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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.
| 201. |
If the molecular weight of `H_(3) PO_(3)` is M, its equivalent weight will beA. MB. M/2C. M/3D. 2M |
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Answer» Correct Answer - B Dibasic acid `E=M//2` |
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| 202. |
STATEMENT -1 : `H_(3)PO_(4)` is a tribasic acid. and STATEMENT -2 : In `H_(3)PO_(4)`, only two H-atoms are replaceable.A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for Statmenet -1.B. Statement -1 is True, Statement -2 is True, Statement-2 is NOT a correct explanation for Statement -1C. Statement -1 is True, Statement -2 is FalseD. Statement -1 is False, Statmenet -2 is True |
| Answer» Correct Answer - C | |
| 203. |
`74.5 g` of a metallic chloride contain `35.5 g` of chlorine. The equivalent weight of the metal isA. `19.5`B. `35.5`C. `39.0`D. 78.0` |
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Answer» Correct Answer - C Equivalent weight of metal `=("Weight of metal")/("Weight of chlorine combined")xx35.5` `therefore` Equivalent weight of metal `=((74.5-35.5))/(35.5)xx 35.5=39.0` |
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| 204. |
In the reaction `CA(OH)_(2)+H_(3)PO_(4) rarr Ca_(3)(PO_(4))_(2)+H_(2)O`, the equivalent mass of `H_(2)`O is: (M molecular mass)A. (a)MB. (b)M/2C. (c )M/3D. (d)M/6 |
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Answer» Correct Answer - A In acid-base neutralisation `1 H^(+)` combines with `1 OH^(-)` to produce `1 H_(2) O` therefore its equivalent mass=molecular mass. |
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| 205. |
If `M` is the molecular weight of `KMnO_(4)`, its equivalent weight will be when it is converted into `K_(2)MnO_(4)`A. `M`B. `M//3`C. `M//5`D. `M//7` |
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Answer» Correct Answer - A `2KMnO_(4)+2KOHto2K_(2)MnO_(4)+H_(2)O+[O]` or `Moverset(+7)(n)O_(4)^(-)+e^(-)to overset(+6)(Mn)O_(4)^(2-)` `:.` Eq. mass `=("Mol. Mass")/("Change in O.N.")=(M)/(1)=M` |
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| 206. |
Equivalent weight of `H_(3) PO_(2)` is (`M rarr` molecular weight)A. `(M)/(1)`B. `(M)/(2)`C. `(M)/(3)`D. `(M)/(4)` |
| Answer» Correct Answer - A | |
| 207. |
STATEMENT - 1 : Equivalent weight of Mohr salt is `(M)/(4)`, where M is molecular weight. STATEMENT - 2 : Mohr salt contains only one metallic cation i.e., `Fe^(2+)` STATEMENT - 3 : Equivalent weight of a salt `= ("Molecular weight")/("Total positive charge")`A. TTTB. FTFC. TFTD. FFT |
| Answer» Correct Answer - A | |
| 208. |
STATEMENT - 1 : Law of conservation of mass is valid in nuclear reaction. STATEMENT - 2 : On dilution, normality of the solution changes. STATEMENT - 3 : Molarity of the solution does not depend on amount of solution.A. TTTB. FTTC. FFTD. TFF |
| Answer» Correct Answer - B | |
| 209. |
One mole of any substance contains `6.022xx10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100mL of 0.02M `H_2SO_(4)` solution isA. `12.044xx10^(20)` moleculesB. `6.022xx10^(23)` molecuelsC. `1xx10^(23)` molecuelsD. `12.044xx10^(23)` molecules |
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Answer» One mole of any substance contains `6.022xx10^(23)` atoms/molecuels. Hence, Number of millimoles of `H_(2)SO_(4)` =molarity xx volume in mL `=0.02xx100=2` millimoles `=2xx10^(-3)` mol Number of molecuels =number of moles `xxN_(A)` `=2xx10^(-3)xx6.022xx10^(23)` `=12.044x10^(20)` molecuels |
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| 210. |
A notation that uses atomic symbols with numerical subscipts to convey the relative proportions of atoms of differenct elements in the substance is called the ...... of a substance.A. molecular formulaB. enpirical formulaC. chemical formulaD. simplest formula |
