InterviewSolution
Saved Bookmarks
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.
| 1. |
Calculate the formula unit masses of ZnO, `Na_(2)O, K_(2)CO_(3)`, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u. |
|
Answer» (i) Formula mass of `ZnO =` Mass of Zn atom + Mass of O atom `= 65 + 16` `= 81 u` (ii) Formula mass of `Na_(2)O =` Mass of 2Na atoms + Mass of O `= 2 xx 23 + 16` `= 46 +16` `= 62 u` (iii) Formula mass of `K_(2)CO_(3)` (See Sample Problem on page `138` of this book). |
|
| 2. |
Calculate the formula masses of following compounds : (i) Sodium oxide, `Na_(2)O` (ii) Aluminium oxide, `Al_(2)O_(3)` (Given : Atomic masses : `Na = 23 u , O = 16 u , Al = 27 u`) |
| Answer» Correct Answer - (i) `62 u` , (ii) `102 u` | |
| 3. |
(a) Define mole. What are the two things that a mole represents. (b) What weight of each element is present in `1.5` moles of sodium sulphate , `Na_(2)SO_(3)` ? (Atomic masses : `Na = 23 u , C = 12 u , O = 16 u`) |
| Answer» Correct Answer - (b) `Na = 69 g ; S = 48 g ; O = 72 g` | |
| 4. |
Express each of the following in kilograms a) `5.84 xx 10^(-3)`mg b) `58.34`g c) `0.584`g d) `5.873 xx 10^(-21)`g |
|
Answer» `(a)5.84xx10^(-9)"kg"` (B)`5.834xx10^(-4)"kg"` `(c)0.584xx10^(-3)"kg"` (d)`5.873xx10^(-26)"kg"` |
|
| 5. |
Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. a) Whose container is heavier? b) Whose container has more number of atoms? |
|
Answer» Molar atomic masss of carbon=12g No. of moles of carbon carried by Raunak=5 Mass of 5 moles of carbon=`(12xx5)=60g` Molar atomic mass of sodium=23g ltbrlt No. of moles of sodium carried by Krish=5 Mass of 5 moles of sodium =`(23xx5)=115g` This shows that the container carried by Krish is heavier. (b)Since both the containers have same number of moles, the number of atoms present in these containers are also same i.e. `5xxN_(A)` atoms. |
|
| 6. |
What is the SI prefix for each of the following multiples and submultiples of a unit? a) `10^(3)` b) `10^(-1)` c) `10^(-2)` d) `10^(-6)` e) `10^(-9)` f) `10^(-12)` |
| Answer» (A)kilo (b)deci (c )centi (d)micro (e)nano (f)pico. | |
| 7. |
The visible universe is estimated to contain `10^(22)` stars. How many moles of stars are present in the visible universe? |
|
Answer» No. of stars present in visible universe=`10^(22)` No. of moles of stars present=`=(10^(22))/(6.022xx10^(23))=0.166"mol"` |
|
| 8. |
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution areA. `6.68xx10^(23)`B. `6.09xx10^(22)`C. `6.022xx10^(23)`D. `6.022xx10^(21)` |
|
Answer» Correct Answer - A a no. of moles of sucrose in 3.442 g `=((3.42g))/((342 g "mol"^(-1)))` =0.01 mol No. of oxygen atoms in 1 mole of sucrose `(C_(12)H_(22)O_(11))=11xxN_(A)` No. of oxygen atoms in 0.01 moles of sucrose `=0.01 xx11 xxN_(A)=0.11 N_(A)` No. of moles of `H_(2)O` in 18 g =`((18g))/((18 g"mol"^(-1))) =1 ` mol No. of oxygen atoms in 1 mole of water =`N_(A)` Total no. of oxygen atoms =`(0.11+1) =1.11 N_(A)` `=1.11xx6.022xx10^(23)` `=6.68xx10^(23)` |
|
| 9. |
How many moles of calcium carbonate `(CaCO_(3))` are present in `10 g` of the substance ? `(Ca = 40 u , C = 12 u, O = 16 u)` |
| Answer» Correct Answer - `0.1` mol | |
| 10. |
What is the number of molecules in 1.5 moles of ammonia ? |
| Answer» Correct Answer - `9.033 xx 10^(23)` molecules | |
| 11. |
How many moles are `9.033 xx 10^(24)` atoms of helium (He) ? |
|
Answer» We know that : `6.022 xx 10^(23)` atoms of helium `=1` mole So, `9.033 xx 10^(24)` atoms of helium `= (1)/(6.022 xx 10^(23)) xx 9.033 xx 10^(24)` `= 15` moles , (or `15` mole) Thus, `9.033 xx 10^(24)` atoms of helium are `15` moles of atoms. |
