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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.
| 351. |
Three metals of alkaline earth metal group `(A,B` and `C`) When reacted with a fixed volume of liquid `Br_(2)` separately gave a product (metal bromides) whose mass is plotted against the mass of metals taken as shown in the figure. From the plot, predict what relation can be concluded between the atomic weight of `A, B` and `C` a. `C gt B` b. `B gt A`, `C lt A lt B` Data is insufficient to predict |
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Answer» (a,b) Since mass of `Br_(2)` taken is constant, same moles of `MBr_(2)` will be produced in each ease. Since, `"moles" = ("Mass")/(Mw)` The heaviest metal will be produce maxmium of product. Hence, the correct order of atomic mass of `A, B, C` is `A lt B lt C`. |
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| 352. |
What is the number of significant figures in each of the following ? (i) Mass of electron `= 9.108 xx 10^(-31) kg` (ii) Mass of proton `= 1.672 xx 10^(-27) kg` (iii) `0.00564`. |
| Answer» `{:("(i)","Mass of electron ="9.108xx10^(-31)"kg.","(Four significant figures)"),("(ii)","Mass of proton ="1.672xx10^(-27)"kg","(Four significant figures)"),("(iii)",0.00564,"(Three significant figures)"):}` | |
| 353. |
The plastic industry uses large amounts of phthalic anhydride `C_(8)H_(4)O_(3)`, made by the controlled `C_(10) H_(8) + (9)/(2) O_(2) rarr C_(8) H_(4) O_(3) + 2 CO_(2) + (5)/(2) H_(2) O` Since some of the naphthalene is oxidised to other products, 80% yield is obtained. What weight of phthalic anhydrid would be produced by the oxidation of `256 g` of `C_(10)H_(8)`. `[Mw "of" C_(10) H_(8) = 128, Mw "of" C_(8) H_(4) O_(3) = 148]` |
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Answer» `128 g of C_(10 H_(8)` gives `= 148 g of C_(8) H_(4) O_(3)` `256 g of C_(10) H_(8)` gives ` = (148 xx 256)/(128) = 296 g` Actual weight of `C_(8) H_(4) O_(3) = (296 xx 80)/(100) = 236.8 g` `C_(8) H_(4) O_(3)` produced |
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| 354. |
Express the following to four significant figure: i. `6.58768 xx 10^(5)` ii. 8.35783` iii. 98.2350 iv. 0.003586 v. 90000 |
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Answer» i. `6.587 xx 10^(5)` and after rounding off `6.588 xx 10^(5)`. ii. 8.357 and after rounding off 98.24. iii. 98.23 and after rounding of 98.24. The righmost digit to be removed is 5, and the preceding number is 3 (odd number), so it is increased by one. iv. 0.003586 (zeros after decimal point and to the left of a number are not significant). v. `9.000 xx 10^(4)` |
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| 355. |
How many significant figures are there in each of the following numbers? a. `pi` b. The sum of `16.4 + 0.3254` c. The product of `12 xx 7.435` d. 0.0075 e. `5.033 xx 10^(22)` f. 7.007 g. 6000 h. The subtractin of 19.3 - 0.4567 |
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Answer» a. As `pi = (22)/(7) = 3.1428571`…, hence it has infinite number of significant figures. b. The result should have upto one decimal point in 16.4. The sum is 16.7254. So, the result is 16.7. So, it has three significant figure. c. Two significant figures. The product is 89.22. The result is reported with least number of significant figures involved in the calculation d. Two significant figures because the zeros on the left of the first non-zero number are not significant. e. Four significant figures becuase the first term gives the significant figures and exponential term is not considered. f. Four significant figures because the zeros between the non-zero digits are significant figures. Four significant figures. But in scientific of exponential notation significant figure vary, e.g., `6.0 xx 10^(3)` has two significant figures, `6.00 xx 10^(3)` or `6.000 xx 10^(3)` has three of four significant numbers, respectively. Three significant figures becuase the number with least significant figures involved in the calculation has three siginficant figures. |
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| 356. |
