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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.
| 351. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - In a gaseous reaction, the ratio by volumes of reactant and gaseous products is in agreement with their molar ratio Reason (R) - Volume of gas is inversely proportional to its number of moles at particular temperature and pressure.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 352. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Relative atomic mass of boron is 10.8 Reason (R) - Boron has two isotopes B-10 and B-11 with percentage abundance of 19.6% and 80.4% respectively.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 353. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Both 32 g of `SO_(2)` and 8g of `CH_(4)` have same number of molecules Reason (R) - Equal moles of substances have equal number of moleculesA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 354. |
Assertion : Empirical and molecular formulae of `NaHCO_(3)` are the same Reason : Upon heating, `NaHCO_(3)` evolves `CO_(2)` gas.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 355. |
In which of the following compounds the oxidation number of carbon is maximumA. `HCHO`B. `CHCl_(3)`C. `CH_(3)OH`D. `C_(12)H_(22)O_(11)` |
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Answer» Correct Answer - B Oxidation number of C in `HCHO=0` `CHCl_(3)=+2` `CH_(3)OH=-2` `C_(12)H_(22)O_(11)=0`. |
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| 356. |
In which of the following compounds iron has lowest oxidation state?A. `FeSO_(4).(NH_(4))_(2)SO_(4). 6H_(2)O`B. `K_(4)Fe(CN)_(6)`C. `Fe(CO)_(5)`D. `Fe_(2)O` |
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Answer» Correct Answer - C Iron has zero oxidation state in carbonyl complexes. |
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| 357. |
Oxidation number of nitrogen in `NaNO_(2)` isA. `+2`B. `+3`C. `+4`D. `-3` |
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Answer» Correct Answer - B Let the oxidation number of N in `NaNO_(2)` be x `+1+x+(-2)xx2=0` `1+x-4=0, x=+3`. |
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| 358. |
What is the net charge on ferrous ion?A. `+2`B. `+3`C. `+4`D. `+5` |
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Answer» Correct Answer - A Net charge on ferrous ion, `Fe^(2+) = +2` |
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| 359. |
The maximum number of molecules is present inA. 10 g of `O_(2)` gasB. 15L of `H_(2)` gas at S.T.P.C. 5L of `N_(2)` gas at S.T.P.D. 0.5 g of `H_(2)` gas |
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Answer» Correct Answer - B More the volume, more is the molecules. Converted the choices (A) and (D) into volumes. (A) 32 g of `O_(2)` at S.T.P = 22.4 L `:.` 10 g `O_(2)` at S.T.P. `= (22.4)/(32)xx10L=7L` (Molar mass of `O_(2) = 32 g mol^(-1)`) (D) 2 g of `H_(2)` at S.T.P. = 22.4 L 0.5 g of `H_(2)` at S.T.P. `= (22.4)/(2)xx0.5L=5.6L` Thus we have 7L of `O_(2), 15 L` of `H_(2)`, 5 L of `N_(2)` and 5.6 L of `H_(2)` (at S.T.P.) Clearly 15 L of `H_(2)` has maximum number of molecules |
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| 360. |
1 mole of hydrocarbon on complete combustion give 4 mole of `CO_(2)`. Calculate no of carbon in hydrocarbon. |
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Answer» 1 mole of hydrocarbon gives n mole of `CO_(2)` where n is no. of carbon atoms in hydrocarbon. 1 mole of hydrocarbon = 4 mole of `CO_(2)` . So, value of n = 4. |
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| 361. |
