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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.
| 651. |
Which of the following has the maximum number of atoms?A. (a)`24 " g of" C(12)`B. (b)`56 " g of" Fe(56)`C. (c )`27 " g of o Al" (27)`D. (d)`108 " g of Ag" (108)` |
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Answer» Correct Answer - A (a) `24 g` of `C(12)` has maximum number of atoms. Number of atoms in `24 g` of `C=24/12xx6.023xx10^(23)` `=2xx6.023xx10^(23)` atoms (b) Number of atoms in `56 g` of `Fe =56/56xx6.023xx10^(23)` `=1xx6.023xx10^(23)` atoms (c ) Number of atoms in `27 g` of `Al=27/27xx6.023xx10^(23)` atoms `=1xx6.023xx10^(23)` atoms (d) Number of atoms in `108 g` of `Ag=108/108xx6.023xx10^(23)` `=1xx6.023xx10^(23)` atoms |
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| 652. |
How many secounds are there in 4 days?A. `172800 s`B. `345600 s`C. `259200 s`D. `216000 s` |
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Answer» Correct Answer - B `(4 "days") xx ((24h)/(1 "day")) xx ((60 min)/(1 h)) xx ((60 s)/(1 min))` `= 345600 s` |
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| 653. |
What mass of hydrochloric acid is needed to decompose 50 g of limestone?A. 36.5 gB. 73 gC. 50 gD. 100 g |
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Answer» `underset(100g)(CaCO_3)+underset(2xx36.5=73g)(2HCl) to CaCl_2 + H_2O +CO_2` 100 g of `CaCO_3` required 73 g of HCl `therefore` 50 g of `CaCO_3 " requires " 73/100xx50=36.5` g of HCl |
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| 654. |
How many seconds are there in 3 days?A. 259200 sB. 172800 sC. 24800 sD. 72000 s |
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Answer» Second in 3 days can be calculated as `3 day xx (24 h )/( 1 day )xx(60 "min")/(1 h)xx(60s)/(1 "min")=3xx24xx60xx60s = 259200s` |
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| 655. |
How much mass of silver nitrates will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrates if law of conservation of mass is followed?A. 22.85 gB. 108 gC. 17.0 gD. 28.70 g |
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Answer» `underset(x)(AgNO_3)+underset(5.85g)(NaCl) to underset(8.5g)(NaNO_3)+ underset (14.35g)(AgCl)` x+5.85=8.5+14.35 `rArr` x=17g |
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| 656. |
4.88 g of `KClO_3` when heated produced 1.92 g of `O_2` and 2.96 g of KCl. Which of the following statements regarding the experiment is correct?A. The result illustrates the law of conservation of mass.B. The result of illustrates the law of multiple proportions.C. The result illustrates the law of constant proportion.D. None of the above laws is followed. |
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Answer» `underset(4.88g)(2KClO_3)tounderset(2.96g)(2KCl)+underset(1.92g)(3O_2)` Since, mass of the products (2.96+1.92) is equal to mass of the reactant, this illustrates the law of conservation of mass. |
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| 657. |
Law of constant composition doesnot hold good forA. Exothermic compoundsB. Endothermic compoundsC. Non stoichiometric compoundsD. Stoichiometric compounds |
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Answer» Correct Answer - C Law of costant composition does not hold good for non-stoichoimetric compounds. |
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| 658. |
Law of constant composition doesnot hold good forA. Endothermic compoundsB. exothermic compoundsC. stoichiometric compoundsD. non-stoichiometric compounds |
| Answer» d) Law of constant composition does not hold good for non-stoichiometric compounds. | |