The following chemical reactions used to be untilized to rapidly produce large amounts of `N_(2)` gas inside an automobile air bag: `2NaN_(3) rarr 2 Na + 3 N_(2) (g)` `10 Na + 2 KNO_(3) rarr K_(2) + 5 Na_(2) O + N_(2) (g)` `K_(2) O+ Na_(2) O + SiO_(2) rarr` Alkaline silicate (glass) How many grams of `KNO_(3)` are needed to produce enough `N_(2)` to fill a `12.3 L` air baG AT `27^(@)C` and 4 atm. a. `202 g` b. `81.25 g` c. `404 g` d. `25.25 g` |
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Answer» d. `PV = nRT` `n_(N_(2)) = (PV)/(RT) = (4 xx 12.3)/(0.082 xx 300) = 2 mol` 2 mol produces `= 3 "mol" N_(2)` a. 10 mol `Na` produces ` = (3)/(2) xx 10 = 15 "mol" N_(2)` b. 10 mol `Na = 2 "mol" KNO_(3) -= "mol" N_(2)` `(15 "mol" N_(2))` Total mol of `N_(2)` produced `= 15 + 1` `= 16 "mol" N_(2))` `:. 16 "mol" N_(2) = 2 "mol" KNO_(3)k (Mw KNO_(3) = 101 g)` `= 2 xx 101 g KNO_(3)` `2 "mol" N_(2) = (2 xx 101 xx 2)/(16) = 25.25 g` `:.` Weight of `KNO_(3) = 25.25 g` |
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| 357. |
Fil in the blanks a. The mass of 1 molecule of water `(H_(2) O)` is ……… b. The number of molecules in `16 g` of sulphur dioxide `(SO_(2))` are ………. c. The weight of one mole of sodium carbonate `(Na_(2) CO_(3))` is…………….. d. Moles and g equivalent in `196 g` of `Ca (OH)_(2)` are ...... and ....... e. Moles and `g` equivalent in `196 g` of `H_(3) PO_(4)` are ............. and ............ f. `g` atoms in `62 g` of `P_(4)` are ....... f. `g` atoms in `24 g` of magnesium are........... |
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Answer» Correct Answer - A::B::C::D a. `(18)/(6.023 xx 10^(23)) = 3.0 xx 10^(-23)` b. `n_(SO_(2)) = (16)/(64) = (1)/(4)` Number of molecules `= (1)/(4) xx 6.023 xx 10^(23) = 1.5 xx 10^(23)` c. `Na_(2) CO_(3) = 2 xx 23 + 12 + 3 xx 16 = 106 g` d. `n_(Ca(OH)_(2)) = (196)/(74) = 2.648, ~~ 2.65` `Eq_(Ca(OH)_(2)) = 2.65 xx 2 = 5.3` (valency factor = 2) e. `n_(H_(3)PO_(4)) = (196)/(98) = 2, Eq = 2 xx 3` (tribasic acid) = 6 f. `g` atom or mol `= (62)/(31 xx 4) = 0.5` g. `g` atom or mol `= (24)/(24) = 1` |
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| 358. |
What is the difference between `2.0 m` and `2.00 m`. |
| Answer» They are scientifically different althouhg they seems to be same. `2.0 m` has two significant figures, and hence its precision is 0.1 part in 2, i.e, 50 ppt (part per thousand). `2.00 m` has three significant figures and its precision is 0.01 parts in 2, i.e., 5 ppt. Hence, `2.00 m` is more precise measurement than `2.0 m`. | |
| 359. |
Objective question (single correct answer). i. The molarity of a aqueous solution of glucose `(C_(6) H_(12) O_(6))` is 0.01 To `200 mL` of the solution, which of the following should be carried out to make it `0.02 M`? I. Evaporate `50 mL` of solution III. Add `0.180 g` of glucose and then evaporate `50 mL` of solution III. Add `50 mL` of water The correct option is: a. I b. III c. II d. I, II, III ii. The atomic mass of `Cu` is 63.546. There are only two naturally occuring isotopes of copper `Cu^(63)` and `Cu^(65)`. The percentage of natural abundance of `Cu^(63)` in nearly a. 30 b. 10 c. 50 d. 73 iii. An aqueous solution of urea `(NH_(2) COHN_(2))` is 3.0 molal. The mole fraction of urea is a. 0.33 b. 0.25 c. 0.66 d. 0.05 iv `0.2 M H_(2) SO_(4)` `(1 mL)` is diluted to 1000 times of its initial volume. the final normality of `H_(2) SO_(4)` is: a. `2 xx 10^(-3)` b. `2 xx 10^(-4)` c. `4 xx 10^(-4)` d. `2 xx 10^(-2)` v. Which of the following question are dependant on temperature? a. Molarlity b. Normality c. Mole fraction d. Molality vi. A sample of `H_(2) SO_(4)` density `1.85 mL^(-1)` is 90% by weight. What is the volume of the acid that has to be used to make `1 L` f `0.2 M H_(2) SO_(4)`? a. `16 mL` b. `18 mL` c. `12 mL` d. `10 mL` vii. The hydrated salt `Na_(2)SO_(4)`. `nH_(2)O` undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of `n` will be a. 5 b. 7 c. 3 d. 10 viii. 0.2 mol of `HCl` and 0.1 mol of barium chloride is dissolved in water to produce a `500 mL` solution. The molarity of `Cl^(ɵ)` is. a. `0.06 M` b. `0.12 M` c. `0.09 M` d. `0.80 M` ix. The density of `1 M` solution of `NaCl` is `1.055 g mL^(-1)`. The molality of the solutions is. a. 1.0585 b. 1.00 c. 0.0585 d. 0.10 x. Hydrochloric acid solution `A` and `B` have concentration of `0.5 N`, and `0.1 N`, respectively. The volume of solutions `A` and `B` required to make `2 L` of `0.2 N` hydrochloric acid are a. `0.5 L of A + 1.5 of B` b. `1.0 L of A + 1.0 L of B` c. `0.75 L of A + 1.25 L of B` d. `1.5 L of A + 0.5 L of B` |