|
| 12. |
A flask P contains `0.5` mole of oxygen gas. Another flask `Q` contains `0.4` mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms ? |
|
Answer» `"1 molecule of oxygen"(O_(2))="2 atoms of oxygen"` `"1 molecule of oxygen"(O_(3))="3 atoms of oxygen"` In flask P: `"1 molecule of oxygen gas"=6.022xx10^(23)"molecules"` `"0.5 molecule of oxygen gas"=6.022xx10^(23)xx0.5"molecules"` In flask Q: `"1 molecule of oxygen gas"=6.022xx10^(23)"molecules"` `"0.4 molecule of oxygen gas"=6.022xx10^(23)xx0.4"molecules"` `=6.022xx10^(23)xx0.4xx3"atoms"=7.23xx10^(23)"atoms"` |
|
| 13. |
What weight of calcium has the same number of atoms as are present in 3.2g of sulphur ? |
|
Answer» Gram atomic mass of S=32g 32g of sulphur contain atoms=`6.022xx10^(23)` 3.2g of sulphur contains atoms=`(6.022xx10^(23))/(32g)=6.022xx10^(22)` Step-II: Weight of `6.022xx10^(22)` atoms of calcium Gram atomic mass of Ca=40g `6.022xx10^(23)` atomic of Ca weigh=40g `6.022xx10^(22)` atoms of Ca weigh=`(40g)/(6.022xx10^(23))xx6.022xx10^(22)=4g` |
|
| 14. |
When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer? |
|
Answer» 3.0 g of carbon cimbines with 8.0 g of oxygen to given 11.0 of carbon dioxide. If 3 g carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11 g of carbon dixide will be formed. The above answer is governed by the law of constant proportions. |
|
| 15. |
When 5g of calcium is burnt in 2g of oxygen, then 7 g of calcium oxide of produced. What mass of calcium oxide will be produced when 5g of calcium is burnt in 20 g of oxygen ? Which law of chemical combination will govern your answer ? |
| Answer» Correct Answer - 7g ; Law of constant proportions | |
| 16. |
An element which can exhibit valencies of 2, 4 and 6 can be :A. copperB. ironC. mercuryD. sulphur |
| Answer» Correct Answer - D | |
| 17. |
Two elements X and Y have valencies of 5 and 3, and 3 and 2, respectively. The element X and Y are most likely to be respectively :A. copper and calciumB. sulphur and ironC. phosphorus and nitrogenD. nitrogen and iron |
| Answer» Correct Answer - D | |
| 18. |
(a) An element X exhibits variable valencies 3 and 5. Write the formulae of the chlorides of the element. (b) What is the ratio by mass of the elements present in the chemical formula of magnesium oxide |
|
Answer» (a) The formulae of the hlorides of the element X = XCl and XCl. (b) The chemical formula of magnesium oxide is MgO. The elements are present in the ratio of 24: 16 or 3 2. |
|
| 19. |
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) |
|
Answer» Correct Answer - `6.022 xx 10^(20)` ions mole of aluminium oxide `(Al_(2)O_(3))` = 2 × 27 + 3 × 16 = 102g i.e., 102g of `Al_(2)O_(3) = 6.022 × 10^(23)` molecules of `Al_(2)O_(3)` Then, 0.051 g of `Al_(2)O_(3)` contains `=6.022xx10^(23)//102xx0.051` molecules `= 3.011 × 10^(20)` molecules of `Al_(2)O_(3)` The number of aluminium ions `(Al^(3+))` present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions `(Al^(3+))` present in `3.11 × 10^(20)` molecules (0.051g) of aluminium oxide `(Al_(2)O_(3)) = 2 × 3.011 × 10^(20)` `= 6.022 × 10^(20)` |
|
| 20. |
Calculate the mass of 0.72gram molecule of carbon dioxide `(CO_(2))` |
|
Answer» Molecular mass of `CO_(2)="Atomic mass of "C+2xx" Atomic mass of O"` `=(12u+2xx16u)=44u` Gram molecule mass (Molar mass) of `CO_(2)=44g` 1 gram of molecule of `CO_(2)=44g` 0.72gram molecule of `CO_(2)=((44g))/((1g))xx(0.72g)`=31.68g |
|
