Worksheet On Atoms, Molecules And Ions 55nf

3

Cycloalkanes

1 Atoms, Molecules and Ions 8

Type I.

MCQ with only ONE Correct Alternate

1. A sample of CaCO 3 is 50% pure. On heating 1.12 L of CO 2 (at STP) is obtained. Residue left (assuming non-volatile impurity) is: (a) 7.8 g (b) 3.8 g (c) 2.8 g (d) 8.9 g 2. In the decomposition of 10 g of MgCO 3, 0.1 mole CO 2 and 4.0 g MgO are obtained. Hence, percentage purity of MgCO 3 is: (a) 50% (b) 60% (c) 40% (d) 84% 3. Consider the following pairs: I. CH 4 , C 2H 6 II. CO, CO 2 III. NO, NO 2 IV. H 2O, H 2O 2 In which cases, law of multiple proportion is followed? (a) I, II (b) I, II, III (c) I, III, IV (d) I, II, III, IV 4. Two substances of carbon and oxygen have respectively 72.73% and 47.06% oxygen. Hence, they follow: (a) law of multiple proportion (b) law of reciprocal proportion (c) law of definite proportion (d) law of conservation of mass 5. In which case purity of the substance is 100%? (a) 1 mol of CaCO 3 gave 11.2 L CO 2 (at STP) (b) 1 mol of MgCO 3 gave 40.0 g MgO (c) 1 mol of NaHCO 3 gave 4 g H 2O (d) 1 mol of Ca(HCO 3 )2 gave 1 mol CO 2

6. Consider following laws of chemical combination with examples: I. Law of multiple proportion : N 2O, NO, NO 2 II. Law of reciprocal proportion : H 2O, SO 2, H 2S Which is correct with examples? (a) I and II (b) I only (c) II only (d) None of these 7. H 2S contains 94.11% sulphur; SO 2 contains 50% oxygen and H 2O contains 11.11% hydrogen. Thus: (a) law of multiple proportion is followed (b) law of reciprocal proportion is followed (c) law of conservation of mass is followed (d) all of the above 35 37 8. Sodium combines with 17 Cl and 17 Cl to give two samples of sodium chloride. Their formation follows the law of: (a) gaseous diffusion (b) conservation of mass (c) reciprocal proportion (d) none of these 9. According to Dalton’s atomic theory, the smallest particle in which matter can exist, is called: (a) an atom (b) an ion (c) an electron (d) a molecule 10. The nucleus of an atom consists of: (a) neutron (b) proton (c) electron (d) both (a) and (b) 35 37 11. 17 Cl and 17 Cl are two isotopes of chlorine. If average atomic mass is 35.5 then ratio of these two isotopes is: (a) 35 : 37 (b) 1 : 3 (c) 3 : 1 (d) 2 : 1

4

Objective Chemistry

12. Ionic mass of X 3− is 17. If it has 10 electrons, then number of neutrons are: (a) 10 (b) 13 (c) 7 (d) 17 13. M 2+ ion is isoelectronic of SO 2 and has ( Z + 2) neutrons (Z is atomic number of M ). Thus, ionic mass of M 2+ is: (a) 70 (b) 66 (c) 68 (d) 64 + 2+ − 14. X , Y and Z are isoelectronic of CO 2. Increasing order of protons in X + , Y 2+ and Z − is: (a) X + = Y 2+ = Z − (b) X + < Y 2+ < Z − (c) Z − < X + < Y 2+ (d) Y 2+ < X + < Z − 15. X − , Y 2− and Z 3− are isotonic and isoelectronic. Thus, increasing order of atomic number of X, Y and Z is: (a) X < Y < Z (b) Z < Y < X (c) X = Y = Z (d) Z < X < Y 16. Number of atoms in increasing order in 1.6 gram CH 4 , 1.7 g NH 3 and 1.8 g H 2O is: (a) H 2O = NH 3 = CH 4 (b) H 2O < NH 3 < CH 4 (c) CH 4 < NH 3 < H 2O (d) CH 4 = NH 3 < H 2O 17. Which has maximum number of H-atoms per gram of the substance? (a) CH 4 (b) CuSO 4 ⋅ 5H 2O (c) H 2O 2 (d) H 2O 18. If each O-atom has two equivalents, volume of one equivalent of O 2 gas at STP is: (a) 22.4 L (b) 11.2 L (c) 5.6 L (d) 44.8 L 19. Each drop of H 2O has 0.018 mL at room temperature. Number of H 2O molecules in one drop is: (a) 1 × 10−3 (b) 6.02 × 1020 (c) 22.4 × 10−3