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Answer» Correct Answer - A::B::C::D i. c. Since the molarity of glucose has to be increased (from `0.01 M` to `0.02 M`), so this can be carried out either by evaporating the solution or by adding some more glucose. So, the only possible answers is (c ). I. Evaporate `50 mL` of solution mmoles of glucose initially `= 0.01 xx 200 = 2` volume after evaporation `= 200 - 50` `= 150 mL` `M_("glucose") = (("mmoles")/(V_(mL))) = (2)/(150) = 0.013 M` II. mmoles of glucose added `= (0.180)/(180) xx 100 = 1` Total mmoles of glucose `= 2 + 1 = 3` Volumes `= 150 mL` (after evaporation of `50 mL` solution) `M_("glucose") = (3)/(150) = 0.02 M` III. Add `50 mL` of water New volume of solution `= 200 + 50 = 250 mL` `M_("glucose") = (2 mmol)/(250 mL) = 0.008 M` Hence, answer is (c ) ii. d. `63.546 = (a xx 63 + (100 - a) xx 65)/(100)` `a = 72.7% ~~ 73.%` iii. d. `m = (X_(2) xx 1000)/((1 - x_(2)) xx Mw_(1))` `3 = (X_(2) xx 1000)/((1 - x_(2)) xx 18)` Solve for `x_(2) = 0.05` iv. c. `M_(1) V_(1) = M_(2) V_(2)` `0.2 M xx 1 mL = M_(2) xx 1000 mL` `M_(2) = 2 xx 10^(-4)` `:. N = 2 xx 2 xx 10^(-4) = 4 xx 10^(-4)` vi. c. `M_(1) V_(1) = M_(1) V_(2) (M = (% "by weight" xx 10 xx d)/(Mw_(2)))` `V_(1) xx (90 xx 10 xx 1.8)/(98) = 0.2 xx 1 L` `V_(1) = 0.012 L = 12 mL` vii. d. Loss in weight is due to `nH_(2) O`. `:. (142 + 18 n) g of Na_(2) SO_(4). nH_(2) O = 18 n g` of loss in weight of `H_(2) O` `100 g of Na_(2) SO_(4) . nH_(2) O = (18 n xx 100)/(142 + 18 n)` `:. (18 n xx 100)/(142 + 18n) = 55.9` solve for `n implies n = 9.99 ~~ 10` viii. d. `0.02 "mol" HCl = 0.02 of H^(o+) + 0.2 "mol" Cl^(ɵ)` `0.01 "mol" BaCl_(2) = 0.1 "mol" of Ba^(2+) + 0.1 xx 2 "mol" of Cl^(ɵ)` Total `Cl^(ɵ) = 0.4 "mol"`. Total volume `= 500 mL = (1)/(2) L` `:. [Cl^(ɵ)] = (0.4)/(1//2) = 0.8 M` ix. b. `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `1.0585 = 1 M ((58.5)/(1000) + (1)/(m))` solve for `m` `m = 1.0` x. a. `V_(1) + V_(2) = 2 L` `N_(1) V_(1) + N_(2) V_(2) = N_(3) V_(3)` `0.5 xx V_(1) + 0.1 xx V_(2) = 0.2 xx 2` `0.5 V_(2) + 0.1 V_(2) = 0.4` Solve equations (i) and (ii) `V_(1) = 0.5 L, V_(2) = 1.5 L` |
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| 360. |
The density of copper is `7.8 g cm^(-3)` and its weight is `5.642 g`. Report the volume of copper to correct decimal point. |
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Answer» Volume `= ("Mass")/("Density") = (5.642 g)/(7.8 g cm^(-3))` `= 0.7233 cm^(3) = 0.72 cm^(3)` The result should have two significant figures because the least precise term (7.8) has two significant figures. |
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| 361. |
Objective question (only one correct). I. Which of the following has least mass? a. 1 mol of `S` b. `3 xx 10^(23)` atom of `C` c. `2 g` atom of nitrogen d. `7.0 g of Ag` ii. The simplest formula of a compound containing 50% of element `A` (atomic mass 10) and 50% of element `B` (atomic mass 20) is: a. `AB` b. `A_(2) B` c. `A_(2) B_(3)` d. `AB_(3)` |
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Answer» Correct Answer - A::B a. 1 mol `S = 32 g` b. `(12 xx 3 xx 10^(23))/(6 xx 10^(23)) = 6 g` c. `2 g` atom of `N = 28 g` d. `7.0 g Ag` ii.b. mol of `A = (50)/(10) = 5`, mol of `B = (50)/(20) = 2.5` `A:B = 5:2.5 = 2:1` Formula `A_(2) B`. iii.a. Weight of `H_(2) O = V xx d = 0.0018 mL xx 1 = 0.0081 g` `n_(H_(2)O) = (0.0018)/(18) = 10^(-4) mol = 6.023 xx 10^(23) xx 10^(-4)` `= 6.023 xx 10^(19)` |
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| 362. |
A metal `M` of atomic weight 54.9 has a density of `7.42 g cm^(-3)`. Calculate the volume occupied and the radius of the atom of this metal assuming it to be sphere. |
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Answer» Correct Answer - A::B::C::D Mass of 1 atom of metal `M = (54.94)/(6.023 xx 10^(23))`, `V = (54.94)/(6.023 xx 10^(23) xx 7.42) = 1.3 xx 10^(23) cm^(3)` `V = (4)/(3) pi r^(3) implies 1.23 xx 10^(4)/(3) pi r^(3)` `:.r = 1.42 xx 10^(-8) cm` |
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| 363. |
Express the result of the following data to the appropriate number of significant figures. `(4.84 xx 0.0744)/(0.016)` |