1.5 g of hydrocarbon on combustion in excess of oxygen produces 4.4 g of `CO_(2)` and 2.7 g of `H_(2)O`, the data illustratesA. Law of conservation of massB. Law of multiple proportionsC. Law of constant compositionD. Law of reciprocal proportions. |
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Answer» Correct Answer - A Mass of hydrocarbon CH = 1.5 g changes to `CO_(2)` and H changes to `H_(2)O` C in `4*4` g of `CO_(2) = 1*2g` H in `2*7 g` of `H_(2)O = (2xx2*7)/(18) = 0*3g` Mass of C + Mass of H `= 1*2 + 0*3 = 1*5g` Which is equal to the mass of hydrocarbon taken. |
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| 362. |
A mixture that can be separated by sublimation isA. `AgCl + NaCl`B. `BaCl_(2) + NaCl`C. `HgCl_(2) + NaCl`D. `MgCl_(2) + NaCl`. |
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Answer» Correct Answer - C `HgCl_(2)` sublimes on heating while NaCl donot. |
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| 363. |
Which one of the following best explains the law of conservation of mass ?A. No change in mass is observed when 2.0 g of Mg is heated in vacuumB. 1.2 g of carbon when burnt in excess of oxygen consumes only 3.2 g of it to form 4.4 g of carbon dioxideC. 12.g of carbon when heated in air produces only 20 g of carbon monoxideD. A sample of `Ca CO_(3)` on heating does not show any change in mass of `C_(a) CO_(3)` but volume increases |
| Answer» Correct Answer - B | |
| 364. |
The compound which does not exist as hydrate formA. (a)ferrous sulphateB. (b)copper sulphateC. (c )magnesium sulphateD. (d)sodium chloride |
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Answer» Correct Answer - a Ferrous sulphate `rarr FeSO_(4). 7H_(2)O` Copper sulphate `rarr CuSO_(4). 5H_(2)O` Magnesium sulphate `rarr MgSO_(4). 7H_(2)O` Sodium chloride `rarr NaCl` |
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| 365. |
Following is the graphical presentation of volumes occupied by different gases at S.T.P. Which is/are not placed at correct position? A. `H_(2), He`B. `He, NH_(3)`C. `NH_(3), CH_(4)`D. `CH_(4), H_(2)` |
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Answer» Correct Answer - B Molar mass of `H_(2)`, He, `NH_(3)` and `CH_(4)` are 2,4,17, and 16 g `mol^(-1)` respectively. 2 g of `H_(2)` at S.T.P. = 22.4 L 1 g of `H_(2)` at S.T.P. = 11.2 L 4 g of He at S.T.P. = 22.4 L 17 g `NH_(3)` at S.T.P. = 22.4 L 3g of `NH_(3)` at S.T.P. = 3.95 L 16 g of `CH_(4)` at S.T.P. = 22.4 L 4 g of `CH_(4)` at S.T.P. = 5.6 L |
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| 366. |
A gaseous hydrocarbon gives upon combustion, 0.72 g of water and 3.08 g of `CO_(2)`. The empirical formula of the hydrocarbon isA. (a)`C_(2)H_(4)`B. (b)`C_(3)H_(4)`C. (c )`C_(6)H_(5)`D. (d)`C_(7)H_(8)` |
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Answer» Correct Answer - D `18 g H_(2)O` contain `2 g H`. `:. 0.72 g H_(2)O` contain `0.08 g H` `44 g CO_(2)` contain `12 g C`. `:. 3.08 g CO_(2)` contain `0.84 g C`. `:. C.H.=0.84/12:0.08/1=0.07:0.08=7:8` `:.` Empirical formula`=C_(7)H_(8)` |
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| 367. |
The molecular mass of glucose `(C_(6) H_(12) O_(6))` molecule to three significant figures isA. `180.162 u`B. `180.0u`C. `180.00u`D. `1.80xx10^(2) u` |
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Answer» Correct Answer - D `180.162u` rounded off to three significant figures is `1.80xx10^(2) u`. |
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| 368. |
Assertion: `NaNO_(3)` has no definite molecule. Reason: Its formula mass is 85.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
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Answer» Correct Answer - B `NaNO_(3)` has solid lattice. |
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| 369. |
