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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. |
Vanillin, `C_(8)H_(8)O_(3) (m = 152g//mol)`, is the molecule responsible for the vanilla flavour in food. How many oxygen atoms are present in a `45.0mg` sample of vanillin?A. `1.78 xx 10^(20)`B. `5.35 xx 10^(20)`C. `1.78 xx 10^(23)`D. `5.35 xx 10^(23)` |
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Answer» Correct Answer - B |
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| 552. |
Which of the following has maximum mass?A. 0.1 g atom of CB. 0.1 mole of `NH_(3)`C. `6.02 xx 10^(22)` molecule of `H_(2)` gasD. 1120 ml of `CO_(2)` at 1 atm, 273 K |
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Answer» Correct Answer - D |
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| 553. |
Volume (in ml) 0f `0.7` M NaOH required for complete reaction with 350 ml of `0.3` M `H_(3)PO_(3)` solution is :A. 300 mlB. 450 mlC. 150 mlD. 350 ml |
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Answer» Correct Answer - A |
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| 554. |
Mass of `CO_(2)` is 88 g. The number of atoms of oxygen present is :A. `2.41 xx 10^(24)`B. `1.2 xx 10^(23)`C. `1.4 xx 10^(23)`D. `2.41 xx 10^(23)` |
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Answer» Correct Answer - A 44.0 grams of `CO_(2)` contain oxygen atoms `= 2xx 6.022 xx 10^(23)` atoms 88.0 grams of `CO_(2)` contain oxygen atoms `= 2 xx 2 xx 6.022 xx 10^(23)` `=2.41 xx 10^(24)` atoms. |
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| 555. |
A 12.0 M acid solution that contains `75.0%` acid by mass has a density of `1.57g//mL`. What is the identity of the acid?A. HCl(M=36.5)B. `CH_(3)CO_(2)H(M=60.0)`C. Hbr(M=80.9)D. `H_(3)PO_(4)(M=98.0)` |
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Answer» Correct Answer - D |
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| 556. |
What volume of water (in mL) should be added to 50 ml of `HNO_(3)` having density `1.5 g` `ml^(-)` and `63.0%` by weight to have one molar solution |
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Answer» Correct Answer - 700 |
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| 557. |
The empirical formula of an organic compound carbon & hydrogen is `CH_(2)` The mass of `1"litre"` of organic gas is exactly equal to mass of 1 litre `N_(2)` therefore molecular formula of organic gas is . |
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Answer» Empirical Mass of `CH_(3)=12+2=14` `because` Mass of 1 litre of organic gas = Mass of 1 litre of `N_(2)` Since V,P,T, n are same. Therefore PV=`(m)/(M)RT` implies that molar mass should also be same. `therefore ` Molecular mass of organic compound will be 28 g `n=("Molecular mass")/("Empirical mass")=(28)/(14)=2` So molecular formula `=2xxCH_(2) =C_(2)H_(4)` |
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| 558. |
A compound of carbon, hydrogen, and nitrogen contains the three elements in the respective ratio of `9 : 1 : 3.5` Calculculate the empirical formula. If the molecular weight of the compound is 108, what its molecular formula? |
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Answer» Correct Answer - B::C `C : H : N = 9 : 1 : 3.5` Mole ratio `C : H : N = (9)/(12) : (1)/(1) : (3.5)/(14) = 0.75 : 1 : 10.25` `= (0.75)/(0.25) : (1)/(0.25) : (0.25)/(0.25) = 3 : 4 : 1` Empirial formula `= C_(3) H_(4) N` Empirical formula weight `= C_(3) H_(4) N` `12 xx 3 + 1 xx 4 + 14 = 54` `Mw = 108` `n = (Mw)/(EFw) = (108)/(54) = 2` Molecular formula `= 2 xx C_(3) H_(4) N = C_(6) H_(8) N_(2)` |
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| 559. |
One mole of potassium chlorote is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many mole(s) of aluminium oxide are formed?A. 1B. 1.5C. 2D. 3 |