(c) N 02

(d) 64 g mol −1

21. Mass of one atom of an element is 6.64 × 10 g. This is equal to: (a) 6.64 × 10−23 amu (b) 40.0 amu 1 (c) (d) 6.64 amu amu 40 22. If Avogadro’s number would have been 1 × 10−10, instead of 6.02 × 1023 then mass of one atom of H would be: (a) 1 amu (b) 1 × 1010 amu (c) 6 amu (d) 6 × 1013 amu 23. Mass of one atom of X is 2.66 × 10−23 g, then its 32 g is equal to: (a) 32 × 2.66 × 10−23 mol 32 (b) mol –23 2.66 × 10 × 6.02 × 1023

1 N 0 × 12

26.

27.

28.

29.

30.

31.

(a) 14 amu (b) 7 amu (c) 14 g (d) 7 g If two compounds have same empirical formula but different molecular formula, they must have: (a) same viscosity (b) same vapour density (VD) (c) different molecular weight (d) different percentage composition If the equivalent weight of an element is 32, then the percentage of oxygen in its oxide is: (a) 16 (b) 40 (c) 32 (d) 20 A hydrocarbon has 3 g carbon per gram of hydrogen, hence, simplest formula is: (a) CH 4 (b) C 6H 6 (c) C 3H 8 (d) CH 2 Molar ratio of Na 2SO 3 and H 2O is 1 : 7 in Na 2SO 3 ⋅ xH 2O. Hence, their mass percentage is: (a) 12.5 : 87.5 (b) 87.5 : 12.5 (c) 50 : 50 (d) 75 : 25 Which has maximum percentage of chlorine? (a) C 6H 6Cl 6 (b) C 6H 5Cl (c) CH 3Cl (d) CCl 4 In which of the following pairs do the two species resemble each other most closely in chemical properties? (a) 11 H and 12 H (b) 168O and 168O 2– (c)

−23

(d)

25. Mass of one 147 N-atom is:

(d) 6.02 × 3 × 102

20. 1 g CH 4 and 4 g of compound X have equal number of moles. Thus, molar mass of X is: (a) 16 g mol −1 (b) 32 g mol −1 (c) 4 g mol −1

32 × 2.66 × 10–23 mol 6.02 × 1023 (d) none of the above 24. Number of mol of 126C in 1 amu is: 1 (a) (b) N 0 N0 (c)

24 12 Mg

24 and 12 Mg 2+

(d)

14 7N

and 147 N 3–

32. One isotope of a non-metallic element has mass number 127 and 74 neutrons in the nucleus. The anion derived from the isotope has 54 electrons. Hence, symbol for the anion is: − − (a) 127 (b) 127 54 X 53 X (c)

74 − 53 X

(d)

74 54

X−

33. Which of the following is the richest source of ammonia on a mass percentage basis? (a) NH 4 NO 3 (b) NH 2CONH 2 (c) NH 3 (d) HNC(NH 2 )2 34. Which of the following substances contains greatest mass of chlorine? (a) 5.0 g Cl 2 (b) 0.5 mol Cl 2 (c) 0.10 mol KCl (d) 30.0 g MgCl 2