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. 0.4 g CaD. 12 g He |
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Answer» (a) 1 mol of He = 4g = `N_A` atoms or 4 g of He `4/4mol=1N_A` atom (b) 1 mole of Na = 23g = `N_A` atoms [ `therefore " no. of moles" = ("given mass")/("atomic mass")` `therefore " 46g of Na " = 46/23 " mol " = 46/23 N_A " atoms " = 2N_A` atoms (c) 1 moles of Ca = 40 g =`N_A` atoms `therefore " 0.40g of Ca " =(0.40)/(40) " mol " =(0.40)/(40) N_A` atoms `=0.01 N_A` atoms (d) 1 mole of He = 4g `=N_A` atoms `therefore " 12g of He " =12/4 " mol " =12/4 N_A " atoms " = 3N_A` atoms |
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| 370. |
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 He"`B. 46 g NaC. `0.4` g CaD. 12 g He |
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Answer» Correct Answer - D For comparing number of atoms, first we calculate the moles as all are monoatomic and hence, Number of moles `=("Mass")/("Atomic mass")` Moles of 4 g He `=4/4 = 1` mol 46 g Na `=46/23 =2` mol `0.40` g Ca `=(0.40)/40=0.01` mol 12 g He `= 12/4=3` mol Now, number of atoms `="moles"xx"N"_("A")` Therefore, number of atoms are highest in 12 g He as it has the Masimum number of moles. |
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| 371. |
`4.0 g` of a mixture of `NaCl` and an unknown metal iodide `MI_(2)` was dissolved in water to form its aqueous solution. To this aqueous solution, aqueous solution of AgNO_(3)` was added gradually so that silver halides are precipitated. The precipitates were weighed at regular interval and following curve for the mass of precipitate versus volume of `AgNO_(3)` added was obtained. With the knowledge of the fact that halides are precipitated successively, i.e, when less soluble halide is precipitating, the other halide remain in the solution, answer the following questions: (Molar mass of `Ag = 108, I = 127, Na=23`). What is the approximate mass percentage of `Ml_(2)` ?A. 25B. 40C. 60D. 75 |
| Answer» Correct Answer - D | |
| 372. |
`4.0 g` of a mixture of `NaCl` and an unknown metal iodide `MI_(2)` was dissolved in water to form its aqueous solution. To this aqueous solution, aqueous solution of AgNO_(3)` was added gradually so that silver halides are precipitated. The precipitates were weighed at regular interval and following curve for the mass of precipitate versus volume of `AgNO_(3)` added was obtained. With the knowledge of the fact that halides are precipitated successively, i.e, when less soluble halide is precipitating, the other halide remain in the solution, answer the following questions: (Molar mass of `Ag = 108, I = 127, Na=23`). What is the approximate molarity of `AgNO_(3)` solution?A. `0.1`B. `0.5`C. `1.0`D. `1.5` |
| Answer» Correct Answer - C | |
| 373. |
`4.0 g` of a mixture of `NaCl` and an unknown metal iodide `MI_(2)` was dissolved in water to form its aqueous solution. To this aqueous solution, aqueous solution of AgNO_(3)` was added gradually so that silver halides are precipitated. The precipitates were weighed at regular interval and following curve for the mass of precipitate versus volume of `AgNO_(3)` added was obtained. With the knowledge of the fact that halides are precipitated successively, i.e, when less soluble halide is precipitating, the other halide remain in the solution, answer the following questions: (Molar mass of `Ag = 108, I = 127, Na=23`). What is the approximate molar mass of unknown metal M?A. 20B. 40C. 56D. 60 |
| Answer» Correct Answer - B | |
| 374. |
The number of synthetic elements areA. `14`B. `25`C. `17`D. `20` |