| 21. |
How many particles are represented by 0.25 mole of an element? (b)Out of 4g of element represent particles=`6.022xx10^(23)` |
|
Answer» 1.0mole of the element represents particles `=6.022xx10^(23)` 0.25mole of the elements represents particles =`=((6.022xx10^23)xx(0.25"mol"))/((1.0"mol"))` `=1.505xx10^(23)"particles"` (b) 4g of methane `(CH_(4))` represebts moles `=((1"mol")xx(4g))/((16g))=0.25"mol"` `11"g of "CO_(2)"represents moles"=((1"mol")xx(11g))/((44g))=0.25"mol"` Since both `CH_(4) and CO_(2)` haave same number of moles (0.25moles), they have also same number of molecules. |
|
| 22. |
Hydrogen and oxygen combne in the ratio of 1:8 by mass to form water. What mass of oxygen will be required completely with 4g of hydrogen? |
|
Answer» Accordingly to availble data, Mass of oxygen combine with 1g of hydrogen =8g Mass of oxygen combine with 4g of hydrogen `=((8g)xx(4g))/((1g))=32g` |
|
| 23. |
A flask contains 4.4g of `CO_(2)` gas. Calculate (b) How many moles of `CO_(2)` gas does it contain? (b)How many molecules of `CO_(2)` gas are present in the sample. (c )How many atoms of oxygen are present in the given sample.( Atomic mass of C=12u, O=16) |
|
Answer» `"No. of moles"=("Given mass (m)")/("Gram molar mas (M)")=((4.4g))/(44"gmol"^(-1))=0.1"mol"` (b) `"1 mole of "CO_(2)"has moleuces"=6.605xx10^(23)` `"0.1 mole of "CO_(2)"has moleuces"=6.605xx10^(23)xx(0.1)=6.602xx10^(22)` (c ) `CO_(2)=2O` `"1 mole of "CO_(2)"has oxygen atoms"=2xxN_(A)` `"0.11 mole of "CO_(2)"has oxygen atoms"=2xxN_(A)xx0.1` `=2xx6.022x10^(23)xx0.1=1.204xx10^(24)"atoms"` |
|
| 24. |
Calculate the number of moles in the following (i) 28g of He (ii) He of Na. (ii) 60g of Ca Given gram atomic mass of (i) He=4g (ii) Na=23g (iii)Ca=40g. |
|
Answer» `(i)"28g of He The no.of moles"=("Mass of He in grams")/("Gram atomic mass ")=(m)/(M)=((28g))/((4g))=7"mol"` `(ii)"46g of Na The no.of moles"=("Mass of Na in grams")/("Gram atomic mass ")=(m)/(M)=((46g))/((23g))=2"mol"` `(iii)"60g of Ca The no.of moles"=("Mass of Ca in grams")/("Gram atomic mass")=(m)/(M)=((60g))/((40g))=1.5"mol"` |
|
| 25. |
Classify the following based on atomicity `(a) O_(3)` (b)`P_(4)` (c )Sg. |
| Answer» Atomicity is the number of atoms present in one molecule of an element. Based upon this, the atomicity of different molecules may be expressed as: (a)3 (b)4 (c )8. | |
| 26. |
Does the solubility of a substance change with temperature? Explain with the help of an example. |
| Answer» In most of the cases, the solutbility of a substance in a particular solvent such as water increases with the reise in temperature. For exampe more of sugar can dissolve in a particular volume of water by increasing the temperature. | |
| 27. |
Work out the formula for magnesium hydrogencarbonate. |
| Answer» Correct Answer - `Mg(HCO_(3))_(2)` | |
| 28. |
What do you understand by the word mole? |
| Answer» A mole denotes Avogadro s number of particles i.e.,` 6.022xx10^(23)` These may be electrons, protons, neutro atoms, ions etc. | |
| 29. |
In hydrogen peroxide `(H_(2)O_(2))`, the proportion of hydrogen and oxygen by mass is :-A. `1 : 8`B. `1 : 16`C. `8 : 1`D. `16 : 1` |
| Answer» Correct Answer - B | |
| 30. |
Classify each fo the following on the basis of their automicity. |
| Answer» `{:(,(a)F_(2),(b)NO_(2),(c)N_(2)O,(d)C_(2)H_(6)),(,(e)P_(4),(f)H_(2)O_(2),(g)P_(4)O_(10),(h)O_(3)),(,(i)HCl,(j)CH_(4),(k)He,(l)Ag):}` | |
| 31. |
Magnesium and oxygen combine in the ratio of `3 : 2` by mass to form magnesium oxide. What mass of oxygen gas would be required to react completely with 24 g of magnesium ? |