| 659. |
The maximum number of molecules is present in :A. (a)`15" L of" H_(2)` gas at STPB. (b)`5` L of `N_(2)` gas at STPC. (c )`0.5` g of `H_(2) gas`D. (d)`10` g of `O_(2) gas` |
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Answer» Correct Answer - A Moles of `H_(2)=15/22.4=0.67` Moles of `N_(2)=5/22.4=0.22` Moles of `H_(2)=0.5/2=0.25` Moles of `O_(2)=10/32=0.31` larger is the number of mole, the more is the number of molecule. |
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| 660. |
Which has maximum number of molecules?A. (a)`7 g N_(2)`B. (b)`2 g H_(2)`C. (c )`16 g NO_(2)`D. (d)`16 g O_(2)` |
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Answer» Correct Answer - B `2 g of H_(2)` means one mole of `H_(2)`, hence contains `6.023xx10^(23)` molecules. Others have less than one mole, so have less no. of molecules. |
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| 661. |
In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?A. (a)20 litres ammonia, 25 litres nitrogen, 15 litres hydrogenB. (b)20 litres ammonia, 20 litres nitrogen, 20 litres hydrogenC. (c )10 litres ammonia, 25 litres nitrogen, 15 litres hydrogenD. (d)20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen |
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Answer» Correct Answer - C `underset((1 vol.),(10 litre))(N_(2))+underset((3 vol.),(30 litre))(3H_(2)) rarr underset((2 vol.),(20 litre))(2NH_(3))` It is given that only 50 % of the expected product is formed hence only 10 liters of `NH_(3)` is formed `N_(2) "used" =5 " litres", "left" =30-5=25` litres `H_(2)` used=15 litres, left`=30-15` litres |
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| 662. |
1 g sample of `KClO_(3)` was heated under such conditions that a part of it decomposed according to the equation : `2KClO_(3) rarr 2KCl + 3O_(2)` and the remaining underwent change according to the equation : `4KClO_(3)rarr 3KClO_(4)+KCl` If the amount of `O_(2)` evolved was 146.8 ml at S.T.P., calculate the % by weight of `KClO_(4)` in the reside. |
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Answer» Moles of `O_(2)` evolved `=(146.8)/(22400)=6.55xx10^(-3)` Moles of `KClO_(3)` decomposed in first reaction `=(2)/(3)xx6.55xx10^(-3)=4.36xx10^(-3)` Mass of `KClO_(3)` decomposed in first reaction `= 122.5xx4.36xx10^(-3)=0.534 g` Mole of `KClO_(3)` decomposed in sescond reaction `=(1-0.534)/(122.5)=3.8xx10^(-3)` Moles of `KClO_(4)` formed `=(3)/(4)xx3.8xx10^(-3)=2.85xx10^(-3)` Mass of `KClO_(4)` formed `=2.85xx10^(-3)xx138.5=0.395 g` Total moles of `KCl = 4.36xx 10^(-3)+(1)/(4)xx3.8xx10^(-3)=5.31xx10^(-3)` Mass of `KCl=5.31xx10^(-3)xx74.5=0.395 g` `therefore` % of `KClO_(4)=(0.395)/(0.395+0.395)xx100=50%` . |
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| 663. |
Sulphuryl chloride, `SO_(2)Cl_(2)`, reacts with `H_(2)O` to give mixture of `H_(2)SO_(4)` and HCl. Aqueous solution of 1 mol `SO_(2)Cl_(2)` will be neutralised byA. 3 moles of NaOHB. 2 moles of `Ca(OH)_(2)`C. Both (A) and (B)D. None of these |
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Answer» Correct Answer - B `underset(1mol)(SO_(2)Cl_(2))+underset(2mol)(2H_(2)O)tounderset(1mol)(H_(2)SO_(4))+underset(2mol)(2HCl)` `underset(1mol)(H_(2)SO_(4))+underset(2mol)(2NaOH)toNa_(2)SO_(4)+H_(2)O` `underset(1mol)(HCl)+underset(1mol)(NaOH)toNaCl+H_(2)xx2` `:.` 1 mol of `SO_(2)Cl_(2)` in aq. solution requires 4 mol of NaOH or 2 mol of `Ca(OH)_(2)` |
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| 664. |
Strictly speaking the term molecular mass has no meaning forA. ethylene, `C_(2) H_(2)`B. acetic acid, `(CH_(3) COOH)`C. zinc phosphate, `Zn_(3) (PO_(4))_(2)`D. benzene, `C_(6) H_(6)` |
