hence the composition is uniform throughout. Formation of homogeneous mixture is a physical chage as no chemical reaction occurs between the components and no new compound in formed, hence the components can be separated physical methods. | |
| 87. |
The percentage of nitrogen in `HNO_(3)` isA. `22.22%`B. 0.35C. 0.2857D. 0.45 |
| Answer» Correct Answer - A | |
| 88. |
The correct relationship between molecular mass (M) and vapour (VD) isA. VD = 2 MB. `VD = (M)/(2)`C. VD = `M^(2)`D. M = `(VD)^(1/2)` |
| Answer» Correct Answer - B | |
| 89. |
STATEMENT -1 : Solution is the example of homogeneous mixture. STATEMENT -2 : Homogeneous mixture have uniform composition. STATEMENT -3 : A solution may contain more than one solute.A. TTTB. FTTC. TTFD. FFT |
| Answer» Correct Answer - A | |
| 90. |
The precentage composition of calcium , carbon and oxygen in `C_(a) CO_(3)` respectively areA. 40% , 12% and 48%B. 12% , 40% and 48%C. 40% , 48% and 12%D. 12% , 48% and 40% |
| Answer» Correct Answer - A | |
| 91. |
Most combustion reactions occur in excess of `O_(2)` (i.e., more than enough `O_(2)` to burn the substance completely). Calculate the mass of `CO_(2)` (in grams) produced by buring `4.00 mol` of `CH_(4)` in excess `O_(2)`. Strategy: using the balanced equation `{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}` we find out that 1 mol of `CH_(4)` is chemically equivalent to 1 mol of `CO_(2)` or `44.0g` of `CO_(2)`. |
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Answer» According to the equation, `n_(CO_(2))`, produced is equal to `n_(CH)` used. Thus, 4 mol `CH_(4)` produces 4 mol `CO_(2)`. The molar mass of `CO_(2)` is `44g`. Thus, the mass of 4 mol `CO_(2)` is `4xx44 = 176g`. Alternatively, using the right unit factors, `?g CO_(2) = 4.00 "mol" CH_(4) xx (1 "mol" CO_(2))/(1 "mol" CH_(4)) xx (44.0 g CO_(2))/(1 "mol" CO_(2))` `= 1.76xx10^(2) g CO_(2)` |
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| 92. |
Oxidation state of Fe in `Fe_(3)O_(4)` isA. `3/2`B. `4/5`C. `5/4`D. `8/3` |
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Answer» Correct Answer - D `overset(**)(F)e_(3)O_(4)` `3x+(-8)=0, 3x-8=0` `3x=8 implies x=8/3`. |
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| 93. |
1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?A. (a)`Mg, 0.16 g`B. (b)`O_(2), 0.16 g`C. (c )`Mg, 0.44 g`D. (d)`O_(2), 0.28 g` |
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Answer» Correct Answer - d The balanced chemical equation is `Mg+1/2O_(2) rarr MgO` `24 g 16 g 40 g` From the above equation, it is clear that, `24 g Mg` reacts with `16 g O_(2)`. Thus, `1.0 g Mg` reacts with `16/24xx0.67 g O_(2)=0.67 g O_(2)` But only `0.56 g O_(2)` is available which is less than `0.67 g`. Thus, `O_(2)` is the limiting reagent. Further, `16 g O_(2)` reacts with `24 g Mg`. `:. 0.56 g O_(2)` react with `Mg=24/16xx0.56=0.84 g` :. Amount of Mg left unreacted`=1.0-0.84 g Mg` `=0.16 g Mg`. Hence, Mg is present in excess and 0.16 g Mg is left behind unreacted. |
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| 94. |
1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?A. `Mg, 0.44 g`B. `O2, 0.28 g`C. `Mg, 0.16 g`D. `O2, 0.16 g` |
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Answer» Correct Answer - C 24 g Mg requires 16 h oxygen `:.` 0.56 g oxygen requires 0.84 g `Mg. :. Mg` left `=0.16 g` |
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| 95. |
Formula unit mass of `K_(2) CO_(3)` is (Atomic mass of `K = 39 u , C = 12 u` and O = 16 u)A. 67 uB. 138 uC. 150 uD. 134 u |
| Answer» Correct Answer - B | |
| 96. |
0.56 g of a gas occupies `280 cm^(3)` at N.T.P., then its molecular mass isA. `4.8`B. `44.8`C. `2`D. `22.4` |
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Answer» Correct Answer - B `280 cm^(3)` of the gas at N.T.P. has mass = 0.56g `:. 22400 cm^(3)` of the gas at N.T.P. has mass `= (0.56g)/(280cm^(3))xx22400cm^(3)` `= 44.8 g` `:.` Molecular mass of the gas = 44.8 |
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| 97. |
The mass of oxygen that would be required to produce enough CO which completely reduces `1*6` kg `Fe_(2)O_(3)` (at. mass of Fe = 56)A. 240 gB. 480 gC. 720 gD. 960 g |
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Answer» Correct Answer - B `3C+(3)/(2)O_(2)to3CO`, `Fe_(2)O_(3)+3COto2Fe+3CO_(2)` 1 mol of `Fe_(2)O_(3) -= 3//2` mol of `O_(2)` `160 g` (GMM) of `Fe_(2)O_(3)` required `O_(2) = (3)/(2)=32 = 48g` `1*6` kg of `Fe_(2)O_(3)` required `O_(2) = 480g` |
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| 98. |
1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?A. Mg, `0.16` gB. `O_(2),0.16` gC. Mg, `0.44` gD. `O_(2), 0.28` g |
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Answer» Correct Answer - A The balanced chemical equaiton is `underset (24 " g")(Mg)+underset(16" g")(1/2O_(2))rarrunderset(40g) (MgO)` From the above equation, it is clear that 24 g Mg reacts with 16 g `O_(2)` Thus, `1.0` g Mg reacts with `16/24 O_(2) = 0.67 g O_(2)` But only `0.56 g O_(2)` is available which is less then `0.67` g. Thus, `O_(2)` is the limiting reagent. Further, 16 g `O_(2)` reacts with 24 g Mg. `therefore 0.56 g O_(2)` will react with Mg `+ 24/16xx0.56= 0.84` g `therefore` Amount of Mg left unreacted `= 1.0 - 0.84 = 0.16` g Mg Hence, Mg is present is excess and `0,16` g Mg is left behind unreacted. |
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| 99. |
The number of moles of nitrogen atom in 56 g nitrogen isA. 2 molB. 4 molC. 8 molD. 10 mol |
| Answer» Correct Answer - B | |
| 100. |
The number of particles presents in 1 mol of nitrogen atom areA. `6.022 xx 10^(25)`B. `6.022 xx 10^(24)`C. `6.022 xx 10^(23)`D. `6.022 xx 10^(22)` |
| Answer» Correct Answer - C | |