1. Chemistry

2. Session Opener

3. Mole concept and stoichiometry

4. Session Objectives

5. Session Objective Problems related to 1. Mole concept 2. Stoichiometry

6. Concept of equivalence Equivalent weight can be defined as Therefore, no. of equivalents = no. of moles x n–factor

7. Illustrative problem 1 0.5 mole of lead nitrate is mixed with 0.1 mole of chromium sulphate in water. The maximum number of moles of lead sulphate that can be obtained is (a) 0.6 mole(b) 0.5 mole (c) 0.3 mole(d) 0.1 mole Solution: Hence, the answer is (c).

8. Illustrative problem 2 A compound of iron and chlorine is soluble in water. An excess of silver nitrate was added to precipitate chloride ion as silver chloride. If a 134.8 mg of the compound gave 304. 8 mg of AgCl, what is the formula of the compound (Fe = 56, Ag = 108, Cl = 35.5)

9. Solution

10. Illustrative example 3 Blood haemoglobin contains 0.33% iron. Assuming that there are four atoms of iron per molecule of haemoglobin, its approximate molecular mass is found to be (a) 34,000(b) 17,000 (c) 67,879(d) 85,000 Solution: Hence, the answer is (c).

11. Illustrative example 4 0.5 g of an impure sample of magnesium contains its own oxide as an impurity, when heated with dil. H 2 SO 4 it gave 448 ml of hydrogen at N.T.P. Calculate the percentage purity of magnesium. At wt. Of Mg = 24. Solution:

12. Illustrative example 5 Determine the percentage composition of a mixture of anhydrous sodium carbonate and sodium bicarbonate from the following data: Weight of the mixture taken = 2 g Loss in weight on heating = 0.124 g

13. Solution Let mass of Na 2 CO 3 =x gm Mass of NaHCO 3 =(2–x) gm Loss in weight on heating is due to the decomposition of NaHCO 3. After decomposition, weight of the remaining substance = (2–0.124)g=1.876 g In the mixture,

14. Solution

15. Illustrative example 6 Two oxides of a metal contain 50% and 40% of the metal by mass respectively. The formula of the first oxide is MO. Then the formula of the second oxide is (a) MO 2 (b) M 2 O 3 (c) M 2 O(d) M 2 O 5 Solution: Hence, the answer is (b).

16. Illustrative example 7 A hydrocarbon contains 75% of carbon. Then its molecular formula is. (a) CH 4 (b) C 2 H 4 (c) C 2 H 6 (d) C 2 H 2 Solution: Hence, the answer is (a).

17. Illustrative problem 8 An hourly requirement of an astronanut can be satisfied by the energy released when 34.2 g of sucrose (C 12 H 22 O 11 ) are burnt in his body. How many grams of oxygen would be needed in a space capsule to meet his requirement for one day ? Solution:

18. Illustrative problem 9 60 g of a compound on analysis gave 24 g C, 4 g H and 32 g O. The empirical formula of the compound is: (a) C 2 H 4 O 2 (b) C 2 H 2 O 2 (c) CH 2 O 2 (d) CH 2 O Solution: % of C =% of H =% of O =

19. Solution ElementPercentageAtomic ratio Simplest ratio C40 H6.66 O53.33 Hence, answer is (d). Hence empirical formula CH 2 O.

20. Simple Titrations Find out the concentration of a solution with the help of a solution of known concentration. For mixture of two or more substances N 1 V 1 + N 2 V 2 + ……= NV Where V=(V 1 + V 2 + …..)

21. Normality of mixing two acids Normality of mixing acid and bases

22. Questions

23. Illustrative example 10 Find the molality of H 2 SO 4 solution whose specific gravity(density) is 1.98 g/ml and 95% mass by volume H 2 SO 4. 100 ml solution contains 95 g H 2 SO 4. Moles of H 2 SO 4 = Mass of solution = 100 × 1.98 = 198 g Mass of water = 198 – 95 = 103 g Molality == 9.412 m Solution:

24. Illustrative example 11 A sample of H 2 SO 4 (density 1.787 g/ml) is 86% by mass. What is molarity of acid? What volume of this acid has to be used to make 1 L of 0.2 M H 2 SO 4 ? Let V 1 ml of this H 2 SO 4 are used to prepare 1 L of 0.2 M H 2 SO 4.M 1 V 1 = M 2 V 2 15.68 × V 1 = 0.2 × 1000 Solution:

25. Illustrative example 12 A mixture is obtained by mixing 500ml 0.1M H 2 SO 4 and 200ml 0.2M HCl at 25 0 C. Find the normality of the mixture. For the mixture, Solution:

26. Illustrative example 13 500 ml 0.2 N HCl is neutralized with 250 ml 0.2 N NaOH. What is the strength of the resulting solution? Equivalents of HCl Equivalents of NaOH Equivalence of excess HCl Normality of HCl (excess) Strength of HCl =.067 × 36.5 g/litre = 2.44 g/litre Solution:

27. Solution N = 2.44 N Strength of HCl =.067 × 36.5 grams/litre = 2.44 grams/litre Normality of HCl (excess),

28. Alkali metals: very energetic They readily form oxides and hydroxides which are strongly alkaline. They do not occur free in nature. Li Na K Rb Cs Helen kabre se farar Group 1 elements (Alkali metals)

29. Alkaline earth metals: Oxides of Ca, Sr and Ba form alkaline hydroxides when dissolved in water and occur in the earth’s crust. Be Mg Ca Sr Ba Ra Why? IE>IE of I Bear mugs can serve bar rats Group 1 elements (Alkaline earth metals)

30. Except Boron, all others are metals. B Al Ga In Tl Bob allen gone indrains jennis lessons Al is longer than Ga Group 13 elements (Boron family)

31. ns 2 np 2 Carbon is a typical non-metal. Si and Ge are metalloids. Sn and Pb are metals. C Si Ge Sn Pb Can sily or Gervans snatch lead Group 14 elements (Carbon family)

32. N and P are non-metals. As and Sb are metalloids. Bi is a true metal. N P As Sb Bi Never put arsence in silver bullet bear Group 15 elements (Nitrogen family)

33. ns 2 np 4 First four elements are called chalcogen meaning ore forming. O S Se Te Po Oh, she sells tie moles Group-16 elements (Oxygen family)

34. ns 2 np 5 2. Diatomic molecule in the elemental form. 1. Astatine is radioactive with very short half-life period. F Cl Br I At Fat Clyde bribed Innocents Sea salt producer Group-17 elements (Halogen family)

35. Thank you