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Answer» Correct Answer - C It is an ionic compound. |
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| 665. |
In the reaction `3 Mg+N_(2) rarr Mg_(3)N_(2)`A. Magnesium is reducedB. Magnesium is oxidizedC. Nitrogen is oxidizedD. None of these |
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Answer» Correct Answer - B In the given reaction oxidation state of `Mg` is changing from 0 to +2 while in nitrogen it is changing from 0 to -3. So oxidation of Mg and reduction of nitrogen takes place. |
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| 666. |
The atomic number of an element which shows the oxidation state of +3 isA. 13B. 32C. 33D. 17 |
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Answer» Correct Answer - A Al shows +3 oxidation state. |
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| 667. |
The percentage of `P_(2)O_(5)` in diammonium hydrogen phosphate is:A. (a)23.48B. (b)46.96C. (c )53.78D. (d)71.00 |
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Answer» Correct Answer - C `underset(2(36+1+31+64)=264)(2(NH_(4))_(2)HPO_(4))-= underset(62+80=142)(P_(2)O_(5))` `% of P_(2)O_(5)=("wt. of" P_(2)O_(5))/("wt. of salt")xx100` `=142/264xx100=53.78%` |
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| 668. |
A compound is in its high oxidation state. Then its will beA. Highly acidicB. Highly basicC. Highest oxidising propertyD. Half acidic, half basic |
| Answer» Correct Answer - C | |
| 669. |
Nitrogen shows different oxidation states in the range:A. 0 to +5B. `+3, +5, +7, 0`C. `+5, +7, -1, 0`D. `-1, -5, -1, 0` |
| Answer» Correct Answer - B | |
| 670. |
The percentage of `P_(2)O_(5)` in diammonium hydrogen phosphate is:A. 23.48B. 46.96C. 53.78D. `71.00` |
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Answer» Correct Answer - C `underset(2(36+1+31+64)=264)(2(NH_(4))_(2))HPO_(4) equiv underset(62+80=142)(P_(2)O_(5))` % of `P_(2)O_(5) =("wt. of "P_(2)O_(5))/("wt of salt")xx100` `=142/264xx100=53.78%` |
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| 671. |
Oxidation number of F in `Mg_(2)P_(2)O_(7)` isA. `+3`B. `+2`C. `+5`D. `-3` |
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Answer» Correct Answer - C `Mg_(2) overset(**)(P_(2))O_(7)` `4+2x-2xx7=0, 2x=14-4=10` `2x=10, x=10/2=+5`. |
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| 672. |
Oxidation no. of `P` in `H_(4)P_(2)O_(5)`, `H_(4)P_(2)O_(6)`, and `H_(4)P_(2)O_(7)` are respectivelyA. `+3, +4, +5`B. `+3, +5, +4`C. `+5, +3, +4`D. `+5, +4, +3` |
| Answer» Correct Answer - A | |
| 673. |
The number of atoms present in `0*1` mole of `P_(4)` (at. Mass = 31) areA. `2*4xx10^(24)` atomsB. same as in `0*05` mol of `S_(8)`C. `6*02xx10^(22)`D. same as in `3*1` g of phosphorus |
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Answer» Correct Answer - B `0*1` mole `P_(4)` contains `=4xx0*1xx6*02xx10^(23)P` atoms `0*05` mole of `S_(8)` contains `=8xx0*05xx6*02xx10^(23)S` atoms |
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| 674. |
The number of atoms presentin 0.1 mole of `P_(4)` (at mass = 31) areA. `2.4 xx 10^(23)` atomsB. same as in 0.05 mole of `S_(8)`C. `6.02 xx 10^(22)` atomsD. same as in 3.1 g of phosphorus |
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Answer» a) `therefore` 1 mole of `P_(4)` contains phosphorus atoms `=4 xx 6.02 xx 10^(23)` `therefore` 0.1 mole of `P_(4)` contains phosphorous atoms `=4 xx 6.02 xx 10^(23) xx 0.1` `=2.4 xx 10^(23)`atoms |
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| 675. |
