calculate the number of moles in 100g of water (atomic mass o=16u,H=1u)
Answers
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.
Sodium carbonate + Ethanoic acid → Sodium ethanoate + Carbon dioxide + Water
5.3 g + 6 g → 8.2 g + 2.2 g + 0.9 g
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.
NCERT TEXTBOOK PAGE 35
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
NCERT TEXTBOOK PAGE 35
1. Write down the formulae of (i) Sodium oxide (ii) Aluminium chloride (iii) Sodium sulphide (iv) Magnesium hydroxideAns. 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) CaCO3Ans. (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
NCERT TEXTBOOK PAGE 40
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
NCERT TEXTBOOK PAGE 42
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 → ?