5

Atoms Molecules and Ions 35. When 0.273 g of Mg is heated strongly in a nitrogen (N 2 ) atmosphere, 0.378 g of the compound is formed. Hence, compound formed is: (a) Mg 3N 2 (b) Mg 3N (c) Mg 2N 3 (d) MgN 36. A certain metal sulphide, MS 2, is used extensively as a high temperature lubricant. If MS 2 is 40.06% by mass sulphur, metal M has atomic mass: (a) 160 amu (b) 64 amu (c) 40 amu (d) 96 amu 37. The molar mass of a compound if 0.372 mol of it has a mass of 186 g, is: (a) 200 g (b) 372 g (c) 500 g (d) 186 g 38. Which of the following has maximum number of C-atoms? (a) 4.4 g CO 2 (b) 3.0 g C 2H 6 (c) 4.4 g C 3H 8 (d) 1.3 g C 6H 6 39. Mg 2C 3( X ) is decomposed by H 2O forming a gaseous hydrocarbon ( Y ). 8.4 g of X gives ………… mol of Y. (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 40. Number of atoms in 20 g Ca is equal to number of atoms in: (a) 20 g Mg (b) 1.6 g CH 4 (c) 1.8 g H 2O (b) 1.7 g NH 3 41. Na 2SO 3 ⋅ xH 2O has 50% H 2O by mass. Hence, x is: (a) 4 (b) 5 (c) 6 (d) 7 42. Mass of one atom of X is 6.66 × 10−23 g. Hence, number of moles of atom X in 40 kg is: (a) 103 mol (b) 10−3 mol 40 × 103 40 × 103 (c) (d) mol mol −23 6.66 × 10 6.022 × 1023 43. To make 0.01 mol, which of the following has maximum mass? (a) NaHCO 3 (b) Na 2CO 3 (c) Na 2SO 4 (d) Na 2C 2O 4 44. In a glass-tube, there is 18 g of glucose. 0.08 mol of glucose is taken. Glucose left in the glass-tube is: (a) 0.10 g (b) 17.92 g (c) 3.60 mol (d) 3.60 g 45. Rest mass of an electron is 9.11 × 10−31 kg. Molar mass of the electron is: (a) 1.50 × 10−31 kg mol −1 (b) 9.11 × 10−31 kg mol –1 (c) 5.5 × 10−7 kg mol −1 (d) 6.02 × 1023 kg mol −1

8

Type II.

46. A sample of ammonium phosphate (NH 4 )3 PO 4 contains 3.18 mol of H-atoms. The number of moles of oxygen atoms in the sample is: (a) 0.265 (b) 0.795 (c) 1.06 (d) 3.18 47. A sample of CuSO 4 ⋅ 5H 2O contains 3.782 g of Cu. How many grams of oxygen are in this sample? (Cu = 63.5) (a) 0.952 g (b) 3.80 g (c) 4.761 g (d) 8.576 g 48. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is 69.98%. Its molar mass is: (a) 176.5 (b) 252.2 (c) 287.6 (d) 360.1 49. In a hydrocarbon C is 3 g per gram of hydrogen. Hence, hydrocarbon is: (a) CH 4 (b) C 2H 6 (c) C 3H 8 (d) C 4 H10 50. One equivalent of magnesium oxide weighs 20 g then one equivalent of magnesium chloride weighs: (a) 29.75 g (b) 47.5 g (c) 95.0 g (d) 20.0 g 51. A spherical ball of radius 7 cm contains 56% iron. If density is 1.4 g /cm 3, number of mol of Fe present approximately is: (a) 10 (b) 15 (c) 20 (d) 25 52. In the following final result is …0.1 mol CH 4 + 3.01 × 1023 molecules CH 4 − 9.6 g CH 4 = x mol H atoms: (a) 0 mol H atom (b) 0.2 mol H atom (c) 0.3 mol H atom (d) 0.4 mol H atom 53. If we assume that N 0 = 1.2 × 1023 mol −1, then molar mass of O 2 will be taken as: 32 (a) 32 g mol −1 (b) g mol −1 6 1 × 1023 (c) 32 × 1023 g mol −1 (d) g mol −1 32 54. CO, CO 2, C 2O 3 follow: (a) law of definite proportion (b) law of multiple proportion (c) law of conservation of mass (d) all of the above 55. In a gas S and O are 50% by mass, hence, their mol ratio is: (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1