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Answer» Correct Answer - A `KClO_(3) rarr KCl + (3)/(2) O_(2)` `2Al + (3)/(2) O_(2) rarr Al_(2)O_(3)` 1 mol of `KClO_(3) -= 1 "mol of" Al_(2) O_(3)` |
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| 560. |
`3.4 g` sample of `H_(2)O_(2)` solution containing `x% H_(2)O_(2)` by weight requires `x mL of a KMnO_(4)` solution for complete oxidation under acidic condition. The normality of `KMnO_(4)` solution isA. 1B. `0.5`C. `0.4`D. `0.2` |
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Answer» Correct Answer - 3 Eq. mass of `H_(2)O_(2)=(34)/(2)=17` Eq. of `H_(2)O_(2)=(3.4xxx)/(100xx17)` `=Eq.` of `KMnO_(4)=x xx Nxx10^(-3)" "rArr" "N=2` Molarity of `KMnO_(4)=(2)/(5)=0.4` |
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| 561. |
Find the sum of molarity of all the ions present in an aqueous soultion of 5 M `NaNO_(3)` and 3 m `BeCl_(2)?` The specific gravity of the given solution is 1.665. Assume `100%` dissociation of each salt. `[Be=9]` |
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Answer» Correct Answer - 19 |
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| 562. |
A solvent X (mol mass 50) contains solute A (mol. Mass 125) and solute B (mol mass 100). If solution is 4 M A and 6 M B, then find simplest ratio of moles of A:B:X[Given: `d_(solution)=1.3 gm//ml`]. If your answer is A:B:C, then fill your answer is A+B+C. |
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Answer» Correct Answer - 7 |
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| 563. |
If 150 g of carbon reacts with 250 g of `Cl_(2)` and the reaction has an 85% yield, how many grams of `CCl_(4)` are produced? Given answer excluding decimal places. |
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Answer» Correct Answer - 230 |
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| 564. |
What weight of CO is required to form `Re_(2)(CO)_(10)` from 2.50 g fo f`Re_(2)O_(7)` according to the unbalanced reaction: `Re_(2)O_(7) + CO rarr Re_(2)(CO)_(10)+CO_(2)` (Re =186 ,C =12 and O=16).Give your answer to the nearest integar. |
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Answer» Correct Answer - 2 |
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| 565. |
2.5 of a mixture of BaO and CaO when heated with `H_(2)SO_(4)` , produced 4.713 g of the mixed sulphates . Find the percentage of BaO present in the mixture. |
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Answer» Correct Answer - 60 |
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| 566. |
Which of the following have equal mass of `Cl^(ɵ)` ions in `1.0 L` of each of the following solution?A. `5% NaCl` (density `= 1.07 g mL^(-1)`)B. `5% KCl (d = 1.06 g mL^(-1))`C. `58.5 g NaCl`D. `55.5 g BaCl_(2)` |
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Answer» Correct Answer - C::D Use, `M = (% "by weight" xx 10 xx d)/(Mw_(2))` a. `M = (5 x 10 xx 1.07)/(58.5) = 0.91 M (Cl^(ɵ) = 0.91 M = 0.91 xx 35.5 = 32.56 g)` b. `M = (5 xx 10 xx 1.06)/(74.5) = 0.77 M (Cl^(ɵ) = 0.77 M = 0.77 xx 35.5 = 25.5 g)` c. `M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)("in" mL))` `= (58.5 xx 1000)/(58.5 xx 1000) = 1 M` `Cl^(ɵ) = 1 M = 1 xx 35.5 = 35.5 g)` d. `M = (55.5 xx 1000)/(111 xx 1000) = 0.5 M` `BaCl_(2) rarr Ba%(2+) + 2 Cl^(ɵ)` `:. Cl^(ɵ) = 2 xx M = 2 xx 0.5 = 1 M = 1 xx 35.5 = 35.5 g` |
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| 567. |
Determine the molecular formula of the compound which contains `H = 6.67 %, C = 40 %` and the rest is oxygen. 0.6 g of the compound occupy 224 cc at N.T.P |