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Answer» Correct Answer - C It is general term acceptable for elements and all types of compounds. |
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| 211. |
One mole of any substance contains `6.022xx10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100mL of 0.02M `H_2SO_(4)` solution isA. `12.044xx10^20`B. `6.022xx10^23`C. `1xx10^23`D. `12.044xx10^23` |
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Answer» Molarity `=("No. of moles of solute" xx 1000)/("Volume of solution (in mL)")` `therefore " 1M " H_2SO_4` contains 1 mole of `H_2SO_4` in 1000 mL of solution `therefore " 0.02M " H_2SO_4` in 100 mL of solution contains `(0.02xx100)/(1000)=2xx10^(-3)` mol `=2xx10^(-3) xx 6.022xx10^23=12.044xx10^20` molecules |
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| 212. |
STATEMENT - 1 : Equivalent weight of a substance can never be greater than its molecular weight. STATEMENT - 2 : Equivalent weight of boric acid is `(M)/(1)` where M is the molecular weight. STATEMENT - 3 : Equivalent weight of `H_(3)PO_(2)` is `(M)/(3)`.A. FTFB. TFTC. FTTD. TFF |
| Answer» Correct Answer - A | |
| 213. |
`27 g` of `Al` will react completely with…… `g` of `O_(2)`A. (a)8 gB. (b)16 gC. (c )32 gD. (d)24 g |
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Answer» Correct Answer - D `4 Al+3O_(2)rarr2Al_(2)O_(3)` At. Wt. of `Al=27` Thus `4xx27 g` of Al reacts with oxygen`=3xx32 g` `:. 27 g` of Al reacts will oxygen`=(3xx32)/(4xx27)xx27=24 g`. |
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| 214. |
A substance that is composed of molecules all of which are alike isA. an elementB. a molecular substanceC. a covalent compoundD. an ionic compound |
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Answer» Correct Answer - B Ionic compounds consists of ions but all covalent compounds fo molecules. |
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| 215. |
Which of the following is a polyatomic ion?A. Oxide ionB. Peroxide ionC. Superoxide ionD. Both(1) and (2) |
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Answer» Correct Answer - D Peoxide in `(O_(2)^(2-))` and superoxide ion `(O_(2)^(2-))` contain more than one atom. |
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| 216. |
A molecule is a definite group of atoms that are .....bonded together, that is, tightly connected by attractive forces.A. ionicallyB. metaliicallyC. physicallyD. convalently |
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Answer» Correct Answer - D Atoms of a molecule are always connected by convalent bonds. |
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| 217. |
In the reaction, `4NH_(3)(g)+5O_(2)(g) rarr 4NO(g)+6H_(2)O(g)`, when 1 mole of ammonia and 1 mole of `O_(2)` are made to react to completionA. (a)1.0 mole of H_(2)O is producedB. (b)1.0 mole of NO will be producedC. (c )All the oxygen will be consumedD. (d)All the ammonia will be consumed |
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Answer» Correct Answer - C `4NH_(3(g))+5O_(2(g))rarr 4NO_((g))+ 6H_(2)O_((g))` `t=0" "1" " 1" "0 " "0` `t=" "t" "1-4x" " 1-5x" "4x " " 6x` Oxygen is limiting reagent So, `X=1/5=0.2` all oxygen consumed Left `NH_(3)=1-4xx0.2=0.2.` |
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| 218. |
In the reaction, `4NH_(3)(g)+5O_(2)(g) rarr 4NO(g)+6H_(2)O(g)`, when 1 mole of ammonia and 1 mole of `O_(2)` are made to react to completionA. 1.0 mole of `H_(2)O` is producedB. 1.0 mole of NO will be producedC. All the oxygen will be consumedD. All the ammonia will be consumed |
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Answer» Correct Answer - C `{:(,4NH_(3(g)),+,5O_(2(g)),rarr,4NO_((g)),+,6H_(2)O_((g))),(t=0,1,,1,,0,,0),(t=t,1-4x,,1-5x,,4x,,6x):}` Oxygen is limiting reagent So, `X=1/5=0.2` all oxygen consumed Left `NH_(3)=1-4xx0.2=0.2` |