MCQ with ONE or more than ONE Correct Alternates

1. Which is/are correct about 4.25 g NH 3? (a) It contains 0.25 mol of NH 3 (b) It contains 0.75 mol of H-atoms (c) It contains total of 1.0 mol of N and H atoms (d) It contains 1.5 × 1023 molecules of NH 3

2. Which of the following are isoelectronic of O 2− ? (a) N 3− (b) F− (c) Ti + (d) Na + 3. Among the following groupings which represents the collection of isoelectronic species? − (a) NO, CN − , N 2, O −2 (b) NO + , C 2− 2 , O 2 , CO 2− + (c) N 2, C 2 , CO, NO (d) CO, NO , CN − , C 2− 2

6

Objective Chemistry

4. The atomic number of an element is always equal to: (a) number of neutrons in the nucleus (b) half of the atomic weight (c) electrical charge of the nucleus (d) number of protons 5. A bivalent metal ion has equivalent mass of 12. Then: (a) equivalent mass of its oxide is 28 (b) molar mass of its oxide is 40 (c) equivalent mass of its hydride is 13 (d) molar mass of its hydride is 14 6. Volume of a gas at NTP is 1.12 × 10−7 cc. The number of molecules is thus equal to: (a) 3.01 × 1012 (b) 3.01 × 1018 (c) 3.01 × 1024 (d) 3.01 × 1030 7. Which of the following may contain one proton and one neutron? (a) H +2 (b) 42 He

(c)

2 1D

(d)

3 1T

37 and 17 Cl differ in: (a) atomic number (b) number of neutrons (c) number of electrons (d) atomic mass 9. Isoelectronic species are represented by pairs:

8.

35 17 Cl

(a) N 3− , O 2−

(b) CO, CN −

(c) O 2− 2 , F2

(d) O −2 , CN −

10. X − is isoelectronic of CO and has ( Z + 2) neutrons (Z = atomic number of X − ). Thus, (a) ionic mass of X − is 28 (b) ionic mass of X − is 30 (c) atomic number of X − is 13 (d) atomic number of X − is 14

8 Type III. Testing of Numerical Skill 1. 1021 molecules are removed from 200 mg of CO 2. Thus, number of moles of CO 2 left is 2. Al 2(SO 4 )3 ⋅ xH 2O has 8.20 per cent Al by mass. Calculate value of x. 3. In addition to carbon monoxide (CO) and carbon dioxide (CO 2 ), there is a third compound of carbon called carbon suboxide. If a 2.500 g sample of carbon suboxide contains 1.32 g of C and 1.18 g of O, show that the law of multiple proportions is followed. What is possible formula of carbon suboxide? 4. XHCO 3 and YCO 3 are two pure substances of equal molar masses decomposing as shown

8 Type IV

∆

YCO 3 → YO + CO 2 16.8 g of XHCO 3 gave 6.2 g of mixture of H 2O and CO 2. Identify the substances. 5. Cesium atoms are the largest naturally occurring atoms. The radius of cesium atom is 2.62 Å. How many cesium atoms would have to be laid side by side to give a row of cesium atoms 2.50 cm long? Assume that the atoms are spherical.

Comprehension Based MCQ

Example 1. Study following observations and answer the questions at the end of it. The following is a crude but effective method for estimating the order of magnitude of Avogadro’s number using stearic acid (C18H 36O 2 ). When stearic acid is added to water, its molecules collect at the surface and form a monolayer; that is, the layer is only one molecule thick. The cross-sectional area of each stearic acid molecule has been measured to be 0.21 nm 2. In one experiment it is found that 1.4 × 10−4 g of stearic acid is needed to form a monolayer over water in a dish of diameter 20 cm. (the area of a circle of radius r is πr 2. ) Q. I.