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Answer» `(4.84 xx 0.0744)/(6.016) = 0.0598885` As 4.84 has least number of three significant figures, the result should contain three signigicant figures only. Hence the result after rounding off is 0.0599. |
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| 364. |
Calculate the amount of calcium oxide required when it reacts with `852 g` of `P_(4) O_(10)`. |
| Answer» Correct Answer - `1008g` | |
| 365. |
Calculate the amount of calcuium oxide required when it reacts with `852 g` of `P_(4) O_(10)`. |
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Answer» Correct Answer - A `6CaO + P_(4) O_(10) rarr 2Ca_(3) (PO_(4))_(2)` `6CaO -= P_(4)O_(10)` `(6 (40 + 16))/(W) = ((31 xx 4) + (16 xx 10))/(852)` or `W = 1008 g` |
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| 366. |
Which of the following statements `is//are` correctA. Mass of `Al_(2) (SO_(4))_(3).18H_(2) O` needed ot make up `100 mL` of an aqueous solution of concentration `27.0 mg` of `Al^(3+)` per `mL` is `33.3 g`B. Mass of `CrCl_(3).6H_(2)O (Mw = 266.5 g)` needed ot prepare `1.0 L` solution containing `26.0 g Cr^(3+)` per litre is `133.25 g`. (Atomic weight of `Cr = 5 g`)C. Mass of `NH_(4) Cl` needed to prepare 100ml for solution containing `80 mg` `NH_(4)Cl` per `mL` is `8.0 g`D. Mass of `NH_(3)` per `mL` of solution needed for solution of `NH_(3)` in water containing 20% `NH_(3)` by weighter (density `= 0.8 g mL^(-1)_` is `0.16 g mL^(-1)` |
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Answer» Correct Answer - A::B::C::D a. Weight `g Al^(3+)` in `100 mL` of solution `= 100 xx 27` `= 2700 mg = 2.7 g` 1 mol `Al^(3+) (27 xx 2 g) -= 1 "mol of" Al_(2) (SO_(4))_(3) . 18 H_(2) O` `2 "mol" Al^(3+) (27 xx 2 g) -= 1 "mol of" Al_(2) (SO_(4))_(3). 18 H_(2) O` `= 666 g` `2.7 g "of" Al^(3+) = (666 xx 2.7)/(54) = 33.3 g` b. moles of `Cr^(3+) = (26)/(52) = 0.5 "mol"` Weight of `CrCl_(3). 6H_(2)O = 0.5 "mol" Cr^(3+) xx 266.5 g` `= 133.25 g` c. Weight of `NH_(4) Cl = (100 mL) ((80 mg)/(mL))((10^(-3) g)/(mg)) = 8.0 g` d. Weight of `NH_(3) = ((0.8 g)/(mL "solution"))((20 g NH_(3))/(100 g solution))` `= 0.16 g mL^(-1)` |
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| 367. |
An excellent solution for cleaning grease stains from cloth of leather consists of the folllowing components: `C Cl_(4)` (80% by volume), ligroin (16%) and amyl alcohol (4%) How many `mL` of each should to taken to make up `80 mL` of solution?A. `64 mL C Cl_(4)`B. `12.8 mL` ligroinC. `32 mL` of amy alcoholD. `3.2 mL` of amyl alcohol |
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Answer» Correct Answer - A::B::D a. Volume of `C Cl_(4) = 80 xx (80)/(100) = 64 mL` b. Volume of ligroin `= 80 xx (16)/(100) = 12.8 mL` c. Wrong d. volume of amyl alcohol `= 80 xx (4)/(100) = 3.2 mL` |
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| 368. |
`A_(2)+2B_(2)rarrA_(2)B_(4)` `(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)` Two substance `A_(2) & B_(2)` react in the above manner when `A_(2)` is limited it gives `A_(2)B_(4)` in excess gives `A_(3)B_(4^(.) A_(2)B_(4)` can be converted to `A_(3)B_(4)` when reacted with `A_(2)`. Using this information calculate the composition of the final mixture when the mentioned amount of `A % B` are taken `:c` `(a) 4` mole `A_(2) & 4`mole `B_(2)` `(b) (1)/(2)` moles `A_(2) & 2` moles `B_(2)` `(c ) 1.25` moles `A_(2) & 2` moles `B_(2)` |
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Answer» Correct Answer - `(a) A_(3)B_(4)=2&A_(2)=1 ; (b) A_(2)B_(4)=(1)/(4)&B_(2)=1(c )A_(2)B_(4)=0.5&A_(3)B_(4)=0.5` `{:((a)A_(2)+2B_(2)rarrA_(2)B_(4)),("Initial" 4" "4),("After"4-2" " "4-4 "2),(" "2" "0" "2):|(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)` `2A_(2)B_(4) " "+" "A_(2)" "rarr " "2A_(3)B_(4)` `2" "2` `2-2" "2-1" "2` `1" "2` `therefore A_(2)=1,A_(2)B_(4) =2` `{:(A_(2)+2B_(2)rarrA_(2)B_(2)),("initial"(1)/(2)" "2),("After"1.25-1-" "1),(0.25 " "-" "1):}:|{:(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4):}` `therefore A_(2)B_(4)=0.5,B_(2)=1` `{:(A_(2)+2B_(2)rarrA_(2)B_(4)),("Initial"1.25" "2),(0.25" "-" "1):}:|(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)` `2A_(2)B_(4)+A_(2)rarr2A_(3)B_(4)` `1" "0.25` `1-0.5" "-" "0.5` `therefore A_(2)B_(4)=A_(3)B_(4)=0.5` |
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| 369. |
What is no of `O_(2)` molecules in `3.2 xx 10^(-15)g` sample of oxygen ? . |