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Answer» Correct Answer - `C_(2)H_(4)O_(2)` Step I. Percentage of oxygen `= 100 - (6.67 + 40) = 100-46.67 = 53.33` Step II. Empirical formula of the compound `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",40.00,12,(40.0)/(12)=3.33,(3.33)/(3.33)=1.0,1),("H",6.67,1,(6.67)/(1)=6.67,(6.67)/(3.33)=2.0,2),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(3.33)=1.0,1):}` Empirical formula of the compound `= CH_(2)O` Step III. Molecular formula of the compound 224 cc of the vapours of the compound at N.T.P occupy mass = 0.6 g 22400 cc of the vapours of the compound at N.T.P occupy mass `= 0.6//224 xx 22400 = 60.0 g` `:.` Molecular mass of the compound = 60 g = 60 u Empirical formula mass `= 12 + 2 xx 1 + 16 = 30 u , n = ("Molecular mass")/("Empirical formula mass")=((60u))/((30u))=2` `:.` Molecular formula `= 2 xx CH_(2)O=C_(2)H_(4)O_(2)`. |
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| 568. |
In a compound, C, H and N are present in the ratio `9 : 1: 3.5` by weight. If the molecular weight. If the molecular weight of the compound is 108, what is the number of hydrogen atoms present in the molecular formula of the compound ? |
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Answer» Correct Answer - `=8` Step I. Empirical formula of the compound `{:("Element","Percentage/weight (ratio)","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",9,(9)/(12)=0.75,(0.75)/(0.25)=3,3),("H",1,(1)/(1)=1.0,(1.00)/(0.25)=4,4),("N",3.5,(3.5)/(14)=0.25,(0.25)/(0.25)=1,1):}` Empirical formula of the compound `= C_(3)H_(4)N` Step II. Molecular formula of the compound Empirical formula mass `= 3 xx 12 + 4 xx 1 +14 = 54 u` Molecular mass = 108 u (given) `n=("Molecular mass")/("Empirical formula mass")=((108u))/((54u))=2` Molecular formula of the compound `= n xx` Empirical formula `:.` Number of hydrogen atoms present = 8. |
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| 569. |
Calculate the weight of iron which will be converted into its oxide by the reaction of `18 g` of steam. |
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Answer» `underset(3 xx 56)(3 Fe) + underset(4 xx 18)(4 H_(2) O) rarr Fe_(3) O_(4) + 4 H_(2)` `4 xx 18 g "of" H_(2) O = 3 xx 56 "of" Fe` `18g "of" H_(2) O = (3 xx 56)/(4 xx 18) xx 18 = 42 g "of" Fe` |
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| 570. |
The amount of arsenic pentasulhide that can be obtained when `35.5g` arsenic acid is treated with axess `H_(2)S` in the presence of conc. `HCl` `("assuming" 100% "conversion")` is :A. `0.25` molB. `0.50` molC. `0.125` molD. `0.333` mol |
| Answer» Correct Answer - C | |
| 571. |
A compound contains `10^(2)%` of phosphorous . If atomic mass of phosphorouus is 31, the moleular mass of the compound having one phosphorous atom per molecules is :-A. 31B. `3.1xx10^(3)`C. `3.1xx10^(5)`D. `3.1xx10^(4)` |
| Answer» Correct Answer - C | |
| 572. |
One commercial system removes `SO_(2)` emmission from smoke at `95^(@)`C by the following set of reactions- `SO_(2)(g)+ Cl_(2)(g) to SO_(2)Cl_(2)(g)` `SO_(2)Cl_(2) + 2H_(2)O to H_(2)SO_(4) + 2HCl` `H_(2)SO_(4) + Ca(OH)_(2) to CaSO_(4) + 2H_(2)O` Assuming the process to be 95% efficient . How many moles of `CaSO_(4)` may be produced from 128 g `SO_(2).[Ca=40,S-32,O-16]`A. 1.9 molesB. 2 molC. 3.8 molD. 0.95 mol |
| Answer» Correct Answer - A | |
| 573. |
If 12 g of water is formed during complete combustion of pure propene `(C_(3)H_(6))`, the mass of propene burnt is :A. 42 gB. 21 gC. 14 gD. 56 g |
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Answer» Correct Answer - B `underset(underset(=42g)((3xx12+1xx6)))(C_(3)H_(6)(g))+9//2O_(2)(g)rarr3CO_(2)(g)+underset(3xx18=54g)(3H_(2)O(l))` 54 g of `H_(2)O` is formed by the combustion of `C_(3)H_(6) = 42g` 27 g of `H_(2)O` is formed by the combustion of `C_(3)H_(6)=((42g))/((54g))xx(27g)=21g`. |
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| 574. |
When 2.46 of hydrated salt `(MSO_(4)x H_(2)O)` is completely dehydrated, 1.20 g of anhydrous salt is obtained. It molecular weight of anhydrous salt is `120 g mol^(-1)`, the value of x isA. 2B. 4C. 5D. 7 |
