Class IX Science
IN-TEXT QUESTIONS SOLVED NCERT TEXTBOOK PAGE 32
1. In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass carbonate.
Ans.
LHS RHS 11.3 g = 11.3 g (Mass of reactant) (Mass of product) This shows, that during a chemical reaction mass of reactact = mass of product.
2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans. Ratio of H : O by mass in water is: Hydrogen : Oxygen → H2O 1 : 8 = 3 : x x = 8 × 3 ∴ 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.
3. Which postulate of Dalton's atomic theory the result of the law of conservation of mass?
Ans. The postulate of Dalton's atomic theory that is the result of the law of conservation of mass is-the relative number and kinds of atoms are constant in a given compound. Atoms cannot be created nor destroyed in a chemical reaction.
4. Which postulate of Dalton's atomic theory can explain the law of definite proportions?
Ans. The relative number and kinds of atoms are constant in a given compound.
1. Define the atomic mass unit.
Ans. One atomic mass unit is equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon-12.
2. Why is it not possible to see an atom with naked eyes?
Ans. Atom is too small to be seen with naked eyes. It is measured in nanometres. 1 m = 109 nm
1. Write down the formulae of (i) Sodium oxide (ii) Aluminium chloride (iii) Sodium sulphide (iv) Magnesium hydroxide Ans. The formulae. are (i) Formula bf Sodium Oxide Symbol → Na O Charge → +1 –2 Formula → Na2O (iii) Formula of Sodium Oxide Symbol → Na S Charge → + 1 –2 Formula → Na2S (ii) Formula of aluminium chloride Symbol → Al Cl Charge → +3 –1 Formula → A1C13 (iv) Formula of magnesium hydroxide Symbol → Mg OH Charge 1 Formula → Mg(OH)2
2. Write down the names of compounds represented btthe following formulae: (i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3 Ans. (i) A12(SO4)3 → Aluminium sulphate (ii) CaC12 → Calcium chloride (iii) K∴SO4 → Potassium sulphate (iv) KNO3 → Potassium nitrate (v) CaCO3 → Calcium carbonate
3. What is meant by the term chemical formula? Ans. The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.
4. How many atoms are present in a (i) H2S molecule and (ii) PO43– ion? Ans. (i) H2S → 3 atoms are present (ii) PO43– → 5 atoms are present
1. Calculate the molecular masses of H2, O2, C12, CO2, CH4, C2H6, C2H4, NH3, CH3OH. Ans. The molecular masses are: H2 ⇒ 1 × 2 → 2u O2⇒ 16 × 2 → 32u Cl2 ⇒ 35.5 × 2 → 71 u CO2 ⇒ 1 × 12 + 2 × 16 = 12 +32 = 44 u CH4 ⇒ 1 × 12 + 4 × 1 = 16 u C2H6 ⇒�n2 × 12 + 6 × 1 = 30 u C2H4 ⇒ (2 × 12) + (4 × 1) = 28 u NH3 ⇒ (1 × 14) + (3 × 1) = 17 u CH3OH ⇒ 12 + (3 × 1) + 16 + 1 = 32 u
2. Calculate the formula unit masses of ZnO, Na2O, K2Co3 given atomic masses of Zn=65 u, Na = 23 u, K = 39 u, C = 12u, and O = 16 u. Ans. The formula unit mass of (i) ZnO = 65 u + 16 u =.81 u (ii) Na2O = (23 u × 2) + .16 u : = 46u+ 161.u. = 62 u (iii) K2CO3 = (39 u × 2) + 12 u + 16 u × 3 = 75 u + 12u = 48u = 138u
1. If one mole of carbon atoms weigh 12 grams, what is the mast (in grams) of 1 atom of carbon? Ans. 1 mole of carbon atoms 6.022 × 1023 atoms = 12 g Mass of 1 atom = ? ∴ Mass of 1 atom of carbon  =1.99 × 10–23 g
2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u)? Ans. 23 g of Na = 6.022 × 1023 atoms (1 mole). 100 g of Na = ?
= 26.182 ×1023 = 2.6182×1024 atoms 56 g of Fe = 6.022 × 1023 atoms 100 g o Fe = ? 100 g of Na contain → 2.618 × 1024 atoms 100 g of Fe contain → 1.075 × 1024 atoms ∴ 100 g of Na contains more atoms.
QUESTIONS FROM NCERT TEXTBOOK
1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight. Ans. Boron and oxygen compound → Boron + Oxygen 0.24 g → 0:096g + 0.144g Percentage composition of the compound For boron: 0.24g → 0.096 g 100 g → ?