| Answer» Moles of `O_(2)=(3.2xx10^(-15))/(32)=10^(-16)` Molecules of `O_(2) = 10^(-16) N_(A) = 6.023 xx 10^(7)` . | |
| 370. |
Find mass of `12.046 xx 10^(23)` atoms `.^(12)C_(6)` sample ? . |
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Answer» Mass of `.^(12)C_(6)` atom `=12 xx 1.66 xx 10^(-24)g` Mass of `12.046 xx 10^(23)` atoms `=12 xx 1.66 xx 10^(-24) xx 12.046 xx 10^(23) = 24gm` . |
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| 371. |
How many atoms of each type present in 143 g of washing soda `(Na_(2)CO_(3).10H_(2)O)` ? |
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Answer» Molar mass of washing soda `(Na_(2)CO_(3).10H_(2)O)=2xx23+12+3xx16+10xx18` `=46+12+4180=286` g 286 g of washing soda = 1 gram mol 143 of washing soda `= (1)/(286) xx 143 = 0.5` gram mol (i) No. of sodium atoms present 1 gram mole of washing soda contains Na atoms `= 2xx 6.022 xx 10^(23)` 0.5 gram mole of washing soda contains Na atoms `= 2xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(23)` (ii) No. of carbon atoms present 1 gram mole of washing soda contains C atoms `= 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains C atoms `= 6.022 xx 10^(23)` 0.5 gram mole of washing soda contains C atoms `= 6.022 xx 10^(23) xx 0.5 = 3.011 xx 10^(23)` (iii) No. of oxygen atoms present 1 gram mole of washing soda contains O atoms `= 13 xx 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains O atoms `= 13 xx 6.022 xx 10^(23) xx 0.5 = 3.914 xx 10^(24)` (iv) No. of hydrogen atoms present 1 gram mole of washing soda contains H atoms `= 20 xx 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains H atom `= 20 xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(24)`. |
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| 372. |
What weight of calcium the same number of atoms as are present in 3.2g of sulphur ? |
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Answer» Step I : No. of atoms in 3.2 g of sulphur Gram atomic mass of `S = 32` g 32 g of sulphur atoms `= 6.022 xx 10^(23)` 3.2 g of sulphur contain atoms `= (6.022xx10^(23))/((32g))xx(3.2g)=6.022xx10^(22)` Step II. Weight of `6.022 xx 10^(22)` atoms of calcium Gram atomic mass of Ca = 40 g `6.022 xx 10^(23)` atoms of Ca weigh = 40 g `6.022 xx 10^(22)` atoms of Ca weigh `= ((40g))/(6.022xx10^(23))xx6.022xx10^(22)=4g`. |
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| 373. |
Find out the equivalent weight of `H_(3) PO_(4)`in the reaction: `Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2 H_(2) O` |
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Answer» Correct Answer - D `Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2 H_(2) O` In the above equation, `2H` atom has reacted, so `H_(3) PO_(4)` acts as a dibasic acid. `( :. Mw of H_(3) PO_(4) = 3 + 31 + 16 xx 4 = 98 g)` `:. Ew of H_(3) PO_(4) = (Mw)/(n) = (98)/(2) = 49 g` |
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| 374. |
`N_(2) + 3 H_(2) rarr 2NH_(3)` Molecular weight of `NH_(3)` and `N_(2)` and `x_(1)` and `x_(2)`, respectively. Their equivalent weights are `y_(1)` and `y_(2)`, respectively. Then `(y_(1) - y_(2))`A. `((2x_(1) - x_(2))/(6))`B. `(x_(1) - x_(2))`C. `(3x_(1) - x_(2))`D. `(x_(1) - 3x_(2))` |
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Answer» Correct Answer - A `N_(2) -= 2NH_(3) -= 3H_(2) -= 6H` `Ew "of" N_(2) = (x_(2))/(6) = y_(2)` `Ew "of" NH_(3) = (2x_(1))/(6) = y_(1)` `:.y_(1) = y_(2) = ((2x_(1))/(6) - (x_(2))/(6)) = ((2x_(1) - x_(2))/(6))` |
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| 375. |
Which of the following has the same mass ?A. 0.1 mole of `O_(2)` gasB. 0.1 mole of `SO_(2)` gasC. `6.02 xx 10^(22)` molecule of `SO_(2)` gasD. `1.204 xx 10^(23)` molecule of `O_(2)` gas |
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Answer» Correct Answer - B::C::D All the above options correspond to same mass of `SO_(2)` which is 0.1 mol. |
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| 376. |
Which of the following have the same number of significant figures ?A. `0.050`B. `0.50`C. `5.0`D. 50 |
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Answer» Correct Answer - A::B::C All have the same number of significant figure (2) |
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| 377. |
8 g of `O_(2)` has the same number of molecules as in :A. 11 g `CO_(2)`B. 22 g `CO_(2)`C. 7 g COD. 14 CO |
