Class 11 Some basic concept of chemistry by Tarang part 01.pptx

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1. Class XI Chemistry Some Basic Concept of Chemistry Lecture by Dr. Tarang Tomar PGT Chemistry AVV Khargone (M.P.) Part I

2. Matter : Substances which has both mass and volume and also where particles occupy space. It has define volume except for gases though they still contain molecules.  Non-  Matter : Substances which has neither mass nor volume and thus having no spaces (obviously) or we can say that doesn't have any appearance but exists .

3. System of unit qualitative properties Qualitative properties -----  Difficult to measure -------- eg . Malleability, ductility, lustre , odour , etc. Quantitative properties -----  Easy to measure -------- eg . Length, volume, height etc . It requires a no. and units Fundamentalunits

4. Prefixes used in SI system Derived units

5. The uncertainty of a measured value is an interval around that value, so each repetition of the measurement produces a new result which falls within that interval. Uncertainty Accuracy Precision The accuracy indicates how close the measurement is to the actual measurement The precision determined how close a single measurement is to others.

6. Q. The volume of a liquid is 26 mL. A student measures the volume and finds it to be 26.2 mL, 26.1 mL, 25.9 mL, and 26.3 mL in the first, second, third, and fourth trial, respectively. Which of the following statements is true for his measurements ? a. They are neither precise nor accurate. b. They have poor accuracy. c. They have good precision. d. They have poor precision . Q. The volume of a liquid is 20.5 mL. Which of the following sets of measurement represents the value with good accuracy ? a.18.6 mL, 17.8 mL, 19.6 mL, 17.2 mL b.19.2 mL, 19.3 mL, 18.8 mL, 18.6 mL c.18.9 mL, 19.0 mL, 19.2 mL, 18.8 mL d.20.2 mL, 20.5 mL, 20.3 mL, 20.1 mL

7. Scientific notation Scientific notation is a way to express numbers as the product of two numbers: a coefficient and the number 10 raised to a power. It is a very useful tool for working with numbers that are either very large or very small. As an example, the distance from Earth to the Sun is about 150,000,000,000 meters—a very large distance indeed. In scientific notation, the distance is written as  1.5×10 11 m. To write in scientific notation, follow the general form N x 10 m where N is a number between 1 and 10, but not 10 itself, and m is any integer (positive or negative number).

8. Q. Express the following in the scientific notation: 0.0048 234,000 8008 500.0 6.0012 3.2 602,000,000 0.0000428 35700

9. All nonzero digits are significant . 211.8 = 4 significant figures. All zeros that are found between nonzero digits are significant . 20,007 = 5 significant figures. Leading zeros (to the left of the first nonzero digit) are not significant . 0.0085 = 2 significant figures Trailing zeros for a whole number that ends with a decimal point are significant . 320 = 2 significant figures Trailing zeros to the right of the decimal place are significant . 12.000 = 5 significant figures Exact numbers, and irrationally defined numbers like  Euler’s number  ( e ) and  pi  (π), have an  infinite  number of significant figures Rules for significant figures

10. Q. Find significant no. in the following:- 0.020 2.341 2.10 x 10 6 0.0123 0.004200 26001 3100 120 130 x 10 4 43.00 0.00648 7.0005 208 5.00 500.0 2.0034 312.84 6.45 x 10 6 6.00009 985 80000

11. Q. Convert scientific notation to decimal values . 1.92 x 10 3 3.05 x 10 1 -4.29 x 10 2 6.251 x 10 9 8.317 x 10 6 1.03 x 10 -2 8.862 x 10 -1 9.512 x 10 -8 -6.5 x 10 -3 3.519 x 10 2

12. Rounding off Q. Round off the following upto two significant figures:- 2.43 2.26 2.255 2.358 2.252 2.253 2.25 2.250 2.35 2.350 Q. Round off the following upto three significant figures :- 0.04234 2.345 7.123 3.1352 23510

13. Multiplication and division of significant figures:- Addition and subtraction of significant figures:-

14. Metric system unit conversion

15. Conversion of Day to Hours, Week to Days, Year to Days Conversion of Units of Time

16. Q. What mass of silver nitrate will react with 5.85 gm of Sodium Chloride to produce 14.35 gm of Silver Chloride and 8.5 gm of Sodium nitrate is formed if the law of conservation of mass is followed ?

17. Q. 1.5 g of ethane (C2H6) on complete combustion gave 4.4 g of CO2 and 2.7 g of H2O. Show that the results are in accordance to the law of conservation of mass.

18. Q. If  6.3g of NaHCO3 are added to 15.0g of CH3COOH solution, the residue is found to weigh 18.0g. What is the mass of CO2 released in the reaction ? Ans = 3.3 g

19. Q. In an experiment, 2.4 g of iron oxide on reduction with hydrogen yields 1.68 g of iron. In another experiment, 2.9 g of iron oxide gives 2.03 g of iron on reduction with hydrogen. Prove that it illustrates the law of definite proportion.

20. Law of Multiple P roportion

21. Let us study the following example Here Nitrogen combines with oxygen to form…………. Let us study the following example Here Hydrogen combines with oxygen to form………….

22. Q. Hydrogen and oxygen combine to form two compounds. The hydrogen content in one of these is 5.93 % and in other it is 11.2 %. Show that this explain law of multiple proportion.

23. Q. On analysis it was found that the black oxide of copper and red oxide of copper contain 80% and 89% of copper respectively. Show that this data is in accordance with law of multiple proportion.

24. Q. Sulphur and Oxygen are known to form two compounds. The sulphur content in one of these is 51% while in the other is 41 %. S how that this data is in agreement with the law of multiple proportions.