| Answer» Correct Answer - 16 g | |
| 32. |
What is meant by atomicity ? Explain with two examples. |
|
Answer» The atomicity of an element is the number of atoms present in one molecule of the element. For example, Atomicity of oxygen in `O_(2)=2` Atomicity of oxygen in `O_(3)=3`. |
|
| 33. |
What is meant by the term chemical formula ? Write the chemical formula of calcium oxide. Calculate its formula unit mass. `("Atomic mass of Ca"=40u, O=16u)`. |
|
Answer» Chemical formula of calcium oxide =CaO. Formula unit mass Aromic mass of Ca +Atomic mass of O= `(40+16)=56u.` |
|
| 34. |
What do you understand by 1 amu? |
| Answer» 1 amu (or 1u) is 1/12th of the mass of one atom of carbon taken as 12. | |
| 35. |
An element X has a valency of 4 whereas another element Y has a valency of 1. What will be the formula of the compound formed between X and Y ? |
| Answer» Correct Answer - `XY_(4)` | |
| 36. |
A particle P has 18 electrons, 20 neutrons and 19 protons. This particle must be :A. a moleculeB. a binary compoundC. an anionD. a cation |
| Answer» Correct Answer - D | |
| 37. |
In photosynthesis, 6 molecules of carbon combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula `C_(6)H_(12)O_(6)`. How many grams of water would be requried to produce 18g of glucose? Compute the volume of water so consumed assuming the density of water to be `1g cm^(-3)`. |
|
Answer» The chemical reaction involved is: `6CO_(2)+underset((6xx18g))(6H_(2)O) underset("Sunlight")overset("Chlorophyll")to underset((180g))(C_(6)H_(12)O_(6))+6O_(2)` To produce 180g of glucose, mass of water consumed=108g To produce 18g of glucose, mass of water consumed=`=((108g))/((180g))xx(18g)=10.8g` `"Density of water"=1"g cm"^(-3)` `"Volume of water"=("Mass of water")/("Density of water")=((10.8g))/((1"g cm"^(-3)))=10.8cccm^(3)` |
|
| 38. |
Calculate the molecular mass of nitric acid, `HNO_(3)`. (Atomic masses :` H = 1 u, N = 14 u , O = 16 u`) |
| Answer» Correct Answer - 63 u | |
| 39. |
Fill in the following blanks with suitable words : (a) The particle which is formed by the gain of electrons by an atom is called `"……….."`. (b) The particle which is formed by the loss of electrons by an atom is called `"……….."`. (c ) The particles which is formed by the loss or gain of electrons by an atom is called `"……….."`. (d) A potassium ion has positive charge because it contains less `"……….."` than `"……….."`. (e) A sulphide ion has negative charge because it contains less `"……….."` than `"……….."`. |
| Answer» Correct Answer - (a) anion , (b) cation , (c ) ion , (d) electron ; protons (e ) protons ; electrons | |
| 40. |
(a) Calculate the relative molecular mass of water `(H_(2)O)`. (b) Calculate the molecular mass of `HNO_(3)`. |
|
Answer» (a) Atomic mass of hydrogen = 1u, oxygen = 16 u So the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is = 2 × 1+ 1×16 = 18 u (b) The molecular mass of `HNO_(3)` = the atomic mass of H + the atomic mass of N+ 3 × the atomic mass of O = 1 + 14 + 48 = 63 u |
|
| 41. |
Calculate the numbe of oxygen atoms present in 120g of nitric acid `(HNO_(3))` |
|
Answer» Molar mass of `HNO_(3)==1xx"Atomic mass of H"+1xx"Atomic mass of N"+3xx"Atomic mass of O"` `=1xx1+1xx14xx3xx16=63u` Gram molar mass of `HNO_(3)`=63g `63"g of "HNO_(3) "contains oxygen atoms"=3xx6.022xx10^(3)"atoms"` `120"g of "HNO_(3) "contains oxygen atoms"=3xx6.022xx10^(230)xx((120g))/((63g))` `=3.44xx10^(24)"atoms"` |
|
| 42. |