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Answer» Correct Answer - A::C 8g of `O_(2)` (0.25 mol) have the same no. of moles in 11g of `CO_(2)` (0.25 mol) and 7g of CO (0.25 mol). Both of them have same number of molecules. |
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| 378. |
Which of the following represent 32 g fo the substance ?A. one mole of oxygen moleculesB. one gram atom of sulphurC. `22.4 dm^(3)` of `O_(2)` gas at S.T.PD. 1 gram mole of sulphur molecules |
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Answer» Correct Answer - A::B::C Correspond to 32 g. |
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| 379. |
Assertion : 22 carat gold is a mixture Reason : A compound has fixed composition of the elements present in it.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
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Answer» Correct Answer - B 22 carat gold is a homogeneous mixture un which gold is mixed with small amount of copper or silver. |
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| 380. |
Assertion : Empirical and molecular formulae of `NaHCO_(3)` are the same Reason : Upon heating, `NaHCO_(3)` evolves `CO_(2)` gas.A. Both assertion and reason are correct and reason is correct explanation for assertionB. Both assertion and reason are correct but reason is not correct explanation for assertionC. Assertion is correct but reason is incorrectD. Both assertion and reason are incorrect |
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Answer» Correct Answer - B The value of n in this case is 1. Therefore, empirical and molecular formulae are the same |
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| 381. |
A substance used as a water softener has the following mass percentage composition : `42.07% Na, 18.9% P`, and `39.04%` of oxygen. Determine the empirical formula of the compound. `(Na = 23, P = 31, 0 = 16)` |
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Answer» `{:("Element", "Moles", "Least ratio"),(Na, (42.07)/(23) = 1.829, (1.829)/(0.6) = 3),(P, (18.9)/(31) = 0.6, (0.6)/(0.6) = 1),(O, (39.04)/(16) = 2.44, (2.44)/(0.6) = 4):}` Empirical formula `= Na_(3) PO_(4)` |
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| 382. |
An orgainc compound contains `43.98% C`, `2.09% H`, and `37.2% Cl`. Calculate its empirical formula. |
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Answer» Oxygen `= 100 - 43.98 - 2.09 - 37.2 = 1673`. `{:("Element", "Mole", "Least ratio", "Whole number ratio"),(C , (43.98)/(12) = 3.66, (3.66)/(1.04) = 3.5, 7),(H, (2.09)/(1) = 2.09, (2.09)/(1.04) = 2, 4),(Cl, (37.2)/(35.5) = 1.04, (1.04)/(1.04) = 1, 2),(O, (16.73)/(16) = 1.04, (1.04)/(1.04), 2):}` Empirical formula `= C_(7) H_(4) Cl_(2) O_(2)` |
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| 383. |
The density of 3 M sodium thiosulphate is 1.25 g/ml . Identify the correct statements among the following.A. % by weight fo sodium thiosulphate is 37.92.B. The mole fraction of sodium thiosulphate is 0.065.C. The molarity of `Na^(+)` is 2.53 and `S_(2)O_(3)^(2-)` is 1.25.D. `S_(2)O_(3)^(2-)` contains S-O-S linkage. |
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Answer» Correct Answer - A::B |
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| 384. |
632 g of sodium thiosulphate `(a_(2)S_(2)O_(3))` reacts with copper sulphate to form cuproc thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothisosulphate `(Na_(4)[Cu_(6)(S_(2)O_(3))_(5)])`, (MW=1033) `CuSO_(4) + Na_(2)S_(2)O_(3) rarr CuS_(2)O_(3) + Na_(2)SO_(4)" "` [very fast] `2CuSO_(4)+Na_(2)S_(2)O_(3)+Na_(2)S_(2)O_(3) rarr Cu_(2)S_(2)O_(3) + Na_(2)S_(4)O_(6)` `3Cu_(2)S_(2)O_(3) + 2Na_(2)S_(2)O_(3) rarr Na_(4)[Cu_(6)(S_(2)O_(3))_(5)]` `" "` (Sodium cuprothisoulphate ) In this process , 0.2 mole of sodium cuprothiosulphate is formed .(O=16 , Na=23 , S=32) Moles of sodium thiosulophate reacted and unreacted after the reaction are respectively,A. 3 and 2B. 2 and 3C. 2.2 and 1.8D. 1.8 and 2.2 |
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Answer» Correct Answer - C |
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| 385. |
Calculate the molality `(m)` of `3 M` solution of `NaCl` whose density is `1.25 g mL^(-1)`. |