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Answer» Correct Answer - D There are a total 118 elements of which 98 occur in nature. |
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| 375. |
Which of the following has the minimum number of atoms?A. `0.5g` atom of `Cu`B. `0.635g` of `Cu`C. `0.25` mol of `Cu`D. `6.35xx10^(20) u` of `Cu` |
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Answer» Correct Answer - D It has the minimum number of moles, and hence, the number of atoms. One gram atoms means 1 mol of atoms, thus `0.5` gram atoms `Cu` means `0.5` mol `Cu`. `0.635g Cu` contains `(0.635g//63.5g mol^(-1)) = 0.01 mol Cu` `(63.5xx10^(20)u)/(63.5 u//"atom") rarr 10^(9)` atoms `rarr 10^(19)//6.022xx10^(23) rarr 1.661xx10^(-5) mol` |
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| 376. |
Law of reciprocal proportions was establish byA. LavoisierB. ProustC. DaltonD. Richter |
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Answer» Correct Answer - D (Lavoiser) law of conservation of mass, (Proust) law of constant composition, and (Dalton) law of multiple proportions |
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| 377. |
What is the normlity of a 1 M solution of `H_(3)PO_(4)`?A. (a)`0.5 N`B. (b)`1.0 N`C. (c )`2.0 N`D. (d)`3.0 N` |
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Answer» Correct Answer - b `H_(3)PO_(4)` is tribasic so `N=3M=3xx1=3` |
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| 378. |
What is the volutme of `CO_(2)` liberted in litres at 1 atmosphere and `0^(@)C` when 10% of 100 pure calcium carbonate is treated with excess dilute sulphuric acid? (at mass of Ca=40, C=12, O=16)A. `0.224`B. `2.24`C. `22.4`D. `224` |
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Answer» Correct Answer - B `underset(=100g)underset((40+12+3xx16))(CaCO_(3)+)2HCl to CaCl_(2)+H_(2)Ounderset("1 atm and "0^(@)C)underset(224Cat)(+CO_(2))` `:.` 10 g of `CaCO_(3)` will produce `CO_(2)` `=(22400)/(100)xx10=2.24L`. |
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| 379. |
Use data given in the following table to calculate the molar mass of naturaly occuring argon isotopes: `{:("Isotope",,"Isotopic molar mass",,"Abundance"),(.^(36)Ar,,35.96755 g mol^(-1),,0.337%),(.^(38)Ar,,37.96272 g mol^(-1),,0.063%),(.^(40)Ar,,39.9624 g mol^(-1),,99.600%):}` |
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Answer» Correct Answer - `39.948 g mol ^(-1)` Molar mass of argon `=[(35.96755xx(0.337)/(100))+(37.96282xx(0.063)/(100))+(39.9624xx(90.60)/(100))]gmol^(-1)` `=[0.121+0.021+39.802] g mol^(-1)` `=39.947 g mol^(-1)` |
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| 380. |
Which one of the following is not an elementA. DiamondB. GraphiteC. SilicaD. Ozone |
| Answer» Correct Answer - C | |
| 381. |
What will be the molarity of chloroform in the water sample which contains 15 ppm chloroform by mass?A. `1.25xx10^(-4)` mB. `2.5xx10^(-4)` mC. `1.5xx10^(-3)` mD. `1.25xx10^(-5)`m |
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Answer» 15 ppm `=(15)/(10^6)xx10^2=1.5xx10^(-3)g` Molarity of `CHCl_3 " solution " =(1.5xx10^(-3))/(100)xx(1000)/(119.5)` `=1.25xx10^(-4)` |
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| 382. |
One atmosphere is equal toA. 101.325 K paB. 1013.25 K paC. `10^(5)` NmD. None of these |
| Answer» Correct Answer - A | |
| 383. |
A picometre is written asA. (a)` 10^(-9) m`B. (b)` 10^(-10) m`C. (c )` 10^(-11) m`D. (d)` 10^(-12) m` |
| Answer» Correct Answer - D | |
| 384. |
One atmosphere is equal toA. (a)` 101.325 KPa`B. (b)`1013.25 Kpa`C. (b)`10^(5) Nm `D. (d)None of these |
| Answer» Correct Answer - A | |
| 385. |