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Answer» Correct Answer - D `underset(underset(2.46g)(2.46))(MSO_(4)xH_(2)O)rarrunderset(underset(120g)(1.2g))(MSO_(4))+xH_(2)O` Weight of water `(H_(2)O)=(246 - 120) = 126 g` No. of moles of water `(x) = (126 g)/(("18 g mol"^(-1))) = 7` mol. |
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| 575. |
Consider the following chemical reaction `Pb(NO_(3))_(2) + Na_(2)SO_(4) rarr PbSO_(4)+2NaNO_(3)` . If a series of experiments are run maintaining sum of the weights of two reactant constant but varying the weights of reactants , which of the following statements is (are) true? `[M_(Pb(NO_(3))_(2))=394 , M_(Na_(2)SO_(4))=142]`A. Maximum weight of the ppt `(PbSO_(4))` will be formed if equal weights of reactant are taken.B. Maximum weight fo the ppt `(PbSO_(4))` will be formed if equal moles of reactants are takenC. In the experiments , as the weight of `Pb(SO_(4))` increase , weight of ppt`(PbSO_(4))` increases.D. In the experiment , as the weight of `Pb(NO_(3))_(2)` increases, weight of ppt `(PbSO_(4))` increases and than decreases. |
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Answer» Correct Answer - B::D |
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| 576. |
The mole fraction of NaCl in aqueous soluition is 0.2 . The solution is :A. 13.9 mB. mole fraction of `H_(2)O` is 0.8C. acidic in natureD. neutral |
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Answer» Correct Answer - A::B::D |
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| 577. |
What is the mole fraction of `CH_(3)OH` in an aqueous solution that is 12.0m in `CH_(3)OH`?A. 0.178B. 0.216C. `0.400`D. 0.667 |
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Answer» Correct Answer - A |
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| 578. |
If a piece of iron gains 10% of its weight due to partial rusting into `Fe_(2)O_(3)` the percentage of total iron that has rusted is:A. 23B. 13C. 23.3D. 25.67 |
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Answer» Correct Answer - C |
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| 579. |
Which of the following statements is/are correct?A. Chloropicrin `(C Cl_(3).NHO_(2))` can made cheapy for use as an insectide by the following reaction: `CH_(3) NO_(2) + Cl_(2) rarr C Cl_(3) NO_(2) + HCl`B. In a rocket motor fueled with butane `(C_(4) H_(10))`, 0.1 mol of butane requires `14.56 L` of `O_(2)` at `STP` for complete combustion.C. A portable hydrogen generator utilises the reaction: `(CaH_(2) + H_(2)O rarr Ca (OH)_(2) + H_(2)), 2.1 g "of" CaH_(2)` would produce `2.24 L` "of" `H_(2)` at `STP`D. In the Mond process for purifying nickel, the volatile nickel carbonyl `[Ni(CO)_(4)]` is produced by the reaction. `Ni + Co rarr Ni (CO)_(4). 58.8 7g` of `Ni` utilises `89.6 L` of `CO` at standard conditions. |
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Answer» Correct Answer - A::B::C a. `CH_(3) NO_(2) + underset(3 "mol")(3Cl_(2)) rarr underset(1 "mol")(C Cl_(3). NO_(2)) + 3HCl` Volume of `Cl_(2)` at `STP = 3 xx 0.54 xx 22.4 = 33.6 L` b. `underset(1 "mol")(C_(4) H_(10)) + underset(13//2 "mol")((13)/(2) O_(2)) rarr 4 CO_(2) + 5H_(2) O` c. `underset((1 "mol" = 42 g))(CaH_(2) + 2H_(2) O) rarr Ca (OH)_(2) + underset(2H_(2))(2H_(2))` d. `underset({:(1 "mol"),(= 58.7 g):})(Ni) + underset(4 "mol")(4CO) rarr Ni (CO_(4))` Volume of `CO` standard conditions `= 4 xx 24.4` `= 976 L` Hence, (d) is wrong. |
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| 580. |
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 `0.00006 g cm^(-3)` , then the volume occupied by water molecules in `1 L` steam at this temperature isA. `6 cm^(3)`B. `60 cm^(3)`C. `0.6 cm^(3)`D. `0.06 cm^(3)` |
