1. (a) Calculate the number of moles and number of grams of Na₂SO₄ which contain 1 ×10²³ ion.
    (b)  Calculate the number of formula units of Na₂SO₄ which contains 0.2 moles of oxygen atoms.

(a)  1 mol Na₂SO₄ containes 1 mol of ions.  6.023 × 10²³ × 2.78 ×10^-3 ions are contained in 1 mol of Na₂SO₄

10²³  ions are present in mol Na₂SO4

= 1.66 × 10^-1 mol Na₂SO₄

1 mol of Na₂SO₄ = 142 g

1.66 × 10^-1 mol of Na₂SO₄ = 142 × 1.66 × 10^-1

= 23.57 g Na₂SO₄ .

(b) 4 mols of O in Na₂SO₄  = 1 formula unit

0.2 mol of O in Na₂SO₄  =

= 0.05 formula units.

2. A sample of a pure compound contains 2.04 g of sodium, 2.65 ×10²² atoms of carbon, and 0.132 mol of oxygen atoms.  Find the empirical formula.

Moles of Na atom =

Moles of C atom =

Moles of O atoms = 0.132

The mole ratio of Na, C, O is

  0.087 : 0.044 : 0.132

Dividing by the least common factor (0.044).

or 2 : 1 : 3.

Hence, the empirical formula is Na₂CO₃.

3. Welding fuel gas contains carbon and hydrogen only.  Burning a small sample of it in oxygen given 3.38 g carbon dioxide, 0.690 g of water and no other products.  A volume of 100 L (measured at STP) of this welding gas is found to weigh 11.69 g.  Calculate (i) empirical formula (ii) molar mass of the gas and (iii) molecular formula.

Number of moles of CO₂ =

number of moles of C = 0.0768 mol

number of moles of H₂O =

number of moles of H = 2 × 0.0383 = 0.0766 mol

(i)  The ratio of moles of C to H is 0.0768 : 0.0766

or 1 : 1

Therefore, the empirical formula is CH

(ii)  10.0 L of fuel gas weighs 11.6 g

22.4 l of fuel gas at STP weighs

Molar mass of gas = 26 g mol^-1

Molecular formula = (EF)_n = (CH)₂ = C₂ H₂

4.  What are the essentials of a chemical equation ?

Essentials of a chemical equation are :

• It should represent a true chemical change that is. if a reaction is not possible between certain substance, if cannot be represented by a chemical equation.
• It should be balanced.
• It should be molecular that is all the species should be represented in their molecular form.  For example, elementary gases like hydrogen, nitrogen,and the like, should be represented as H₂ and N₂.

5.  What is the information conveyed by chemical equations ?

A chemical equation gives the following information :

• The substances which take part in the reaction and those which are produced.
• The relative number of molecules of the reactants and the products.
• The relative masses of the reactants and the products.
• The relative volumes of the gaseous reactant and products, if any.

6.  What mass of copper (II) sulphate can be obtained by the action of 2.941 g of hot concentrated Sulphuric acid on an excess of copper ?

The balanced equation for the reaction is

Cu + 2H₂SO₄ →CuSO₄ + 2H₂O + SO₂

Now 2.941 g H₂SO₄ = 

From the balanced equation

2 mol H₂SO₄ gives 1 mol CuSO₄

0.030 mol H₂SO₄ will give = 0.015 mol CuSO₄

1 mol  CuSO₄ = 159.5 g

0.015 mol CuSO₄ = 159.5 × 0.015 = 2.39 g CuSO₄

7. What mass of zinc is required to produce hydrogen by reaction with HCI which is enough to produce 4 mol of ammonia according to the reactions ?
  Zn + 2HCI → ZnCI₂ + H₂
  3H₂ + N₂ → 2NH₃

From the equation, it is clear that, 2 mol of NH₃ require 3 mol of H₂ and 3 mol of H₂ require 3 mol of Zn

2 mol NH₃ require 3 mol of Zn or 3 × 65g Zn

4 mol NH₃ require 3 ×65 × 4/2 = 390 g Zn

8. A mixture containing 100g of H₂ and 100 g O₂ is ignited so that water is formed according to the reaction.
           2 H₂ + O₂ → 2 H₂O
How much water is formed ?  Also, calculate the volume of the gas left unreacted at STP.