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| 219. |
How much time would it take to distribute one Avogadro number of wheat grains, if `10^(10)` grains are distributed each second?A. 0.1673B. 1.673C. 16.73D. 167.3 |
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Answer» Correct Answer - B If `10^(20)` grains are distributed in one sec, `6.023xx10^(23)` grains will be distributed in |
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| 220. |
Element X forms five stable oxides with oxygen of formula `X_(2) O, XO , X_(2) O_(3) , X_(2) O_(5)` . The formation of these oxides explainsA. Law of definite proportionsB. Law of partial pressuresC. Law of multiple proportionsD. Law of reciprocal proportions |
| Answer» Correct Answer - C | |
| 221. |
Vapour density of a metal chloride is `6.6`. Its oxide contains `53%` metal. The atomic weight of metal is:A. (a)`21`B. (b)`54`C. (c )`27.06`D. (d)`2.086` |
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Answer» Correct Answer - C Let wt. of metal oxide`=100 g` wt. of metal`=53 g` wt. of oxygen`=47 g` Equivalent weight of oxygen `=("wt. of metal")/("wt. of oxygen")xx8=53/47xx8=9.02` Valency`=(2xxV.D)/(E+35.5)=(2xx66)/(9+35.5)=132/44.5=2.96~~3` `:.` Atomic weight=Equivalent weight `xx`Valency `=9.20xx3=27.06` |
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| 222. |
Barnett Roseberg is the discover of cisplatin a leading anti-cancer drug. The substance contains the metalA. plutoniumB. platinumC. palladiumD. plonium |
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Answer» Correct Answer - B It is `[Pt(NH_(3))_(2) Cl_(2)]`. |
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| 223. |
`AZT` (azidothymindine) is used for helping .......victims.A. arthritisB. thalassemiaC. AlDSD. tuberculosis |
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Answer» Correct Answer - C Also called zidvidudine `(ZDV), AZT` is a nuclesoside anolge reverse transcitase inhibitor `(NRTI)`, a type of antiretoviral durg used for the treatment of `HIV//AIDS` inferction. It was the first breakthrough in `AIDS` therapy. |
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| 224. |
The equivelent weight of an element is `4`. Its chloride has a `V.D 59.25`. Then the valency of the element isA. (a)`4`B. (b)`3`C. (c )`2`D. (d)`1` |
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Answer» Correct Answer - B Valency of the element`=(2xxV.D)/(E+35.5)` `=(2xx59.25)/(4+35.5)=118.50/39.5=3`. |
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| 225. |
The equivalent weight of a metal is 9 and vapour density of its chloride is 59.25. The atomic weight of metal isA. 23.9B. 27.3C. 36.3D. 48.3 |
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Answer» Correct Answer - A Given equivalent weight of metal `=9` Vapour density of metal chloride `=59.25` `:.` molecular weight of metal chloride `=2xxV.D=2xx59.25=118.5` `:.` valency of metal `=("molecular weight of metal chloride")/("equivalent weight of metal + 35.5")` Valency of metal `=118.5/(9+35.5)=118.5/44.5=2.66` Therefore atomic weight of the metal = equivalent weight `xx` valency `=9xx2.66=23.9`. |
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| 226. |
The value of x in the partial redox equation `MnO_(4)^(-)+8H^(+)+xe^(-) hArr Mn^(2+)+4H_(2)O` isA. 5B. 3C. 1D. 0 |
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Answer» Correct Answer - A `MnO_(4)^(-)+8H^(+)+5 e^(-) hArr Mn^(++)+4 H_(2)O`. |
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| 227. |
1 g of a mixture of `NaHCO_(3)` and `Na_(2)CO_(3)` is heated to `150^(@)C`. The volume of the `CO_(2)` produced at STP is 112.0 mL. Calculate the percentage of `Na_(2)CO_(3)` in the mixture `(Na=23, C=12, O=16)`A. 20B. 46C. 84D. 16 |