What mass of sodium chloride would be decomposed by 9.8 g of sulphuric acid if 12 g of sodium bisulphate and 2.75 g of hydrogen chloride were produced in a reaction?A. 14.75 gB. 3.8 gC. 4.95 gD. 2.2 g |
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Answer» `underset(x)(NaCl)+underset(9.8g)(H_2SO_4) to underset (12g)(NaHSO_4)+underset(2.75g)(HCl)` x+9.8=12+2.75 `rArr` x=4.95g |
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| 676. |
Which of the following is not an element but a compound?A. GraphiteB. OzoneC. DiamondD. Dry ice |
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Answer» Correct Answer - D Dry ice is solid `CO_(2)`. It is so called because it sublimes at `-78^(@) C` without forming a liquid. It is used as a refrigerant. |
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| 677. |
A compound `(80 g)` on analysis gave `C=24 g, H=4 g, O=32 g`. Its empirical formula isA. (a)`C_(2)H_(2)O_(2)`B. (b)`C_(2)H_(2)O`C. (c )`CH_(2)O_(2)`D. (d)`CH_(2)O` |
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Answer» Correct Answer - D `C=24 g, H=4 g, O=32 g`, So, Molecular formula `=C_(2)H_(4)O_(4)` So, Empirical formula`=CH_(2)O` (Simplest formula) |
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| 678. |
Two oxides of a metal contain `50%` and `40%` metal `M` respectively. If the formula of the first oxide is `MO_(2)`, the formula of the second oxide will beA. `MO_(2)`B. `MO_(3)`C. `M_(2)O`D. `M_(2)O_(5)` |
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Answer» Correct Answer - B `{:(,"Oxide I","Oxide II"),("Metal",M" " 50%,40%),("Oxygen",O" "50%,60%):}` As first oxide is `MO_(2)` Let atomic mass of M = x `:. % O = (32)/(x+32)xx100` or `(50)/(100) = (32)/(x+32)` or `0.5 = (32)/(x+32)` or `0.5x+16 = 32` `0.5 x =16` `x = 32` At. mass of metal M, x = 32 Let formula of second oxide `M_(2)O_(n)` `%M = (2x)/(2x+16n)xx100 = (64)/(64+16n)xx100` `(40)/(100) = (64)/(64+16n) or (100)/(40) = (64+16n)/(64)` 2.5 = 1 +0.25n 2.5n = 2.5-1=1.5 `n = (1.5)/(0.25) = 6` Therefore, formula of second oxide `= M_(2)O_(6) or MO_(3)` |
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| 679. |
Two oxides of a metal contain `50%` and `40%` metal `M` respectively. If the formula of the first oxide is `MO_(2)`, the formula of the second oxide will beA. (a)`MO_(2)`B. (b)`MO_(3)`C. (c )`M_(2)O`D. (d)`M_(2)O_(5)` |
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Answer» Correct Answer - B `underset(50%)M underset(50%) (O_(2)) 50 g.......1 "metal" underset(40)(M_(x)) underset(60)(O_(y))` `1 g…….(1)/(50) :. 40 g…….(4)/(5)` metal and `50 g…….2 `oxygen `1 g…….(2)/(50)` oxygen For second oxide Atoms of metal `(M)=(1xx40)/50=0.8` similarly Atoms of oxygen`=(2xx60)/50=12/5=2.4` Hence, ratio of `M:O=0.8:2.4` or `1:3` `:.` Formula of the same metal oxide is `MO_(3)` |
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| 680. |
In the following reaction: `3Fe+4H_(2)OrarrFe_(3)O_(4)+4H_(2)`, if the atomic weight of iron is `56`, then its equivalent weight will beA. (a)`42`B. (b)`21`C. (c )`63`D. (d)`84` |
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Answer» Correct Answer - B `3overset(0)(Fe)+4H_(2)OrarrFe_(3)overset(+8//3)(O_(4))+4H_(2)` `3Fe+4H_(2)OrarrFe_(3)O_(4)+8H^(+)+8e^(-)` V.F. of `Fe=8/3`. `E_(Fe)("Atomic mass")/(V.F)=56/(8//3)=21` |
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| 681. |
(i) Add `73.24 mL` and `20.3 mL` (ii) Subtract `21.2342 g` foru `27.87 g` Strategy: We first ensure that the quatities to be added or subtracted are expressed in the same units. We carry out the adition or subtraction. Then we follow rule 1 to express the answer to the correct number of significant figures, i.e., in the final answer, the number of right of the decimal point should be equal to the number of digits after the decimal point. |
