Std 9 Science and Technology · Chapter 4

4. Measurement of Matter

Complete solutions to the Exercises. Maharashtra State Board (Balbharati), Standard Nine Science and Technology.

Exercises

1Give examples.

(a) Positive radicals: Na⁺, K⁺, Ag⁺, Ca²⁺, Mg²⁺, Al³⁺, NH₄⁺

(b) Basic radicals: Na⁺, K⁺, Cu²⁺, Zn²⁺, Fe³⁺, NH₄⁺

(d) Metals with variable valency: copper (1 and 2), mercury (1 and 2), iron (2 and 3), tin (2 and 4)

(e) Bivalent acidic radicals: O²⁻, S²⁻, SO₄²⁻, SO₃²⁻, CO₃²⁻, CrO₄²⁻, Cr₂O₇²⁻

(f) Trivalent basic radicals: Al³⁺, Fe³⁺ (ferric), Cr³⁺, Au³⁺

Positive radicals and basic radicals are the same thing — cations. They are called basic radicals because they combine with the hydroxide radical to form bases, such as NaOH and KOH.

2Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals. (Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen)

Mercury — HgHg⁺ (mercurous), charge +1
Hg²⁺ (mercuric), charge +2

Potassium — KK⁺ (potassium), charge +1

Nitrogen — NN³⁻ (nitride), charge −3
Composite: NO₃⁻ nitrate, NO₂⁻ nitrite, NH₄⁺ ammonium

Copper — CuCu⁺ (cuprous), charge +1
Cu²⁺ (cupric), charge +2

Sulphur — SS²⁻ (sulphide), charge −2
Composite: SO₄²⁻ sulphate, SO₃²⁻ sulphite

Carbon — CComposite: CO₃²⁻ carbonate, charge −2; HCO₃⁻ bicarbonate, charge −1

Chlorine — ClCl⁻ (chloride), charge −1
Composite: ClO₃⁻ chlorate

Oxygen — OO²⁻ (oxide), charge −2
Composite: OH⁻ hydroxide

Metals (mercury, potassium, copper) give basic radicals by losing electrons. Non-metals (nitrogen, sulphur, chlorine, oxygen) give acidic radicals by gaining electrons. Carbon forms no simple ion in the chart; it appears in composite radicals.

3Write the steps in deducing the chemical formulae of the following compounds. (Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide)

Step 1: Write the symbols of the radicals — the basic radical on the left, the acidic radical on the right.

Step 2: Write the valency of each radical below its symbol.

Step 3: Cross-multiply: the valency of each radical becomes the number (subscript) of the other radical.

Step 4: Write the formula. Drop the subscript 1, divide both numbers by any common factor, and put a composite radical in brackets when it is taken more than once.

Sodium sulphate

Potassium nitrate

Ferric phosphate: Fe₃(PO₄)₃ simplifies to FePO₄

Calcium oxide: Ca₂O₂ simplifies to CaO

Aluminium hydroxide: OH is a composite radical, so it goes in brackets

Formulae: sodium sulphate Na₂SO₄, potassium nitrate KNO₃, ferric phosphate FePO₄, calcium oxide CaO, aluminium hydroxide Al(OH)₃.

4Write answers to the following questions and explain your answers.

(a) Explain the monovalency of the element sodium.

The atomic number of sodium is 11, and its electronic configuration is 2, 8, 1. It has one electron in its outermost orbit.
While forming an ionic bond, a sodium atom gives away this one electron and forms the sodium ion Na⁺ (2, 8), which has a complete octet.
The valency of an element is the number of electrons its atom gives away or takes up while forming an ionic bond.
Sodium gives away only one electron, so its valency is 1 — sodium is monovalent. For example, one Na⁺ combines with one Cl⁻ to form NaCl.

(b) M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals: sulphate and phosphate.

M is bivalent, so its basic radical is M²⁺ (valency 2). Sulphate is SO₄²⁻ (valency 2) and phosphate is PO₄³⁻ (valency 3).