Which of the following contains maximum number of molecules?A. 1g `CO_(2)`B. 1g `N_(2)`C. 1g `H_(2)`D. 1g `CH_(4)` |
|
Answer» Correct Answer - C (c) (a) `((1g))/((44g))xxN_(A) =0.023 xx N_(A)` (b)`((1g))/((28g))xxN_(A)=0.036xxN_(A)` (c) `((1g))/((2g)) xxN_(A)=0.5 xxN_(A)` (d) `((1g))/((16g))xxN_(A)=0.063 xxN_(A)` |
|
| 43. |
Which of the following has maximum number of atoms?A. 18g of `H_(2)O`B. 18g of `O_(2)`C. 18g of `CO_(2)`D. 18g of `CH_(4)` |
|
Answer» Correct Answer - D (d) `("Mass of substance"xx "No. of atoms in one molecule")/("Molar mass of substance") xxN_(A)` (a) ` ((18g)xx3)/((18g)) xxN_(A) =3N_(A)` `(b) ((18g)xx2)/((32g))xxN_(A)=1.23 N_(A)` (c) `((18g)xx3)/(44g))xxN_(A)=1.23 N_(A)` (d) ((18g)xx5)/((16g))xxN_(A) =5.63 N_(A)` |
|
| 44. |
(a)What do the following observations stand for? (i)2O (ii)`3O_(2)` (b)Which amongst the following has more number of atoms and how much? (i)11.5g of sodium (ii)15.0g of calcium |
|
Answer» Gram atomic mass of sodium (Na)=23g 23g of sodium have atoms=`N_(A)` 11.5g of sodium have atoms `=0.5xxN_(A)=0.5xx6.022xx10^(23)` `=3.011xx10^(23)` Gram atomic mass of calcium (Ca)=40g 40g of Calcium have atoms =`N_(A)` 15g of calcium have atoms `=(15)/(40)xxN_(A)=0.375xx6.022xx10^(23)` `=2.258xx10^(23)` 11.5g of sodium has more atoms=`(3.011-2.258)xx10^(23)` `=7.53xx10^(22)"atoms:` |
|
| 45. |
Which of the following would weigh the maximum?A. 0.2 mole of sucrose `(C_(12)H_(22)O_(11))`B. 2 moles of `CO_(2)`C. 2 moles of `CaCO_(3)`D. 10 moles of `H_(2)O` |
|
Answer» Correct Answer - C c (a) `0.2xx342 g =68.4 g ` (b) 2xx44 g =88g` (c) `2xx100g =200 g` (d) `10xx 18 g =80 g ` |
|
| 46. |
How many atoms are present in a (i) `H_(2)S` molecule and (ii) `PO_(4)^(3–)` ion? |
|
Answer» (i) There atoms are present in a `H_(2)S` molecule. (ii) Five atoms are present in a `PO_(4)^(3-)` ion. |
|
| 47. |
A student wants to have `3.011 xx 10^(23)` atoms each of magnesium and carbon elements. For this purpose, he will have to weigh :A. 24 g of magnesium and 6 g of carbonB. 12 g of carbon and 24 g of magnesiumC. 20 g of magnesium and 10 g of carbonD. 12 g of magnesium and 6g of carbon |
| Answer» Correct Answer - d | |
| 48. |
Calculate formula unit mass of `CuSO_(4)` |
|
Answer» Formula unit mass of `CuSO_(4)="Atomic mass of Cu"+"Aromic mass of S"+4xx"Atomic mass of O"` `=63.5+32+4xx16=159.4u` |
|
| 49. |
If the law of constant compositon is true, what weights of calcium carbon, and oxygen are present in `1.5 g` of calcium carbonate, if a sample of calcium carbonate from another source contains the following percentage composition: `Ca = 40.0%, C = 12.0%`, and `O = 48.0%`? |
|
Answer» We have learnt that according to the law of constant proportions, the percentage of the elements presents in different sample of pure of substance always remains the same. This means that in the second sample of calcium carbonate, the ration of the different elements present will remain the same or will remain unchanged. Thus, `"Mass of calcium in 1.5g of the sample"=(40)/(100)xx(1.5g)=0.6g` `"Mass of carbon in 1.5g of the sample"=(12)/(100)xx(1.5g)=0.18g` `"Mass of oxygen in 1.5g of the sample"=(48)/(100)xx(1.5g)=0.72g` |
|
| 50. |
(a) Write the symbols/formulae of two simple ions and two compound ions (or polyatomic ions). (b) An element Y has a valency of 4. Write the formula for its : (i) Chloride , (ii) oxide , (iii) sulphate, (iv) carbonate , (v) nitrate |
| Answer» Correct Answer - (b) (i) `YCl_(4)` , (ii) `YO_(2)` , (iii) `Y(SO_(4))_(2)`, (iv) `Y(CO_(3))_(2)` , (v) ` Y(NO_(3))_(4)` | |