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Answer» (i) `{:(73.24mL, larr "Two digits after the decimal point"),(+20.30 mL,larr "One digit after the decimal point"),(bar(93.54 mL),larr "Reported as 93.5 mL"):}` (ii) `{:(27.8700 g, larr "Two digits after the decimal point"),(-21.2342 g,larr "One digit after the decimal point"),(bar(6.6358 g),larr "Reported as 93.5 mL"):}` |
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| 682. |
Rearrange the following (I to IV) in the order of increasin masses and choose the correct answer from (A), (B), (C) and (D) (At. Mass N=14, O=16, Cu=63) (I) 1 molecule of O (II) 1 atom of nitrogen (III) `1xx10^(-10)`g molecular mass of oxygen (IV) `1xx10^(-7)` atomic mass of copper.A. II lt I lt III lt IVB. IV lt III lt II lt IC. II lt III lt I lt IVD. III lt IV lt I lt II |
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Answer» Correct Answer - A (I) Mass of one O atom `= (16g)/(6.02xx10^(23))` `=2.66xx10^(-23)g` (II) Mass of one N atom `=(14g)/(6.02xx10^(23))` `=2.33xx10^(-23)`g (III) 1g-molecule of oxygen = 32 g `1xx10^(10)` g-molecule of oxygen `=32xx10^(-10)=3.2xx10^(-9)g` (IV) 1 g-atom of copper = 63 g `1xx10^(-7)g` g-atom of copper `=63xx10^(-7)=6.3xx10^(-6)g` Arranging in the increasing order of masses II lt I lt III lt IV |
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| 683. |
Molecule mass: Calculate the molecule mass of vitamin `C` or ascrobic acid `(C_(6) H_(8) O_(6))` using rounded values for atomic masses. Strategy: Add the atomic masses of the elements in the formula, each multipled by the number of times the element occurs. |
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Answer» `{:(underset("of given kind")("Number of atoms"),xx,underset("atom")("Mass of one"),=,underset("element")("Mass due to")),(C=6,xx,12.0 "amu",=,72.0 "amu of C"),(H=8,xx,1.0 "amu",=,8.0 "amu of H"),(O=6,xx,16.0"amu",=,96.0 "amu of O"):}` Molecule mass of ascrobic acid `= 176.0 "amu"` |
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| 684. |
Rearragne the following (I to IV) in the order of increasing masses and choose the correct answer (Atomic masses O=16, Cu=63 and N=4) I. 1 Molecule of oxygen II. 1 atom of nitrogen IIIgt `1xx 10^(-10)`g atomic weight of copper `A. II lt I lt III lt IVB. IV lt III lt II lt IC. II lt III lt I lt IVD. III lt IV lt I lt II |
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Answer» a) I. 1 molecule of `O_(2)` = (32)/(6.022 xx 10^(23)` = 5.3 xx 10^(-23)`g (ii) 1. atom of N= (14)/(6.022 xx 10^(23)`g = `2.3 xx 10^(-23)`g IIIgt `10^(-19)` g molecular weight of oxygen `= (10^(-10) xx 32 = 3.2 xx 10^(-9)`g IV. `10^(-10)`g atomic weight of copper `= 10^(-10) xx 63 =6.3 xx 10^(-9)`g `therefore` Order of increasing mass is II lt I lt III lt IV |
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| 685. |
Arrange the following in the order of increasing mass (at. Mass of O=16, Cu=63, N=14) (I) one atom of oxygen (II) one atom of nitrogen (III) `1xx10^(-10)` mole of oxygen (IV) `1xx10^(-10)` mole of copperA. `II lt I lt III lt IV`B. `I lt II lt III lt IV`C. `III lt II lt IV lt I`D. `IV lt II lt III lt I` |
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Answer» Correct Answer - A Mass of one atom of oxygen `=16/(6.023xx10^(23))=2.66xx10^(-23)` Mass of one atom of nitrogen `=14/(6.023xx10^(23))=2.32xx10^(-23)` Mass of `1xx10^(-10)` mole of oxygen `=16xx10^(-10)` Mass of `1xx10^(-10)` mole of copper `=63xx1xx10^(-10)` `=63xx10^(-10)` So, the order of increasing mass is `II lt I lt III lt IV`. |