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Answer» Molar mass `(Mw_(2))` of `NaCl = (23 + 35.5) g = 58.5 g` `M = 3 mol L^(-1)` Mass of weight `(W_(2))` of `NaCl` in `1 L` solution `3 xx 58.5 = 175.5 g` First method: `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `1.25 g mL^(-1) = 3 ((58.5)/(1000) + (1)/(m))` `(1.25)/(3) (58.5)/(1000) = (1)/(m)` `0.416 - 0.058 = (1)/(m)` `0.358 = (1)/(m)` `m = (1)/(0.358) = 2.79 m` second method: `M = 3 mol L^(-1)` Mass of `NaCl` in `1 L` solution `(W_(2)) = 3 xx 5835 = 175.5 g` Mass of `1 L` solution `= V_(sol) xx d_(sol)` `= 1000 xx 1.25 = 1250 g` Mass of `H_(2) O` in solution `(W_(1))` Mass of solution - Mass of slute `= W_(sol) - W_(2)` `= 1.250 - 175.5 = 1074 g = 1.0754 kg` `:. m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))` `= (175.5 xx 1000)/(58.5 xx 1074.5) = 2.79 m` or `m = ("Number of moles of solute")/("Mass of solvent in kg")` `= (3 mol)/(1.0745 kg) = 2.79 m` |
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| 386. |
The density of 3 M solution of NaCl is `1.25 "g L"^(-1)`. The molality of the solution is :A. 2.79B. 1.79C. 0.79D. 2.98 M |
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Answer» Correct Answer - A Density (d) `= ("Mass of solution")/("Volume of solution")` Mass of solution `= (1.25 gL^(-1)) xx(1000 mL) = 1250 g` Mass of solvent = Mass of solution - Mass of solute `= (1250 - 3 xx 58.5) = 1074.30 g` Molality of solution `= ("No. of moles of solute")/("Mass of solvent in kg")=(("3 mol"))/(("1.074 kg"))` `=2.79"mol kg"^(-1)=2.79` m. |
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| 387. |
The molality of a `H_(2)O_(2)` solution of density 1.068` gm/ml is 2m. The only incorrect concentration of the same solution is :A. molarity =2MB. volume strength =22.7 vol at STPC. 6.8`%`(w/w)D. 6.8 `%` (w/w) |
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Answer» Correct Answer - D |
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| 388. |
The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following elements contains the greatest number of atoms ?A. 4 g HeB. 46 g NaC. 0.40 g CaD. 12 g He |
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Answer» Correct Answer - D 4 g He = 1 mol 46 g Na = 2 mol 0.4 g Ca = 0.01 mol 12 g He = 3 mol Therefore (d) is the correct option. |
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| 389. |
An oleum sample contains `10 g SO_(3)` and `15 g H_(2) SO_(4)` Answer the following questions on the basis of above information : `%` labelling of oleum sample is .A. `27.25%`B. `106%`C. `109%`D. `118%` |
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Answer» Correct Answer - (C ) `%` labelling of oleum sample `= (100 + x) %` `{:(SO_(3),+,H_(2)O,rarr,H_(2)SO_(4)),(10g,,,,),(1//8mol,,1//8mol,,):}` `2.25g` `:. %` labelling `= (100 + 9) % = 109 %` . |
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| 390. |
Atomic mas of an element is :A. actual mass of one atom of the elementB. relative mass of an atom of the elementC. average relative mass of different atoms of the elementD. always a whole numberEx |
| Answer» Correct Answer - C | |
| 391. |
An oleum sample contains `10 g SO_(3)` and `15 g H_(2) SO_(4)` Answer the following questions on the basis of above information : Find new `%` labeling of `0.45g` of `H_(2)O` is added to the above oleum sampleA. `100%`B. `102.83%`C. `107.07%`D. `109%` |
| Answer» Correct Answer - (C ) | |
| 392. |
That the atom is indivisible was proposed by .A. RuterfordB. DaltonC. BohrD. Einstein |
| Answer» Correct Answer - B | |
| 393. |
200 mL of water is added to 500 mL of 0.2 M solution. What is the molarity of the diluted solution ?A. 0.5010 MB. 0.2897 MC. 0.7093 MD. 0.1428 M |
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Answer» Correct Answer - D `M_(1)V_(1)-=M_(2)V_(2)` `0.2Mxx500-=M_(2)xx700` `M_(2)=((0.2M)xx500)/(700)=0.1428M` |
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| 394. |
Copper `(Cu)` and `(Zn)` react differently with `HNO_(3)` as follows: `Cu + 4H^(o+) (aq) + 2NO_(3)^(ɵ) (aq) rarr 2NO_(2)(g) _ Cu^(2+) + 2H_(2)O` `4Zn +10 H^(o+) (aq) + 2NO_(3)^(ɵ) (aq) rarr NH_(4)^(o+) + 4Zn^(2+) + 3H_(2)O` What volume of `2.0 M HNO_(3)` would react with `10.0g` of a brass `(90.% Cu, 10.0% Zn)` according to the above equation?A. `~~ 100 mL`B. `~~ 150 mL`C. `~~ 200 mL`D. `~~ 300 mL` |