∆

2XHCO 3 → H 2O + CO 2 + X 2CO 3

Based on these measurements value of Avogadro’s number is: (a) 3 × 1023 (b) 6 × 1023 (c) 4 × 1023 (d) 1 × 1023

Q. II. What is equivalent of 1g H in amu for this value of Avogadro’s number?

(a) 1.66 × 10−24 g (c) 2.5 × 10−24 g

(b) 3.33 × 10−24 g (d) 1 × 10−23 g

Example 2. Read the following age regarding fertiliser K 2 O and answer the questions at the end of it. Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride (KCl) and potassium sulphate (K 2SO 4 ). Potash production is often reported as the potassium oxide (K 2O) equivalent or the amount of K 2O that could be made from a given mineral. KCl costs Rs. 50 per kg. Q. I. What is the cost of K per mol of the KCl sample? (a) Rs. 13.42 mol −1 (b) Rs. 3.73 mol −1 −1 (c) Re. 1.00 mol (d) Rs. 2.00 mol −1 Q. II. For what price must K 2SO 4 be sold in order to supply the same amount of potassium as in KCl? (a) Rs. 58.40 kg −1 (b) Rs. 50.00 kg −1 (c) Rs. 42.82 kg −1 (d) Rs. 25.00 kg −1

7

Atoms Molecules and Ions Q. III. What mass (in kg) of K 2O contains the same number of moles of K atoms as 1.00 kg KCl? (a) 0.158 kg (b) 0.315 kg (c) 1.262 kg (d) 0.631 kg

Relative abundance

100 80 60

Q. I. Which isotope has maximum number of atoms per gram of each? (a) Mg-24 (b) Mg-25 (c) Mg-26 (d) equal Q. II. Number of atoms in one mole of each isotope is placed in increasing order: (a) 24 Mg < 25Mg < 26Mg (b)

26

Mg < 25Mg < 24 Mg

(c)

26

Mg =

25

Mg =

24

Mg

(d) given data is insufficient Q. III. Average atomic mass of Mg is approximately: (a) 25.0 (b) 24.3 (c) 25.2 (d) 25.8

40 20

24 25 26 Atomic mass (amu)

Example 3. The mass spectrum (given below) of magnesium has three peaks, which indicates that magnesium has three isotopes. Questions given below are based on this mass spectrum.

8

Type V. Miscellaneous Exercise

➥ A. Assertion/Reason Type Questions

1.

2.

3.

4.

Codes : (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. Assertion (A) : When 10 g of CaCO 3 is decomposed, 5.6 g of residue is left and 4.4 g of CO 2 escapes. Reason (R) : Law of mass conservation is followed. Assertion (A) : SO 2 and SO 3 obey “law of multiple proportion”. Reason (R) : Every sample of SO 2 contains 1 part sulphur and 1 part oxygen by mass. Assertion (A) : SO 2 from oxidation of sulphur or H 2S contains sulphur and oxygen in the mass ratio of 1 : 2. Reason (R) : Different samples of a pure chemical substance always contain the same proportions of elements by mass. Assertion (A) : In Fe2+ , there are 24 electrons and 30 neutrons and thus, ionic mass is 56. Reason (R) : Ionic mass = neutron + proton and proton = electrons in neutral species

➥ B. True/False 1. If there are 10 electrons and 8 neutrons in X 2− , then its ionic mass is 18. 2. SO 2 contains sulphur and oxygen in 1 : 1 mass ratio.

3. 20 g CaCO 3 (50% pure) on strongly heating gives 15.6 g residue. 4. Mass of 1 mol of electrons is 5.48 × 10−7 kg. 5. Number of atoms in 1.6 g CH 4 , 1.7 g NH 3 and 1.8 g H 2O are equal. 6. If 1 mol of H 2 is taken out from 2 g H 2, 1 g H 2 is left. 7. 1 amu is equal to 1.66 × 10−24 g. 8. Out of CH 3Cl, C 6H 6Cl 6, and CCl 3CHO, maximum percentage of chlorine is in C 6H 6Cl 6. 9. 1 ppm is equal to 1 mg per kg. 10. Equivalent mass of NH 3 is 17.