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| 686. |
Arrange the following in the order of increasing mass (at. Mass of O=16, Cu=63, N=14) (I) one atom of oxygen (II) one atom of nitrogen (III) `1xx10^(-10)` mole of oxygen (IV) `1xx10^(-10)` mole of copperA. II lt I lt III lt IVB. I lt II lt III lt IVC. III lt II lt IV lt ID. IV lt II lt III lt I |
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Answer» Correct Answer - A (i) 1 atom of oxygen `=(16)/(6.02xx10^(23)g)` `=2.65xx10^(-23)g` (ii) 1 atom of nitrogen `=(14)/(6.02xx10^(23))g` `=2.326xx10^(-23)g` (iii) `1xx10^(-10)` mol of oxygen `=32xx10^(-10)` `=3.2xx10^(-9)`g (iv) `1xx10^(-10)` mol of `CO=63.5xx10^(-10)g` `=6.35xx10^(-9)`g Thus, the increaseing order of mass is II lt I lt III lt IV |
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| 687. |
A metal M of equivalent mass E forms an oxide of molecular formula `M_(x)O_(y)`. The atomic mass of the metal is given by the correct equation.A. 2E(y/x)B. xy/EC. E/yD. E/2(x/y) |
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Answer» Correct Answer - A If m is the atomic mass of the metal M, then 16y g of oxygen combine with xm g of the metal `:.` 8 g of oxygen will combine with metal `= (xm)/(16y)xx8 = (xm)/(2y)` `:. (xm)/(2y)=E` or `m=2E//(y//x)` |
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| 688. |
The percentage of element M is 53 in its oxide of molecular formula `M_(2)O_(3)`. Its atomic mass is aboutA. `45`B. `9`C. `18`D. `36` |
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Answer» Correct Answer - C If m is the atomic mass of the element M, then 2m + 48g of `M_(2)O_(3)` contain O = 48 g `:.` % of O in `M_(2)O_(3) = (48)/(2m+48)xx100=53` `:. 2m + 48 = (4800)/(53)=90.56` or 2m = 42.56 or m = `21.28 ~~ 22` |
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| 689. |
A 400 mg iron capsule contains 100 mg of ferrous fumarate, `(CHCOO)_(2)`Fe. The percentage of iron present in it is approximatelyA. 0.33B. 0.25C. 0.14D. 0.08 |
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Answer» Correct Answer - D Molecular weight of `(CHCOO)_(2) Fe=170` Fe present in 100 mg of `(CHCOO)_(2)Fe` `=56/170xx100mg =32.9 mg` This is present in 400 mg of capsule % of Fe in capsule `=32.9/400xx100=8.2`. |
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| 690. |
The mass of `112 cm^(3)` of `CH_(4)` gas at STP isA. 0.16 gB. 0.8 gC. 0.08 gD. 1.6 g |
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Answer» Correct Answer - C The mass of `112 cm^(3)` of `CH_(4)` at S.T.P. `= (16xx112)/(22400) = 0.08 g` |
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| 691. |
When 200 g of lime strongly heated , it undergoes thermal decomposition to form 112 g of lime and unknown mass of carbon dioxide gas as `underset(200 g)(CaCO_(3))rarr underset(112 g)(CaO)+underset(?)(CO_(2))` What will be the mass of `CO_(2)` formed ?A. 88 gB. 24 gC. 64 gD. 40 g |
| Answer» Correct Answer - A | |
| 692. |
A 400 mg iron capsule contains 100 mg of ferrous fumarate, `(CHCOO)_(2)`Fe. The percentage of iron present in it is approximatelyA. (a)33%B. (b)25%C. (c )14%D. (d)8% |
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Answer» Correct Answer - D Molecular weight of `(CHCOO)_(2) Fe=170` Fe present in `100 mg` of `(CHCOO)_(2)` Fe `=56/170xx100 mg=32.9 mg` This is present in `400 mg` of capsule `%` of Fe in capsule`=32.9/400xx100=8.2`. |
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| 693. |