For oxygen: 0.24 g → 0.144 g of oxygen 100 g → ?
2. 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 formed when. 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemicai combination will govern your answer? Ans. The reaction of burning of carbon in oxygen may be written as:
It shows that 12 g of carbon burns in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO2. Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 �V 8 = 42 g of 02. The answer governs the law of constant proportion.
3. What are polyatomic ions? Give eaamples: Ans. The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g., OH–, SO2–4, CO32–.
4. Write the chemical formulae of the following: (a) Magnesium chloride (b) Caldum oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate. Ans. (a) Magnesium chloride Symbol → Mg Cl Change → +2 –1 Formula → MgCl2 (b) Calcium oxide Symbol → Ca O Charge → +2 –2 Formula → CaO (c) Copper nitrate Symbol → Cu NO3 Change → +2 –1 Formula → Cu(NO3)2 (d) Aluminium chloride Symbol → Al Cl Change → +3 –1 Formula → A1Cl3 (d) Calcium carbonate Symbol → Ca CO3 Change → +2 –2 Formula → CaCO3
5. Give the names of the elements present in the following compounds: (a) Quick time (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate. Ans. (a) Quick lime → Calcium oxide Elements → Calcium and oxygen (b) Hydrogen bromide Elements → Hydrogen and bromine (c) Baking powder → Calcium hydrogen carbonate Elements → Calcium, hydrogen, carbon and oxygen (d) Potassium sulphate Elements → Potassium, sulphur and oxygen
6. Calculate the molar mass of the following substances. (a) Ethyne, C2H2 (b) Sulphur molecule, S8 (c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31) (d) Hydrochloric acid, HCl (e) Nitric acid, HNO3 Ans. The molar mass of the following: [Unit is ‘g’] (a) Ethyne, C2H2 = 2 × 12 + 2 × 1 = 24 + 2 = 26 g (b) Sulphur molecule, S8 = 8 × 32 = 256 g (c) Phosphorus molecule, P4 = 4 × 31 = 124 g (d) Hydrochloric acid, HCl = 1 × 1 + 1 × 35.5 = 1 + 35.5 = 36.5 g (e) Nitric acid, HNO3 = 1 × 1 + 1 × 14 + 3 × 16 = 1 + 14 + 48 = 63 g
7. What is the mass of (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (c) 10 moles of sodium sulphite (Na2SO3)? Ans. (a) Mass of 1 mole of nitrogen atoms = 14g (b) 4 moles of aluminium atoms Mass of 1 mole of aluminium atoms = 27 g Mass of 4 moles of aluminium atoms = 27 × 4 = 108 g (c) 10 moles of sodium sulphite (Na2SO3) Mass of 1 mole of Na2SO3 = 2 × 23 + 32 + 3 ×16 =46 + 32 + 48 = 126 g Mass of 10 moles of Na2SO3 = 126 × 10 = 1260 g
8. Convert into mole. (a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of Carbon dioxide. Ans. (a) Given mass of oxygen gas = 12 g Molar mass of oxygen gas (O2) = 32 g Mole of oxygen gas 12 (b) Given mass of water = 20 g Molar mass of water (H2O) = (2 × 1) + 16 = 18 g Mole of water (c) Given mass of Carbon dioxide = 22 g Molar mass of carbon dioxide (CO2) = (1 × 12) + (2 × 16) = 12 + 32 = .44 g Mole of carbon dioxide
9. What is the mass of (a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules? Ans. (a) Mole of Oxygen atoms = 0.2 mole Molar mass of oxygen atoms = 16 g Mass of oxygen atoms = 16 × 0.2 = 3.2 g (b) Mole of water molecule = 0.5 mole Molar mass of water molecules = 2 × 1 + 16 = 18 g Mass of H2O = 18 × 0.5 = 9 g
10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur. Ans. Molar mass of S8 sulphur = 256 g = 6.022 × 1023 molecule Given mass of sulphur = 16 g
= 0.376 × 1023 = 3.76 × 1022 molecules
11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. Ans. Molar mass of aluminium oxide Al2O3 = (2 × 27) + (3 × 16) = 54 + 48 = 102g. 102 g of Al2O3 contains = 2 × 6.022 × 1023 aluminium ions
= 0.006022 × 1023 = 6.022 × 1020 Al3+ ions |
What weight of carbon should be used to react completely with 40 g of oxygen consider carbon and oxygen reacts in 3/8 ratio by mass to form CO2?
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