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Answer» Correct Answer - D `2NaHCO_(3) overset(150^(@)C)(rarr) Na_(2)CO_(3)+CO_(2)+H_(2)O` `n_(NaHCO_(3))/n_(CO_(2))=2/1` `n_(NaHCO_(3))=2n_(CO_(2))` `=2xx112/22400=0.01` mole `W_(NaHCO_(3))=0.01xx84=0.84 g` `W_(Na_(2)CO_(3))=1.00-0.84=0.16 g` `% Na_(2)CO_(3)=16`. |
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| 228. |
One mole of potassium dichromate completely oxidises the following number of moles of ferrous sulphate in acidic mediumA. 1B. 3C. 5D. 6 |
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Answer» Correct Answer - D `{:(Cr_(2)O_(7)^(--) rarr Cr^(3+), Fe^(++) rarr Fe^(+++)),(" n=6 n=1"):}` eq. of `K_(2)Cr_(2)O_(7)=` eq. of `FeSO_(4)` `1xx6=x xx1` |
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| 229. |
Calculate number of atoms in 0.2 mole of oxygen atoms. |
| Answer» Correct Answer - `1.2044xx10^(23)` atoms | |
| 230. |
Match the following `{:(,"Column -I",,"Column -II"),((A),underset(3g)(2H_(2))+underset(22.66g)(O_(2))rarr 2H_(2)O,(p),25.5 "g product"),((B),underset(21 g)(N_(2))+underset(5 g)(3H_(2))rarr 2 NH_(3),(q),H_(2) "is present in excess"),((C ),underset(1g)(H_(2))+underset(40 g)(Cl_(2))rarr 2 HCl,(r ),H_(2) "is the limiting reagent"),((D),underset(20g)(C )+underset(6.375g)(2H_(2))rarr CH_(4),(s),36.5"g product"):}` |
| Answer» Correct Answer - A::B::C::D | |
| 231. |
In an acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the gram molecular mass of potassium dichromate is 294 g, its gram equivalent mass is........ Grams.A. `294`B. `127`C. `49`D. `24.5` |
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Answer» Correct Answer - C In acidic medium, `underset(("O.N. of Cr=6"))(Cr_(2)O_(7)^(2-))+14H^(+) to underset(("O.N. of Cr=3"))(2C^(3+)+7H_(2))O-6e^(-)` Eq. mass of `K_(2)Cr_(2)O_(7)` in acidic medium `=("molecular mass")/(6) = (294)/(6) = 49` |
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| 232. |
Calculate number of atoms in 0.5 mole of oxygen atoms. |
| Answer» Correct Answer - `3.01xx10^(23)` atoms | |
| 233. |
Calculate mass of one atom of nitrogen in gram. |
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Answer» Mass of `6.022xx10^(23)` atoms of nitrogen = gram atomic mass of nitrogen = 14 g Mass of 1 atom = `GAM//N_(A)` `therefore` Mass of 1 atom of nitrogen `=(14)/(6.022xx10^(23))=2.32xx10^(-23)g` |
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| 234. |
One atom of an element weight `3.32xx10^(-25)` g. How many number of gram atoms are in 20 kg of the element?A. 2000B. 20C. 200D. 1000 |
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Answer» Atomic mass of an element = Mass of one atom `xx N_A` `=3.32xx10^(-23)xx6.023xx10^23=19.99=20g` No. of gram atoms `=("Mass of element in grams")/("Atomic mass in grams")=(20xx1000)/(20)=1000` |
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| 235. |
STATEMENT -1 : In 32 g of `O_(2)`, two gram of oxygen atom are present. and STATEMENT -2 : Molecular weight of `O_(2)` will be 32 g.A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for Statmenet -1.B. Statement -1 is True, Statement -2 is True, Statement-2 is NOT a correct explanation for Statement -10C. Statement -1 is True, Statement -2 is FalseD. Statement -1 is False, Statmenet -2 is True |
| Answer» Correct Answer - C | |
| 236. |
Calculate mass of one molecule of sulphur dioxide `(SO_(2))` in gram. |
| Answer» Correct Answer - `1.06xx10^(-22)g` | |
| 237. |
Calculate mass of one atom of calcium in gram. |
| Answer» Correct Answer - `6.6xx10^(-23)g` | |
| 238. |
Calculate the molecular mass of glucose `(C_(6)H_(12)O_(6))` molecule in amu. |
| Answer» Correct Answer - 180.162 u | |
| 239. |
Calculate the molecular mass of `CH_(4)` molecule in amu. |
| Answer» Correct Answer - 16.043 u | |
| 240. |