A metal oxide has the formula `A_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. 0.1596 g of this metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic wight of metal?A. `52.3`B. `57.3`C. `55.8`D. `59.3` |
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Answer» Correct Answer - C `underset(0.1596 g)(A_(2) O_(3) )+ underset(0.006 g)(3 H_(2)) rarr 2 A + 3H_(2) O` `0.006 " g of" H_(2)` reduces ` 0. 1596 ` g of ` A_(2) O_(3)` 6 g og `H_(2)` will reduce `(0.1596xx6)/(0.006) = 159.6" g of " A_(2) O_(3)` Thus, molar mass of `A_(2) O_(3) = 159.6` g Let, atomic weight of A = x `therefore 2x + 3 xx 16 = 159.6 ` `2=159.6 - 48 = 111.6` `x = 55.8` |
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| 694. |
The mass of `112 cm^(3)` of `CH_(4)` gas at STP isA. `0.16` gB. `0.8` gC. `0.08` gD. `1.6` g |
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Answer» Correct Answer - C At STP, `22400 cm^(3)` of `CH_(4)= 12 +4 = 16 `g At STP, 112 `cm^(3)` of `CH_(4) = (16xx112)/22400=0.08` g |
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| 695. |
In the conversion `NH_(2)OH rarr N_(2)O`, the equivalent weight of `NH_(2)OH` will be:A. (a)`M//4`B. (b)`M//2`C. (c )`M//5`D. (d)`M//1` |
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Answer» Correct Answer - B `NH_(2)OH rarr underset(+1)(N_(2)O)` `:. V.f. of NH_(2)OH=2` `:. Eq. wt.=M//2` |
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| 696. |
The law reciprocal proportions is obeyed byA. `NaH, NCl, NaCl`B. `NaCl, NaBr, Nal`C. `HCl, HBr, HI`D. `LiH, NaH, KH` |
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Answer» Correct Answer - A The law of reciprocal proportions is obyed if two different elements `A` and `B` combine chemically with one another and also with another elelment `C` separately. |
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| 697. |
In the synthesis of ammonia `N_(2)(g)+3H_(2)hArr2NH_(3)(g)` If the quantity of `N_(2)` reacted is `700mL`, the quantity of `H_(2)` and `NH_(3)` would beA. 300 mL `H_(2)` and 200 mL `NH_(3)`B. 300 mL `H_(2)` and 300 mL `NH_(3)`C. 300 mL `H_(2)` and 100 mL `NH_(3)`D. 100 mL `H_(2)` and 200 mL `NH_(3)` |
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Answer» Correct Answer - A `N_(2)(g) + 3H_(2) (g) hArr 2 NH_(3) (g)` If the quantity of `N_(2) (g)` reacted is 100 mL `therefore` Thequantiy of `H_(2) (g)` is three times, 300 mL and `NH_(3)` produced = 200 mL |
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| 698. |
The volume strength of `1.5 N H_2O_2` solution isA. `4.8`B. `8.4`C. `3.0`D. `8.0` |
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Answer» Correct Answer - B Volume strength of `H_(2)O_(2)` = Normality `xx 5.6 = 1.5 xx 5.6 =8.4 V` |
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| 699. |
The volume strength of `1.5 N H_2O_2` solution isA. (a)4.8B. (b)8.4C. (c )3.0D. (d)8.0 |
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Answer» Correct Answer - B Normality of 10 volume `H_(2)O_(2)=1.78 N` Thus, volume strength of `1.78 N` solution `=10` `:. `Volume strength of `1.5 NH_(2)O_(2)= =1.5xx10/1.78=15/1.78=8.4` |
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| 700. |
In the reaction, `N_(2)+ 3" H"_(2) rarr 2 NH_(3),` the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3 : 2 These ratio illustrate the law ofA. constant proportionsB. Gay-LussacC. multiple proportionsD. reciprocal proportions |
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Answer» Correct Answer - B `N_(2)+3H_(2) rarr 2NH_(3).` The ratio of volumes of nitrogen, Hydrogen and ammonia is 1 : 3 : 2. These ratio illustrate the law of Gay-Lussac. |
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