2 H₂ + O₂ → 2 H₂O

100g H₂ = 50 mol H₂

100 g O₂ = = 3.125 mol O₂

2 mol H₂ + 1 mol O₂ → 2 mol H₂O

2 × 3.125 mol H + 3.125 mol O₂ →2 × 3.125 mol H₂O

More H₂ is present than required.  Therefore, O₂ is the limiting reactant.

Amount of H₂O formed

                              = 2 × 3.125 mol H₂O

                              = 2 × 3.125 × 18

                             = 112.5 g H₂O

Number of moles of H₂ left unreacted

  = (50 - 2 × 3.125) = 43.75 mol H₂ .

Volume occupied by 43.75 mol H₂ at STP = 980 L H₂ at STP.

9.  A sample of sodium hydroxide weighing 0.38 g is dissolved in water and the solution is made to 50.0 cm³ in a volumetric flask.  What is the molarity of the resulting solution ?

Mass in g of NaOH = 0.38 g

Number of moles of NaOH =

Volume of solution = 50 cm³ = 0.05 L

Molarity =

              = 0.190 mol L^-1 = 0.190 M

10.  Calculate the molarity of solution obtained by dissolving 0.212 g of Na₂ CO₃ in 250 cm³ of solution.

Molar Mass of Na₂CO₃ =106 g mol^-1

Number of moles of Na₂CO₃ =

Volume of solution = 250 cm³ = 1/4L

Molarity = 8 × 10^-3 =0.008 M

11. Commercially available HBr solution contains 48% HBr by mass.  What is the molarity of this solution ?  The density of solution is 1.50 g cm^-3 .

Let the volume of solution 1 L = 1000 cm³

Density of solution = 1.50 g cm^-3

Mass of solution = 1.50 g cm^-3 × 1000 cm³

                              = 1500 g

Mass of HBr = 48% of 1500 g

Number of moles of HBr = = 8.89 moles HBr

Molarity = = 8.89 M.

12.  The density of the nitric acid solution which 63% by mass HNO₃ is 1.41 g cm^-3.  What is the molarity of the solution ?

Let the volume of solution = 1000 cm³ = 1 L

Mass of solution = 1.41 × 1000=1414 g

But mass of HNO₃ = 63% of 1410 = 63 × 14.10 g HNO₃

Number of moles of  moles HNO₃

Molarity =

13. 100 ml of 0.1 M NaCI solution is mixed with 100 ml of M AgNO₃ solution.  Find out the mass of AgCI precipitate formed.  Which reagent is acting as limiting reagent ?

AgNO₃ + NaCI →NaNO₃ + AgCI

nAgNO₃ = M × V = 0.2 mol L^-1 × 0.1 L = 0.02 mol

n NaCI = M × V = 0.01 mol L^-1 × 0.1 L = 0.01 mol

More AgNO₃ is present than required.

Hence, NaCI is the limiting reagent.

Number of moles of AgCI obtained = 0.01

Mass of AgCI formed = 0.01 143.5 = 1.435 g AgCI

14. An impure sample of sodium chloride which weighs 0.50 g gave, on treatment with excess of silver nitrate solution, 0.90 g of silver chloride as a white precipitate.  Calculate the percentage purity of the sample.

AgNO₃ + NaCI → AgCI + NaNO₃

                                1 mol    1 mol

143.5 g or 1 mol of AgCI will precipitate from 58.5 g NaCI

0.90 g AgCI will require 58.5 ×

Percentage purity of NaCI =

15.  Chlorophyll, the green colouring matter of plants responsible for photosynthesis, contain 2.68% of magnesium by mass.  Calculate the number of magnesium atoms in 2.00 g chlorophyll.

Chlorophyll contains 2.68% by mass magnesium.

It means 100 g chlorophyll contains 2.68 g Mg

2.00 g chlorophyll contains

Number of moles of Mg =

Number of Mg atoms = 2.23 × 10^-3 ×6.022×10²³ = 1.34 × 10²¹ .