1 amu is equal toA. (a)`1/2` of `C-12`B. (b)`1/14` of `O-16`C. (c )`1 " g of " H_(2)`D. (d)`1.6xx10^(-23) kg` |
| Answer» Correct Answer - A | |
| 241. |
Dilution: How many milliliters of `18.4 M H_(2) SO_(4)` are required to prepare `1L` of `0.940 M` solution of `H_(2) SO_(4)`? Strategy: Since the concentration of the final solution is less than that of the orignal one, this is a dilution process. We are given the molarity of the original solution and the volume `(V_(f))` and molarity `(M_(f))` of the final solution. Thus. Eq (1.21) can be used. |
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Answer» Applying the relationship for dillution `M_(1) V_(1) = M_(f) V_(f)` We have `V_(1) = (M_(f) V_(f))/(M_(i)) = ((1.00L)(0.940M))/((18.4M))` `= 0.0510 L = 51.0 mL` |
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| 242. |
Molarity: A sample of commercial sulphuric acid is `98% H_(2) SO_(4)` by mass and its specfic gravity is `1.84`. Caculate the molartiy of this sulburic acid solution. Strategy: The density of a solution (grams per milliliter) is numercially equal to its specific gravity. Thus, the density of the solution is `98% H_(2) SO_(4)` by mass. Thus, every `100g` of soultuon contains `98g` of pure `H_(2) SO_(4)`. From the mass of `H_(2) SO_(4)`. we calculate its moles and from the density , we calculate its volume. Finally, we calculate molartity using its definition. |
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Answer» Step1: Calculate the moles of solure. Consider `100g` solutiohn, it contains `98g` of pure `H_(2) SO_(4)`. Thus, Number of moles of `H_(2) SO_(4) = ("Mass of" H_(2) SO_(4))/("Molar mass of" H_(2) SO_(4))` `= (98.0g)/(98.0g mol^(-1)) = 1.00 mol` Step 2: Calculate the volume of soltuion. Volume of `H_(2) SO_(4)` Solution `= ("Mass of" H_(2) SO_(4) "solution")/("Density of" H_(2) SO_(4) "solution")` `= (100g)/(1.84g mL^(-1))` `= 54.4 mL = 5.44xx10^(-2) L` Step 3: Calculate the molarity of the solution Molarity of `H_(2) SO_(4)` solution `= ("Number of moles of solute")/("Liters of solution")` `= (1.00 "mol")/(5.44xx10^(-21) L)` `= 18.4 mol L^(-1) = 18.4M` Alternatively, in one setup, considering `1 L` of solution, `(? mol H_(2) SO_(4))/("L soln.") = (1.84 "g soln")/("mL soln.") xx (1000 "mL soln")/("L soln.") xx (98g H_(2) SO_(4))/(100g "soln.")` `xx (1 "mol" H_(2) SO_(4))/(98g H_(2) SO_(4))` `= 18.4 M` Note that a series fo three unit factors is used. |
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| 243. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. (a)`Mn_(2)O_(3)`B. (b)`MnO_(2)`C. (c )`MnO_(4)^(-)`D. (d)`MnO_(4)^(2-)` |
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Answer» Correct Answer - B Equivalent `"weight"=("molecular weight")/("valency factor")` If valency is 2, then equivalent weight will be equal to its molecular weight. In `MnSO_(4)`, the oxidation state of Mn is +II In `Mn_(2)O_(3)`, the oxidation state of Mn is +III In `MnO_(2)`, the oxidation state of Mn is +IV In `MNO_(4)`, the oxidation state of Mn is +VII In `Mn_(4)^(2-)`, the oxidation state of Mn is +VI Thus, when `MnSO_(4)` is converted into `MnO_(2)`, then the valency factor is 2, and the equivalent weight of `MnSO_(4)` will be half of its molecular weight. |
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| 244. |
Sulphur forms the chlorides `S_(2)Cl_(2)` and `SCl_(2)`. The equivalent mass of sulphur in `SCl_(2)` isA. (a)`8 g//mol`B. (b)`16 g//mol`C. (c )`64.8 g//mol`D. (d)`32 g//mol` |
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Answer» Correct Answer - B The atomic weight of sulphur`=32` In` SCl_(2)` valency of sulphur`=2` So equivalent mass of sulphur`=32/2=16` |
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| 245. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(-)`D. `MnO_(4)^(-)` |