➥ C. Fill in the Blanks 1. NO and NO 2 follow the law of ………… 2. SO 3 contains sulphur and oxygen in the mass ratio of ………… 3. Hydrocarbon containing 3 g carbon per gram hydrogen is ………… 4. If X 2+ is isotonic (same number of neutrons) of H 2O and contains 10 electrons then ionic mass of X 2+ is ………… 5. 1 mol of barium nitrate and 1 mol of ammonium phosphate form ………… mol of barium phosphate. 6. Percentage of sodium sulphite (Na 2SO 3 ) in hydrated salt is 50. Hence, molecular formula of hydrated salt is ………… 7. Concentration 1 ppm = ………… ppb. 8. Concentration 1 microgram per gram = ………… ppm. 9. H 2O, H 2S and SO 2 follow law of ………… 10. Atomic mass unit is also called …………

8

Objective Chemistry

➥ D. Matching

B. Law of proportion

multiple 2. 10 mL N 2 combines with 30 mL of H 2 to form 20 mL of NH 3. C. Law of definite 3. S and O 2 combine to proportion form SO 2 and SO 3. D. Law of reciprocal 4. In H 2S and SO 2 mass proportion ratio of H and O w.r.t. sulphur is 1 : 16, hence in H 2O, mass ratio of H and O is 1 : 8. E. Gay Lussac’s law 5. 4.2 g MgCO 3 gives 2.0 g residue on heating.

1. Match % of carbon (in List I) with the compound (in List II). List I A. B. C. D. E.

List II

20% 52.2% 75% 27.3% 12%

1. 2. 3. 4. 5.

CH 4 CO 2 C 2H 6O CN 2OH 4 CaCO 3

2. Match Laws of chemical combinations (in List I) with the examples followed (in List II). List I

List II

A. Law of conservation of 1. CH 4 has carbon and mass hydrogen in 3 : 1 mass ratio.

Type I. MCQ with only ONE Correct Alternate 1. (a) 11. (c) 21. (b)

2. (d) 12. (a) 22. (a)

3. (d) 13. (a) 23. (b)

4. (a) 14. (c) 24. (d)

5. (b) 15. (b) 25. (a)

6. (a) 16. (b) 26. (c)

7. (b) 17. (a) 27. (d)

8. (d) 18. (c) 28. (a)

Type II. MCQ with ONE or more than ONE Correct Alternates

Type V. Miscellaneous Exercise

1. (a), (b), (c), (d) 5. (b), (c) 6. (a) 10. (a), (c)

A. Assertion/Reason Type Questions

2. (a), (b), (d) 3. (d) 4. (c), (d) 7. (c) 8. (b), (d) 9. (a), (b), (c)

1. F 8. T

1. 2.88 × 10−3 2. x = 18 3. C 3O 2 4. XHCO 3 = NaHCO 3 , YCO 3 = MgCO 3 5. 4.77 × 107 atoms

3. (d)

4. (a)

Q. I. (a) Q. I. (b) Q. I. (a)

Q. II. (b) Q. II. (c) Q. II. (c)

2. T 9. T

3. T 10. F

4. T

5. F

6. F

7. T

C. Fill in the Blanks 1. Multiple proportion 2. 2 : 3 3. CH4 4. 20 5. 0.33 6. Na 2SO 3 ⋅ 7H2O 7. 103 8. 1 9. Reciprocal proportion 10. Dalton (Da)

Comprehension Based MCQ

Example 1 Example 2 Example 3

2. (b)

10. (d) 20. (d) 30. (d)

B. True/False

Type III. Testing of Numerical Skill

Type IV.

1. (a)

9. (a) 19. (b) 29. (c)

Q. III. (d) Q. III. (b)

D. Matching 1. (A)—(4); (B)—(3); (C)—(1); (D)—(2); (E)—(5) 2. (A)—(5); (B)—(3); (C)—(1); (D)—(4); (E)—(2)

H ints 8

Type I.