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Answer» Correct Answer - D For water vapours, `P = 0.0006 g c c^(-1)` `:. 0.0006 = ("Mass")/("Volume") = ("Mass")/(1000)` `"Mass" = 1000 xx 0.0006 = 0.6 g` The density of liquid water is `1 g c c^(-1)` So, the volume occupied by water is `("Mass")/("Density"0 = (0.6)/(1) = 0.6 cc` |
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| 581. |
16.26 milligram of sample of an element X contains `1.66 xx 10^(20)` atoms. What is the atomic mass of the element X ? |
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Answer» Correct Answer - 59 g `1.66 xx 10^(20)` atoms X have mass = 16.26 mg `6.022 xx 10^(23)` atoms of X have mass `= ((6.022xx10^(23)"atoms"))/((1.66xx20^(20)"atoms"))xx(16.26mg)` `=59.0xx10^(3)mg=59.0g`. |
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| 582. |
One million of silver atoms weigh `1.79 xx 10^(-16)g`. Calculate the atomic mass of silver. |
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Answer» Correct Answer - 107.8 u `10^(6)` atoms of silver occupy mass `= 1.79 xx 10^(-16)g` `6.022 xx 10^(23)` atoms of silver occupy mass `= (1.79xx10^(-16)g)/(10^(6))xx6.02xx10^(23)=107.8g` Atomic mass of silver `= 107.8 g = 107.8 u`. |
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| 583. |
Calculate the number atoms of oxygen present in 88 g of `CO_(2)` ? What would be the weight of CO having the same number of oxygen atoms ? |
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Answer» 44.0 grams of `CO_(2)` contain oxygen atoms `= 2 xx 6.022 xx 10^(23)` atoms 88.0 grams of `CO_(2)` contain oxygen atoms `= 2 xx 2 xx 6.022 xx 10^(23) = 2.41 xx 10^(24)` atoms `6.022 xx 10^(23)` atoms of oxygen are present in `CO = 28.0g` `2.41 xx 10^(24)` atoms of oxygen are present in `CO=(28.0xx2.41xx10^(24))/(6.022xx10^(23))=112.0g`. |
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| 584. |
Calculate the number of molecules of oxalic acid `(H_(2)C_(2)O_(4).2H_(2)O)` in 100 mL of 0.2 N oxalic acid solition. |
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Answer» Step I. Calcualtion of mass of oxalic acid Normality of solution `= ("Mass of oxalic acid/Equivalent mass")/("Volume of solution in litres")` Normality of solition `=0.2 N=0.2 "equiv. L"^(-1)` Equivalent mass of acid `= ("Molar mass")/("Basicity")=(2xx1+2xx12+4xx16+2xx18)/(2)=(126)/(2)=63 "g equiv."^(-1)` Volume of solution `= 100 mL = (100)/(1000)=0.1 L` `:. ("0.2 equiv. L"^(-1))=("Mass of oxalic acid")/(("63 g equiv"^(-1))xx("0.1 L"))` Mass of oxalic acid `= (0.1 "equiv. L"^(-1))xx("63 equiv."^(-1))xx(0.1 L)=1.26 g` Step II. Calculation of no of molecules of oxalic acid Molar mass of oxalic acid `= 126 g mol^(-1)` Now, 126 g oxalic acid has molecules = Avogadro No. `= 6.022 xx10^(23)` 1.26 g of oxalic acid has molecules `= (6.022xx10^(23))/((126 g))xx(1.26g) = 6.022 xx 10^(21)`. |
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| 585. |
Calculate the molarity and normality of solution containing 3.15 of hydrated oxalic acid (`{:(COOH),(|),(COOH):}.2H_(2)O`) dissolved in 250 mL of the solution. |
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Answer» Step I. Calculation of molarity of the sodium Mass of oxalic acid `=3.15 g` Molar mass of oxalic acid `= 24+64 +2 +2 xx 18 = 126 u = 126 "g mol"^(-1)` No. of moles of oxalic acid `= ((3.15g))/((126"g mol"^(-1)))` Volume of solution `= 250 mL=(250)/(1000)=0.25 L` Molarity of solution `(M)=("No. of moles of oxalic acid")/("Volume of solution in litres")` `= ((3.15g))/((126"g mol"^(-1))xx(0.25 L))` `=0.1 "mol L"^(-1)=0.1 M` Step II. Calculation of normality of the acid solution Basicity of oxalic acid from the formula = 2 Equivalent mass of oxalic acid `= ("Molecular mass")/("Basicity")=((126 "g mol"^(-1)))/(2)=("63 g equiv"^(-1))` No. of equivalent of oxalic acid `= ("Mass of oxalic acid")/("Equivalent mass of oxalic acid")=((3.15g))/(("63 g equiv"^(-1)))` `=0.2 "equiv L"^(-1)=0.2 N` Note : The normality of the acid solution can also be calculated with the help of relation : Normality of acid = Molarity of acid `xx` Basicity `= 0.1 xx 2 = 0.2 N` |