Write the symbols: M on the left, SO₄ (or PO₄) on the right.
Write the valencies below them: M → 2; SO₄ → 2; PO₄ → 3.
Cross-multiply the valencies to get the number of each radical.
Write the formula, simplifying by the common factor and bracketing the composite radical where needed.

With sulphate: M₂(SO₄)₂ simplifies to MSO₄

With phosphate: M₃(PO₄)₂

Formulae: sulphate of M MSO₄; phosphate of M M₃(PO₄)₂.

Need: atoms are extremely tiny, and scientists of the 19th century could not measure the mass of a single atom directly. So they decided to compare the mass of each atom with the mass of one chosen atom, called the reference atom. The mass of an atom expressed in this way is its relative atomic mass.

Hydrogen: hydrogen, the lightest atom, was chosen first. Its relative mass was taken as 1. On this scale a nitrogen atom, 14 times heavier, has relative mass 14. But many elements came out with fractional values, so other reference atoms were tried later.
Carbon: in 1961 the carbon atom (carbon-12) was accepted as the reference atom, with its relative mass taken as 12. On this scale, 1/12 of the mass of a carbon atom is the unit, and the relative mass of hydrogen is 12 × 1/12 = 1. This scale is used even today.

(d) What is meant by Unified Atomic Mass?

Today the mass of an atom can be measured directly by accurate methods, so instead of relative mass a fixed unit of atomic mass has been accepted. This unit is called the unified atomic mass unit, or Dalton, and its symbol is u. It is equal to 1/12 of the mass of one carbon-12 atom.

1 u = 1.66053904 × 10⁻²⁷ kg

Atomic and molecular masses are expressed in this unit — for example, H = 1 u, O = 16 u, H₂O = 18 u.

(e) Explain with examples what is meant by a ‘mole’ of a substance.

A mole is that quantity of a substance whose mass in grams is equal in magnitude to its molecular mass in Daltons. Its SI unit is mol. One mole of any substance contains the same number of molecules, 6.022 × 10²³, called Avogadro’s number (N_A) — just as a dozen always means 12 items.

Molecular mass of water (H₂O) = 18 u, so 1 mole of water = 18 g.
Molecular mass of oxygen (O₂) = 32 u, so 1 mole of oxygen = 32 g.
Molecular mass of carbon dioxide (CO₂) = 44 u, so 1 mole of CO₂ = 44 g.
Molecular mass of sodium chloride (NaCl) = 58.5 u, so 1 mole of NaCl = 58.5 g.

Each of these amounts — 18 g water, 32 g oxygen, 44 g CO₂, 58.5 g NaCl — contains 6.022 × 10²³ molecules.

Number of moles (n) = Mass of substance in grams ÷ Molecular mass of substance

5Write the names of the following compounds and deduce their molecular masses. (Na₂SO₄, K₂CO₃, CO₂, MgCl₂, NaOH, AlPO₄, NaHCO₃)

Atomic masses used: H = 1, C = 12, O = 16, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, K = 39.

Na₂SO₄ — Sodium sulphate
(23 × 2) + (32 × 1) + (16 × 4) = 46 + 32 + 64 = 142 u

K₂CO₃ — Potassium carbonate
(39 × 2) + (12 × 1) + (16 × 3) = 78 + 12 + 48 = 138 u

CO₂ — Carbon dioxide
(12 × 1) + (16 × 2) = 12 + 32 = 44 u

MgCl₂ — Magnesium chloride
(24 × 1) + (35.5 × 2) = 24 + 71 = 95 u

NaOH — Sodium hydroxide
23 + 16 + 1 = 40 u

AlPO₄ — Aluminium phosphate
(27 × 1) + (31 × 1) + (16 × 4) = 27 + 31 + 64 = 122 u

NaHCO₃ — Sodium bicarbonate (sodium hydrogen carbonate)
23 + 1 + 12 + (16 × 3) = 23 + 1 + 12 + 48 = 84 u

The atomic mass of phosphorus (31) is not given in the textbook’s list on page 51; it is one of the values left blank for you to find in the Research chart on page 49.

6Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows: sample m — mass 7 g, oxygen 2 g, calcium 5 g; sample n — mass 1.4 g, oxygen 0.4 g, calcium 1.0 g. Which law of chemical combination does this prove? Explain.

Sample m: Ca : O = 5 g : 2 g = 5 : 2
Sample n: Ca : O = 1.0 g : 0.4 g = 10 : 4 = 5 : 2

This proves the law of constant proportion.

Explanation: the two samples were obtained from two different reactions and have different total masses (7 g and 1.4 g). Yet in both, calcium and oxygen are combined in the same proportion by mass, 5 : 2. The law of constant proportion, stated by Proust in 1794, says that the proportion by mass of the constituent elements in different samples of a compound is fixed, whatever its source or method of preparation. The data shows exactly this.

The same ratio follows from the formula CaO: Ca : O = 40 : 16 = 5 : 2. So the experimental proportion matches the expected one.

The question calls these samples “slaked lime”, but they contain only calcium and oxygen (5 g + 2 g = 7 g), in the ratio of calcium oxide, CaO. That is quicklime (lime). Slaked lime, Ca(OH)₂, also contains hydrogen. The law proved is the same either way.

7Deduce the number of molecules of the following compounds in the given quantities. (32 g oxygen, 90 g water, 8.8 g carbon dioxide, 7.1 g chlorine)

Number of moles n = mass in grams ÷ molecular mass; number of molecules = n × 6.022 × 10²³.

32 g oxygen (O₂) — molecular mass = 16 × 2 = 32
n = 32 ÷ 32 = 1 mol
Molecules = 1 × 6.022 × 10²³ = 6.022 × 10²³

90 g water (H₂O) — molecular mass = (1 × 2) + 16 = 18
n = 90 ÷ 18 = 5 mol
Molecules = 5 × 6.022 × 10²³ = 30.11 × 10²³ = 3.011 × 10²⁴

8.8 g carbon dioxide (CO₂) — molecular mass = 12 + (16 × 2) = 44
n = 8.8 ÷ 44 = 0.2 mol
Molecules = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³

7.1 g chlorine (Cl₂) — molecular mass = 35.5 × 2 = 71
n = 7.1 ÷ 71 = 0.1 mol
Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²²

Oxygen and chlorine gases exist as diatomic molecules, O₂ and Cl₂, so their molecular masses are 32 and 71, not 16 and 35.5.

8If 0.2 mol of the following substances are required, how many grams of those substances should be taken? (Sodium chloride, magnesium oxide, calcium carbonate)

Mass in grams = number of moles × molecular mass.

Sodium chloride (NaCl) — molecular mass = 23 + 35.5 = 58.5
Mass = 0.2 × 58.5 = 11.7 g

Magnesium oxide (MgO) — molecular mass = 24 + 16 = 40
Mass = 0.2 × 40 = 8 g

Calcium carbonate (CaCO₃) — molecular mass = 40 + 12 + (16 × 3) = 100
Mass = 0.2 × 100 = 20 g

Answer: 11.7 g sodium chloride, 8 g magnesium oxide and 20 g calcium carbonate.

PProject: Prepare models of various radicals using cardboard, small circular magnets and araldite. From these make models of molecules of various compounds.

Some ideas for making the models:

Cut a cardboard disc for each radical and write its symbol and charge on it, for example Na⁺, Ca²⁺, Al³⁺, Cl⁻, SO₄²⁻, PO₄³⁻.
Stick as many small magnets on each disc as its valency, using araldite — one for Na⁺, two for Ca²⁺, three for Al³⁺. Use one pole facing out on basic radicals and the other pole on acidic radicals, so that only opposite radicals join.
Join radicals until every magnet is used, and read off the formula: one Na⁺ with one Cl⁻ gives NaCl; one Ca²⁺ with two Cl⁻ gives CaCl₂; two Al³⁺ with three SO₄²⁻ gives Al₂(SO₄)₃.
Use a different colour of cardboard for simple radicals and composite radicals.

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