1.

and Solutions

MCQ with only ONE Correct Alternate CaCO 3 (s )

∆ →

1 mol =100 g 0.05 mol = 5 g pure

CaO(s ) ↓ + CO 2 (g ) ↑ 1 mol = 56 g

0.05 mol = 2.8 g

22.4 L at STP 1.12 L at STP

Impure CaCO 3 taken = 10 g (5 g pure CaCO 3 + 5 g impurity) CaO(s ) left = 2.8 g Impurity = 5.0 g Total residue = 7.8 g

Thus, (a) ∆

2. MgCO 3 (s ) → MgO(s )+ CO 2 (g ) 1 mol = 84 g 8.4 g (pure) in 10 g sample

40 g 4.0 g

1 mol 0.1 mol

Thus, % of pure MgCO 3 = 84% Thus, (d)

9

Atoms Molecules and Ions ∆

5. (a)

CaCO 3 → CaO + CO 2 1 mol

22.4 L

actual = 11.2 L Thus, 50% ∆

(b)

MgCO 3 → MgO + CO 2 1 mol

40 g

Thus, 100% ∆

(c)

2NaHCO 3 → Na 2CO 3 + H2O + CO 2 2 mol 1 mol

18 g 9g

actual = 4 g Thus, 44.4%

Thus, (b) 27. Equivalent weight of element = 32 g and that of oxygen = 8 g Thus, one equivalent of oxide = 40 g 8 Percentage of oxygen in oxide = × 100 40 = 20% Thus, (d) 29. Na 2SO 3 : H2O

∆

(d)

Ca(HCO 3 )2 → CaO + H2O + 2CO 2 1 mol

2 mol

actual = 1 mol Thus, 50% Thus, (b) A1 X 1 + A2 X 2 X1 + X 2 35X 1 + 37 X 2 35.5 = X1 + X 2 X1 3 = X2 1

11. Average atomic mass A =

On solving

19.

Number of moles in

Moles

1

Mass

126

126

50

50

%

Thus (c) 12. X 3− has 10 electrons. Thus, protons (= atomic number) in X 3 − = 7 neutron + proton = ionic mass =17 Thus, neutrons =17 – 7 =10 Thus, (a) 13. M 2+ has electrons = 32 (isoelectronic of SO 2 ) protons = 34 = Z neutrons = 36 Thus, ionic mass =70 Thus, (a) 16.

Thus, (b) 21. Mass of one atom = 6.64 × 10−23 g Thus, atomic mass = 6.64 × 10–23 × 6.02 × 1023 = 40 Thus, (b) 23. Mass of one atom = 2.66 × 10–23 g Mass of N 0 atoms = 2.66 × 10−23 × 6.02 × 1023 g mol −1 = atomic mass Thus, number of moles in 32 g 32 = 2.66 × 10–23 × 6.02 × 1023

Molecules

Atoms

CH4

0.1

0.1 N 0

0.5 N 0

NH3

0.1

0.1 N 0

0.4 N 0

H2O

0.1

0.1 N 0

0.3 N 0

Thus, (b) 0.018 mL = 0.018 g (density of water = 1g/mL) 0.018 = = 0.001 mol 18 = 6.02 × 1020 molecules

:

7

Thus, (c) 44. 18 g glucose = 0.10 mol glucose withdrawn = 0.08 mol left = 0.10 − 0.08 = 0.02 mol = 3.6 g Thus, (d) 45. Rest mass of electron = 9.11 × 10−31 kg mass of one mole of electrons = 9.11 × 10−31 × 6.02 × 1023 = 5.48 × 10–7 kg mol −1 Thus, (c) 4 51. Volume of spherical ball = πr3 3 4 = π (7 )3 3 4 Mass of spherical ball = π (7 )3 × 1.4 g 3 4 56 Iron in spherical ball = π (7 )3 × 1.4 × g 3 100 4π (7 )3 × 1.4 × 56 = mol 3 × 56 × 100 = 20 mol Thus, (c)