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| 586. |
An aqueous solution of urea containing 18 g urea in 1500 `cm^(3)` of solution has a density of 1.5 `g//cm^(3)` . If the molecular weight of urea is 60. Then the molality of solution is:A. 0.2B. 0.134C. 0.064D. 1.2 |
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Answer» Correct Answer - B |
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| 587. |
`A` chemical commonly called "dioxin" has been very much in the news in the past few years. (It is the by product of herbicide manufacture and is thought to be quite toxic.) Its formula is `C_(12)H_(4)Cl_(4)O_(2)`. If you have a sample of dirt `(28.3 g)` that contains `1.0xx10^(-4)%` dioxin, how many moles of dioxin are in the dirt sample? |
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Answer» Correct Answer - `8.8 xx 10^(-8)"mol"` `n=((28.3xx1xx10^(-4))/(100))/([12xx12+4+35.5xx4+16xx2])=8.8xx10^(-8)` mole |
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| 588. |
`A` chemist wants to prepare diborane by the reaction `BF_(3) rarr 6 Li BF_(4)+B_(2)H_(6)` If the starts with `2.0` moles each of `LiH & BF_(3)`. How many moles `B_(2)H_(6)` can be prepared. |
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Answer» Correct Answer - `0.250` `6LiH+8BF_(3)rarr6LiBF_(4)+B_(2)H_(6)` `2" "2" "0.25` |
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| 589. |
Calculate no. of each atom present in `106.5 g` of`NaCIO_(3^(.)` |
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Answer» Correct Answer - `6.023 xx 10^(23)"atom" Na, 6.023 xx 10^(23)"atom"Cl, 18.06 xx 10^(23)"atom" O` `n_(NaClO_(3))=(106.5)/(106.5)=1` mole `NO`. of atom of `{:(Na=1xxN_(A)),(Cl=1xxN_(A)),(O=1xxN_(A)):}` |
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| 590. |
Express the following in scientific notation (i) `0.0048` (ii) `236,000` (iii) `8008` (iv) `600.0` (v) `500` (vi) `783.4`. |
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Answer» (i) `4.8 xx10^(-3)` (ii) `2.36 xx 10^(5)` (iii) `8.008 xx 10^(3)` (iv) `6.0 xx 10^(2)` (v) `5.0 xx 10^(2)` (vi) `7.834 xx 10^(2)`. |
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| 591. |
The volume of 1 mole of a gas at standard temperature and pressure is:A. `11.35` litresB. `22.7` litresC. 100 litresD. 22.4 litres |
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Answer» Correct Answer - B |
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| 592. |
One mole of a gas is defined as:A. the number of molecules in one litre of gasB. the number of molecules in one formula weight of gasC. the number of molecules contained in 12 grams of (12 C) isotopeD. the number of molecules in 22.7 litres of a gas at S.T.P. |
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Answer» Correct Answer - D |
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| 593. |
2.5 litres of 1M NaOH solution is mixed with 3.0 litres of 0.5 M NaOH solution. The molarity of resulting solution is :A. 0.80 MB. 1.0 MC. 0.73 MD. 0.50 M |
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Answer» Correct Answer - C `M_(1)V_(1)+M_(2)V_(2)+M_(3)V_(3)` or `M_(3)=(M_(1)V_(1)+M_(3)V_(3))/(V_(3))` `=(1xx2.5+0.5xx3.0)/(5.5)=0.73M` |
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| 594. |
The weight of one molecule of compound `C_(60)H_(22)` is :A. `1.2 xx 10^(-20) g`B. `1.4 xx 10^(-21) g`C. `5.025 xx 10^(23)g`D. `6.023 xx 10^(23)g` |
