CBSE ADDA

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

Matter in our surrounding solved questions Q. 1. What is a matter? Answer: Anything that occupies space and has mass is called a matter.  For example:- Chalk, Milk, Sugar etc. Q. 2. Sodium chloride and sugar have same appearance. Are they same or different? Answer:  They have different physical and chemical properties. So, they are different Q. 3. All substances around us are alike. How? Answer:  All substances can occupy space and have weight. Q. 4. How can we say that air is a matter? Answer:  Air occupies space and have weight. Hence air is a matter. Q. 5. State the characteristics of matter? Matter has mass, weight and occupies space. Q. 6. What are the intensive properties of matter? Answer:  Density, colour, melting point, boiling point, refractive index etc. are the intensive properties of  matter as it does not depend upon the amount of matter contained in it Q. 7. What are the extensive properties of ma...

1. Q. The description of atomic particles of two elements X and Y is given below Protons neutrons electrons X 8 8 8 Y 8 9 8 (i) What is the atomic number of Y? (ii) What is the mass number of X? (iii) What is the relation between X and Y? (iv) Which element/elements do they represent? (v) Write the electronic configuration of X? (vi) Write the cation/anion formed by the element Ans: (i) Atomic number of y =  8             (ii) Mass number of x – 16           (iii) x and y are isotopes   (iv) x and y represent – oxygen               (v) 2, 6                      ...

Q. Find the gram molecular mass of water (H2O) Calculation: 2(H) = 2 x 1 = 2 . 1(O) = 1 x 16 = 16 ∴ Gram molecular mass of H2O= 18g Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12g of carbon-12 isotope. One mole is also defined as the amount of substance which contains Avogadro number (6.023 x 10^23) of particles. For eg. one mole of oxygen atoms represents 6.023 x 10^23 atoms of oxygen and 5 moles of oxygen atoms contain 5 x 6 . 0 2 3 x 10 ^23 a t oms of oxygen . Q. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. [Hint : The mass of an ion is same as that of an atom of the same element. Sol: 1 mole of Al2O3 contains aluminium ions = 2 × 6.022 × 10^23 = 12.044 × 10^23 Now, 102 g of Al2O3 has number of aluminium ions = 12.044 × 10^23 0.051 g of Al2O3 has number of aluminium ions = (12.044 ×1023 × 0.051)/102= 6.022 × 10^20 ions Q. How many atoms of oxygen are present i...

Q. Write two differences between isobars and isotopes. Ans:  Isobars           (i) Isobars have the same mass number but different atomic numbers (ii) Isobars have different  chemical properties Isotopes Isotopes have the same atomic number but different mass numbers. Isotopes have similar chemical properties. Q. Why do isotopes shows similar chemical properties. Ans: Due to same number of  electron  in valance shell Q. How Rutherford concluded that the center part of any atom not empty? Ans: When Rutherford experimented with  alpha particles , then he realized that few particles bounced back directly. Rutherford’s result lead him to believe that most of the foil was made of empty space, but had  extremely  small, dense lumps of matter inside, which is present only at the center because from center, few particles bounced back. All other particles deflecte...