What is the area of a recantangel `12.34 cm` wide and `1.23 cm` long? Stargetey: The area of a rectangler is its length times its width. First, we must check that the width and length are expressed in the same units. Then we mutiply the length by the width. Finally, we follow rule 3 to find the correct number fo significant figures. The Units for the result are equal to the product of the units for the individual terms in the multiplication. |
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Answer» Area = length `xx` Width `= (1.23 cm)(12.34 cm)` `= 153.1782 cm^(2) larr` Calculate result `= 15.2 cm^(2) larr` Reported answer The reported answer should contain only three significant figures because three is the smallest number of significant figures used in original numbers. |
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| 386. |
In the reaction `Zn(s) + 2H^(+)(aq.) rarr Zn^(2+)(aq.) + H_(2) (g)`, how many liters of hydrogen gas measured at `STP` is produced when `6.54g` of `Zn` is used `(Zn = 65.4 u)` ?A. `22.4 L`B. `11.2 L`C. `2.24 L`D. `1.12 L` |
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Answer» Correct Answer - C According to the equation, 1 mol `Zn` releases 1 mol `H_(2)`. Thus, `n_(H_(2)) = n_(Zn) = (6.54g)/(65.4g) = 0.1 mol` Which occupies `2.24 L` at `STP`. |
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| 387. |
Total number of atoms present in 34 g of `NH_3` isA. `4xx10^23`B. `4.8xx10^21`C. `2xx10^23`D. `48xx10^23` |
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Answer» No. of moles of 34g of `NH_3=34/17=2` No. of molecules `=2xx6.023xx10^23` No. of atoms in one molecules of `NH_3`=4 No. of atoms in `2xx6.023xx10^23 " molecules of NH_3` `=4xx2xx6.023xx10^23=48.18xx10^23` |
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| 388. |
What volume of water is to be a added to `100cm^(3)` of 0.5 M NaOH solution to make it 0.1 M solution?A. 200 `cm^(3)`B. 400 `cm^(3)`C. 500 `cm^(3)`D. 100 `cm^(3)` |
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Answer» `M_(1)V_(1)= M_(2)V_(2)` `0.5xx100=0.1xxV_(2)V_(2)=500cm^(3)` Volume of water to be added to `100cm^(3)` of solution `=500-100 =400 cm^(3)` |
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| 389. |
How many number of molecules and atoms respectively are presetn in 2.8 liters of a diatomics gas at STP ?A. `6.023xx10^23, 7.5xx10^23`B. `6.023xx10^23, 15xx10^22`C. `7.5xx10^22, 15xx10^22`D. `15xx10^22, 7.5xx10^23` |
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Answer» Number of molecules of gas at STP `=(6.023xx10^23xx2.8)/(22.4)=7.5xx10^22` molecules Number of atoms in diatomic molecule `2xx7.5xx10^22=15xx10^22` atoms |
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| 390. |
What will be the molarity of the solution made by dissolving 10 g of NaOH in 100g of water ?A. 2.5 mB. 5 mC. 10 mD. 1.25 m |
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Answer» Molarity `=("Wt. of solute")/("Mol. Wt. of solute")xx(1000)/("wt. of solvent")` `m=10/40xx1000/100=2.5m` |
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| 391. |
Which of the following gases will have least volume if 10g of each gas is taken at same temperature and pressure?A. `CO_2`B. `N_2`C. `CH_4`D. HCl |
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Answer» Number of moles `prop (1)/("Molecular mass")` Molecular mass of `CO_2=44, N_2=28, CH_4=16`, HCl=36.5 `CO_2` will have least volume, as no. of moles is directly propotional to volume at constant P and T. |
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| 392. |
What will be the molality of the solution made by dissolving 10g of NaOH in 100g of water?A. 2.5 mB. 5 mC. 10 mD. 1.25m |
| Answer» Molarity(m)`=("Weight of solute")/("Molar weight of solute")xx(1000)/("Weight of solvent")` | |