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Answer» Correct Answer - B Molecular mass of compound ` (C_(60)H_(122))=60xx12+122=842g` Mass of one molecular `= (842)/(6.022)xx10^(-23)g` `=1.4 xx 10^(-21)g`. |
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| 595. |
Volume occupied by one molecule of water (density `"1 g cm"^(-3)`) is :A. `3.0 xx 10^(-23) cm^(3)`B. `5.5 xx 10^(23) cm^(3)`C. `9.0 xx 10^(-23) cm^(3)`D. `6.023 xx 10^(-23) cm^(3)` |
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Answer» Correct Answer - A Mass of one molecule of water `= ((18g))/(6.022xx10^(23))=3xx10^(-23)g` Since density of water is 1g `:.` Volume occupied by one molecule of water `= 3 xx 10^(-23)` cc. |
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| 596. |
If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximatelyA. `3xx10^(-23)mL`B. `6xx10^(-22)mL`C. `3xx10^(-21)mL`D. `9xx10^(-23)mL` |
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Answer» Correct Answer - 1 `1xx10^(3)kg//m^(3)=1g//mL=1gm//c c" " [` Since ,`1m^(3)=10^(6)cm^(3)=10^(6)mL]` `6.022xx10^(23)H_(2)O` molecule weight `rarr 18g` `1H_(2)O` molecule weight `rarr (18)/(6.022xx10^(23))g=3xx10^(-23)g` `d=(mass)/(volume)," "` So, volume`=(3xx10^(-23))/(1(g//mL))=3xx10^(-23)mL` |
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| 597. |
`DNA` has density `1.1 g//mL`. And its molecular weight is `6xx10^(3) g//"mol"`. Average volume occupied by its single molecule will be :A. `9.1xx10^(-20) cc`B. `9.1xx10^(-21) cc`C. `9.8xx10^(-21) cc`D. `9.6xx10^(-20) cc` |
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Answer» Correct Answer - B `wt`. Of `1` molecule `=(6xx10^(3))/(6.02xx10^(23))` volume occupied by its `=("mass")/("density")=(6xx10^(3)//6.03xx10^(23))/(1.1)mL` `=9.1xx1^(-21)cc` |
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| 598. |
To `50 L` of `0.2 N NaOH, 5 L` of `1 N HCl` and `15 L` of `0.1 NFeCl_(3)` solution are added. What weight of `Fe_(2) O_(3)` can be obtained from the precipitate? Also report the nomality of `NaOH` left in the resultant solution. |
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Answer» Equivalent of `NaOH = 50 xx 0.2 = 10` Equivalent of `HCl = 5 xx 1 = 5` `:.` Equivalent of `NaOH` left after reaction with `HCl` `= 10 - 5 = 5` Also `NaOH` reacts with `FeCl_(3)` to give `Fe (OH)_(3)` which on ignition gives `Fe_(2) O_(3)`. `:.` Equivalent of `NaOH` used fro `FeCl_(3)` = Equvalent of `Fe(OH)_(3)` Equivalent of `Fe_(2) O_(3)` `= 15 xx 0.1 = 1.5` `:.` Equivalent of `NaOH` left finally `= 5 - 1.5 = 3.5` `N_(NaOH)` left `= (3.5)/(70) = 0.05 N` `:.` Total volume `= 70 L` Also, equivalent of `Fe_(2) O_(3) = 1.5 (Mw of Fe_(2)O_(3) = 160)` `:. (W)/(Mw//6) = 1.5` `:. W_(Fe_(2)O_(3)) = (1.5 xx 160)/(6) = 40 g` |
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| 599. |
The molecular mass of an organic acid was determind by the study of its barium salt. `4.290 g` of salt was converted to free acid by the reaction with `21.64 mL` of water of hydration per `Ba^(2+)` ion and the acid is monobasic. What is molecular weight of anhydrous acidgt |
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Answer» mEq of Braium salt = mEq of acid `(4.290)/(Mw..2) xx 1000 = 21.64 xx 0.477 xx 2` Molecular weight of salf = 415.60 Molecular weight of anion `= (("Mw of salt" - Mw of Ba^(2+) - Mw of 2 H_(2) O))/(2)` `= (415.60 - 137 - 36)/(2) = 121.16` `:.` Molecular weight of acid `= 121.6 + 1 = 122.6` |
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| 600. |
How many moles of `O` are present in `4.9 g` of `H_(3) PO_(4)`? (Atomic weight of `P, O` and `H = 31, 16, 1`) |
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Answer» Molecular weight of `H_(3) PO_(4) = 1 xx 3 + 31 + 16 xx 4 = 98 g` `98 g = 1 "mole of" H_(3) PO_(4) = 4` mole of `O` `4.9 g of H_(3) PO_(4) = (4)/(98) xx 4.9 = 0.2` mole of O |
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