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Answer» Correct Answer - B Equivalent weight in redox system is defined as : `E = ("Molar mass")/("n-factor")` Here n-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, n-factor is because equivalent weight is half of molecular weight. Also, `{:("n-factor",MnSO_(4),rarr,(1)/(2)Mn_(2)O_(3),1(+2 rarr +3)),(,MnSO_(4),rarr,MnO_(2),2(+2 rarr +4)),(,MnSO_(4),rarr ,MnO_(4)^(-),5(+2 rarr +7)),(,MnSO_(4),rarr,MnO_(4)^(2-),4(+2 rarr +6)):}` Therefore, `MnSO_(4)`converts to `MnO_(2)`. |
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| 246. |
If 0.5 mole of `BaCl_(2)` are mixed with 0.2 mole of `Na_(3)PO_(4)`,the maximum number of moles. Of `Ba_(3)(PO_(4)_(2))` that can be formed, isA. 0.7B. 0.5C. 0.3D. 0.1 |
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Answer» d) `3BaCl_(2) + 2Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2) + 3NaCl` Limiting reactant is `Na_(3)PO_(4)`. 0.2 mole of `Na_(3)PO_(4)` will give = `1/2 xx 0.2` =0.1 mole of `Ba_(3)(PO_(4))_(2)` |
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| 247. |
If 0.50 mol of `BaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed isA. (a)0.70B. (b)0.50C. (c )0.20D. (d)0.10 |
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Answer» Correct Answer - D (i) Write balanced chemical equation for chemical change. (ii) Find limiting reagent. (iii) Amount of product formed formed will be determined by amount of limiting reagent. The balanced equation is: `underset(No . of "moles")()" "underset(0.5 3 "moles")(3BaCl_(2))+underset(0.2 "moles")(2Na_(3)PO_(4))rarrunderset(1 "mole")(Ba_(3)(PO_(4))_(2)+6"Nacl")` Limiting reagent is `Na_(3)PO_(4)(0.2 mol), BaCl_(2)` is in excess. From the above equation: `2.0` moles of `Na_(3)PO_(4)` yields `Ba_(3)(PO_(4))^(2)= 1` mole `:. 0.2` moles of `Na_(3)PO_(4)` will yield `Ba_(3)(PO_(4))_(2)` `=1/2xx0.2=0.1` mol. |
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| 248. |
If 0.50 mol of `BaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed isA. `0.70`B. `0.50`C. `0.20`D. `0.10` |
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Answer» Correct Answer - D `underset(0.5)(3 BaCl_(2))+ underset(0.2)(2 Na_(3) PO_(4)) rarr Ba_(3) (PO_(4))_(2)+6NaCl`, Limiting reagent in `Na_(3)PO_(4)` (0.2 mol), which gives 0.1 mol of `Ba_(3) (PO_(4))_(2)`. |
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| 249. |
A mixture of `CaCl_(2)` and NaCl weighing 4.44 is treated with sodium carbonate solution to precipitate all the `Ca^(2+)` ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of `CaO`. The percentage of NaCl in the mixture of (atomic mass of Ca=40) isA. 75B. 30.5C. 25D. 69.4 |
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Answer» Correct Answer - A Concerned reaction is `underset(100 g)(CaCO_(3)) overset(Delta)(rarr) underset(56 g)(CaO)+CO_(2)` 56 g of CaO is obtained from 100 g of `CaCO_(3)` 0.56 g of CaO is obtained by `100/56xx0.56=1 g` of `CaCO_(3)` we know that, `underset(111 g)(CaCl_(2))+Na_(2)CO_(3) rarr underset(100 g)(CaCO_(3))+2 NaCl` 100 g of `CaCO_(3)` is obtained by 11 g of `CaCl_(2)` 1 g of `CaCO_(3)` is obtained by `111/100=1.11 g` of `CaCl_(2)` Weight of `NaCl=4.44-1.11=3.33 g` % age of `NaCl=3.33/4.44xx100=75%`. |
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| 250. |
10 g of sample of mixture of `CaCl_(2)` and NaCl are treated to precipitate all the calcium as `CaCO_(3)`. This `CaCO_(3)` is heated to convert all the Ca to CaO and the final mass of `CaCl_(2)` in the original mixture isA. 0.321B. 0.162C. 0.218D. 0.12 |
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Answer» a) `CaCl_(2) + NaCl = 10g` Let, weight of `CaCl_(2)` = xg `CaCl_(2) to CaCO_(3) to CaO` Moleof CaO = `(1.62)/(56) rArr x/111 = 1.62/56 x = 3.21g` `%` of `CaCl_(2)` = (3.21/10 x 100) = 32.1%` |
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