| 393. |
Which of the following is not a mixture but a pure substance?A. CNG (compressed natural gas)B. LPG (liquefied natural gas)C. Distilled waterD. Kerosene |
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Answer» Correct Answer - C It has no dissolved impuritites while others are mixtures of hydrocabons. |
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| 394. |
The stoichiometric coefficients do not representA. relative number of molecules of the rectants and productsB. relative number of moles of the reactants and productsC. relative volumes of each gaseous rectant and product under the indentical condtions of temperature and pressureD. the relative mass of rectants and products |
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Answer» Correct Answer - D 1 mol is a fixed number of formula units but not a fixed mass. |
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| 395. |
What is the stoichiometric coefficient fo Ca in the reaction ? `Ca+Al^(3+) rarr Ca^(2+)+Al`A. 2B. 1C. 3D. 4 |
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Answer» Correct Answer - C `[Ca rarr Ca^(2+)+2e^(-)]xx3` `[Al^(3+)+3e^(-) rarr Al]xx2` `3Ca rarr 3Ca^(2+)+6e^(-)` `2Al^(3+)+6e^(-) rarr 2Al` `bar(3Ca+2 Al^(3+) rarr 3 Ca^(2+)+2 Al)` Therefore, the stoichiometric coefficient of Ca in the given reaction is 3. |
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| 396. |
How many atoms in total are present in 1kg of sugar?A. `7.92xx10^25` atomsB. `6xx10^23` atomsC. `6.022xx10^25` atomsD. 1000 atoms |
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Answer» One molecule of sugar `(C_12H_22O_11)` = 45 atoms No. of moles of sugar `=1000/342=2.92` No. of molecules `=2.92xx6.0223xx10^23` No. of atoms `=45xx17.60xx10^23` `=7.92xx10^25` atoms |
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| 397. |
The number of significant figures in `0.0000349 g` isA. threeB. fourC. sevenD. five |
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Answer» Correct Answer - A Zeroes to the left of the first nonzero digit are never significant |
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| 398. |
One mole of magnesiu nitride on reaction with an excess of water givesA. two moles of ammoniaB. one mole of nitric acidC. one mole of ammoniaD. two moles of nitric acid |
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Answer» Correct Answer - A `underset("1 mole")(Mg_(3)N_(2))+6H_(2)Oto3Mg(OH)_(2)+underset("2 mole")(2NH_(3)` |
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| 399. |
Four one litre flaske are separately filled with the gases `H_(2), He, O_(2) "and" O_(3)` at the same temperature and pressure. The ratio of total number of atoms of these gases present in different flask would be:A. (a)`1:1:1:1`B. (b)`1:2:2:3`C. (c )`2:1:2:3`D. (d)`3:2:2:1` |
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Answer» Correct Answer - C `H_(2): He: O_(2): O_(3)` `{:(,H_(2),: He,:O_(2),:O_(3)),("Ratio of total no. of molecules",= 1,:1,:1,:1),("So ratio of total no. of atoms",= 2,:1,:2,:3):}` |
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| 400. |
A given sample of pure compound contains `9.81 g` of Zn, `1.8xx10^(23)` atoms of chromium, and `0.60` mol of oxygen atoms. What is the simplest formula?A. (a)`ZnCr_(2)O_(7)`B. (b)`ZnCr_(2)O_(4)`C. (c )`ZnCrO_(4)`D. (d)`ZnCrO_(6)` |
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Answer» Correct Answer - B Moles of `Zn=9.81/65.39=0.15` Moles of `Cr=(1.8xx10^(23))/(6.023xx10^(23))=0.3` Moles of `O=0.6` So, whole molar ratio `=0.15:0.3:0.6=1:2:4` Hence, formula`=ZnCr_(2)O_(4)` |
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