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Answer» Correct Answer - D Weight of `Cu = (10 xx 90)/(100) = 9.0 g` Weight of `Zn = (10 xx 10)/(100) = 1.0 g` Mol of `HNO_(3)` with `Cu = (9.0 g Cu)` `((1 "mol" Cu)/(63.5 g))((4 "mol"HNO_(3))/(1 "mol"Cu))` `= (9.0 xx 1 xx 4)/(63.5) = 0.56 "mol" HNO_(3)` Mol of `HNO_(3)` with `Zn = (1.0 g)` `((1 "mol"Zn)/(65.37))((10"mol" HNO_(3))/(4 "mol" Zn))` `= (1.0 xx 10)/(65.37 xx 4)` `= 0.038 "mol" HNO_(3)` Total moles of `HNO_(3) = 0.56 + 0.38 = 0.598 ~~ 0.6` `M xx V_(L) = "moles"` `2.0 xx V_(L) = 0.6` `V_(L) = 0.3 = 300 mL` |
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| 395. |
Assertion (A) : If `30 mL` of `H_(2)` and `20 mL` of `O_(2)` react to form water, `5 mL` of `H_(2)` is left at the end of the reaction. Reason (R ): `H_(2)` is the limiting reagent.A. If both (A) and (R ) are correct and (R ) is the correct explantion for (A)B. If both (A) and (R ) are correct but (R ) is not the correct explantion for (A)C. If (A) is correct but (R ) is incorrect.D. If (A) and (R ) are incorrect. |
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Answer» Correct Answer - D `(A)` is wrong, but `(R )` is correct. `underset(30 mL)(H_(2)) + underset({:(15 mL),("reacted"):})((1)/(2) O_(2)) rarr H_(2) O` volume of `O_(2)` left `= 20 - 15 = 5 mL` `(R )` is correct, since `H_(2)` is completely consumed, and hence is the limiting reagent. |
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| 396. |
If 30 mL of `H_(2)` and 20 mL of `O_(2)` react to form water, what is left at the end of the experiment ?A. 10 mL of `H_(2)`B. 5 mL of `H_(2)`C. 10 mL of `O_(2)`D. 5 mL of `O_(2)` |
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Answer» Correct Answer - D `{:(H_(2)(g),+,1//2O_(2)(g),rarr,H_(2)O(g)),("1 mol",,"1/ mol",,"1 mol"),("1 mL",,"1/2 mL",,"1 mL"),("30 mL",,"15 mL",,"30 mL"):}` Volume of `O_(2)` left uncreacted `=20-15=5 mL` |
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| 397. |
Density of azone relative to oxygen is under the same temperature & pressure :A. 1B. 2C. 1.5D. 2.5 |
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Answer» Correct Answer - C `R.D. = ("Density of" O_(3))/("Density of" O_(2))` at same temp. & pressure of density `alpha` Mw `=("Mw of" O_(3))/("Mw of" O_(2)) = (48)/(32) = (3)/(2) = 1.5` |
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| 398. |
Calculate the mole fraction of benzene in a solution containing 30 % by mass of it in carbon tetrachloride. |
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Answer» Let us start with 100 of the solution in which Mass of benzene = 30 g Mass of carbon tetrachloride = 70 g Molar mas of benzene `(C_(6)H_(6))=6xx12+6xx1=78" g mol"^(-1)` `n_(C_(6)H_(6))=((30g))/((78 " g mol"^(-1)))=0.385` mol Molar mass of carbon tetrachloride `(C Cl_(4))=12+4 xx 35.5 = 154 "g mol"^(-1)` `n_( C Cl_(4))=((70g))/((154"g mol"^(-1)))=0.454` mol `x_(C_(6)H_(6))=(n_(C_(6)H_(6)))/(n_(C_(6)H_(6))+n_(C Cl_(4)))=((0.385" mol"))/((0.385" mol")+(0.454" mol"))=0.459`. |
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| 399. |
How many significant figures should b e present in the answer of the following calculations? `(2.5xx1.25xx3.5)/(2.01)` |
| Answer» The least precise numbers (2.5 and 3.5) have two significant figures. Therefore, the answer should have only two signicant figures. | |
| 400. |
Show that the data given below is in agreement with the law of constant composition `{:("Exp. No.","Mass of iron","Mass of iron oxide"),("(i)","2.7812 g","3.9756 g"),("(ii)","3.0499 g","4.3596 g"),("(iii)","4.4913 g","6.4202 g"):}` |
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Answer» Experiement No. (i) Mass of iron = 2.7812 g , Mass of iron oxide = 3.9756 g Percentage of iron `= ((2.7812g))/((3.9756g))xx100=70%` Percentage of oxygen `= 100 - 70 = 30 %` Experiment No. (ii) Mass of iron = 3.0499 , Mass of iron oxide = 4.3596 g Percentage of iron `= ((3.0499))/((4.3596g))xx100=70%` Percentage of oxygen `= 100-70=30%` Experiment No. (iii) Mass of iron = 4.4913 g , Mass iron oxide = 6.4202 g Percentage of iron `= ((4.4913g))/((6.42402g))xx100=70%` Percentage of oxygen `= 100 - 70 = 30 %` Since the ratio by weights of iron and oxygen in the three oxides is the same, the data is according to Law of constant Composition. |
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