25. Q. A box contains some identical red coloured balls, labelled as 𝐴, each weighting 2 grams. Another box contains identical blue coloured balls, labelled as 𝐵, each weighing 5 𝑔𝑚. Consider the combinations 𝐴𝐵,𝐴𝐵2,𝐴2𝐵 and 𝐴2𝐵3, show that law of multiple proportions is applicable.

26. This law was given by Gay Lussac in 1808. this law states that “ When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.” Note:- Volume of solids and liquids are considered to be zero. For example:- Consider 1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volume of ammonia gas. 1 N2(g) + 3 H2(g) -------------  2NH3 (g) 1 mol 3 mol 1 volume 3 volume Therefore, ratio of volumes of N2 and H2 which combine together bears a simple ratio of 1 : 3 Consider 1 volume of oxygen combines with 2 volumes of hydrogen to form 2 volume of water vapour (steam). 1 H2(g ) + 2 O2(g ) -------------  2H2O (g) 1 mol 3 mol 1 volume 3 volume Therefore, ratio of volumes of O2 and H2 which combine together bears a simple ratio of 1 : 2 Gay-Lussac’s law of gaseous volumes:-

27. Q. Nitric oxide (NO) used as an energy booster. NO is produced by a reaction involving oxidation of ammonia. If 100 litres of ammonia is used what will be amount of NO produced?

28. Q. How much volume of oxygen will be required for complete combustion of 40 ml of acetylene (C2H2) and how much CO2 volume will be formed? All volumes are measured at NTP.

29. LAW OF RECIPROCAL PROPORTIONS The   law of reciprocal proportions  was proposed by German chemist Ritcher in 1792. It states that “When two elements combine separately with a fixed mass of a third element, then the ratio of their masses in which they do so bears a small whole number ratio to the ratio of masses in which they themselves combines”.

30. Example – 01: CO 2  contains 27.27% of carbon, CS 2  contains 15.79% of carbon and SO2 contains 50% of sulphur . Show that the data illustrates the law of reciprocal proportions. Solution: Consider CO 2 % of carbon = 27.27 % of oxygen = 100 – 27.27 = 72.73 27.27 g of carbon combines with 72.73 g of oxygen. Hence, 1 g of carbon combines with 72.73 / 27.27= 2.67 g of oxygen. Consider CS 2 % of carbon = 15.79 % of sulphur = 100 – 15.79 = 84.21 15.79 g of carbon combines with 84.21 g of sulphur . Hence, 1 g of carbon combines with 84.21 / 15.79= 5.33 g of sulphur . The ratio of different masses of sulphur and oxygen combining with fixed mass of carbon is 5.33 : 2.67 i.e. 2 : 1. ………………….. (Ratio 1) Consider SO 2 % of sulphur = 50 % of oxygen = 100 – 50 = 50 50 g of sulphur combines with 50 g of oxygen. The ratio of mass of sulphur to that of oxygen is 50 : 50  i.e. 1 : 1 ……………… (Ratio 2) The second ratio is a simple whole-number is multiple of the first ratio. Hence the data illustrate the law of reciprocal proportions.

31. Example 2 :- CuS contains 33.3 % of sulphur , CuO contains 20.1% of oxygen and SO3 contains 40 % of sulphur . Show that the data illustrates the law of reciprocal proportions. Solution : Consider CuS % of sulphur = 33.3 % of sulphur = 100 – 33.3 = 66.7 66.7 g of copper combines with 33.3 g of sulphur . Hence  1 g of copper combines with 33.3 / 66.7= 0.5 g of sulphur . Consider CuO % of oxygen = 20.1 % of copper = 100 – 20.1 = 79.9 79.9 g of copper combines with 20.1 g of oxygen. Hence 1 g of carbon combines with 20.1 / 79.9= 0.25 g of oxygen. The ratio of different masses of sulphur and oxygen combining with fixed mass of copper is 0.5 : 0.25  i.e. 2 : 1. ………………….. (Ratio 1 ) The ratio of the mass of sulphur to that of oxygen is 40 : 60  i.e. 2 : 3 ……………… (Ratio 2) The second ratio is a simple whole-number multiple of the first ratio. Hence the data illustrate the law of reciprocal proportions. Consider SO 3 % of sulphur = 40 % of oxygen = 100 – 40 = 60 40 g of sulphur combines with 60 g of oxygen.

32. 100 ml of methane gas ------------  Y molecules 100 ml of CO2 gas ------------- Y molecules 50 ml of CO2 gas ----------- Y/2 molecules ** Volume become half so molecules become half ** 200 ml of CO2 gas ------------ 2Y molecules ** Volume become doubled so molecules become doubled ** Also, 100 ml of N2 gas --------------- Y molecules ** as volume of CO2 and N2 gases are same therefore, no. of molecules will also be same.

33. Q. One mole of helium gas fills up an empty balloon to a volume of 1.5 litres . What would be the volume of the balloon if an additional 2.5 moles of helium gas is added? (Assume that the temperature and the pressure are kept constant)

34. Q. A 4.8 L sample of helium gas contains 0.22 mol of helium. How many additional moles of helium gas must be added to the sample to obtain a volume of 6.4 L? Assume constant temperature and pressure.

35. AAM = Atomic mass of isotope 1 x % + Atomic mass of isotope 2 x % + ………………….. 100 Note :- In the periodic table of elements the atomic masses mentioned for different elements actually represent their average atomic masses.

36. Q. Nitrogen occurs in nature in the form of two isotopes of atomic mass 14 and 15 respectively. If average atomic mass of Nitrogen is 14.0067. What is the% abundance of the two isotopes?

37. Q. Boron has two isotopes ,  B−10  and  B−11 , the average atomic mass of boron is found to be  10.80 u . Calculate the percentage of abundance of these isotopes.