Measurement of Matter – Class 9 Science Question & Answers (Maharashtra Board)

• Sodium ion ($\text{Na}^{+}$), Potassium ion ($\text{K}^{+}$), Ammonium ion ($\text{NH}_{4}^{+}$)

• Copper(II) ion ($\text{Cu}^{2 +}$), Magnesium ion ($\text{Mg}^{2 +}$), Aluminium ion ($\text{Al}^{3 +}$)

• Sulphate radical ($\text{SO}_{4}^{2 -}$), Carbonate radical ($\text{CO}_{3}^{2 -}$), Ammonium radical ($\text{NH}_{4}^{+}$)

• Iron ($\text{Fe}^{2 +}$ and $\text{Fe}^{3 +}$), Copper ($\text{Cu}^{+}$ and $\text{Cu}^{2 +}$), Mercury ($\text{Hg}^{+}$ and $\text{Hg}^{2 +}$)

• Sulphate ($\text{SO}_{4}^{2 -}$), Carbonate ($\text{CO}_{3}^{2 -}$), Sulphite ($\text{SO}_{3}^{2 -}$)

• Aluminium ($\text{Al}^{3 +}$), Iron(III) / Ferric ($\text{Fe}^{3 +}$)

(Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen)

• Mercury: Symbol: $\text{Hg}$| Radicals: Mercurous $\text{Hg}^{+}$ or Mercuric $\text{Hg}^{2 +}$

• Potassium: Symbol: $\text{K}$| Radical: Potassium ion $\text{K}^{+}$

• Nitrogen: Symbol: $\text{N}$| Radical: Nitride ion $\text{N}^{3 -}$

• Copper: Symbol: $\text{Cu}$| Radicals: Cuprous $\text{Cu}^{+}$ or Cupric $\text{Cu}^{2 +}$

• Sulphur: Symbol: $\text{S}$| Radical: Sulphide ion $\text{S}^{2 -}$

• Carbon: Symbol: $\text{C}$| Radical: Carbide ion $\text{C}^{4 -}$(Carbon also forms complex radicals like carbonate $\text{CO}_{3}^{2 -}$).

• Chlorine: Symbol: $\text{Cl}$| Radical: Chloride ion $\text{Cl}^{-}$

• Oxygen: Symbol: $\text{O}$| Radical: Oxide ion $\text{O}^{2 -}$

(Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide)

1. Write the symbols of the basic radical ($\text{Na}$) and acidic radical ($\text{SO}_{4}$).

2. Write their valencies below them: $\text{Na}$ has valency $1$, and $\text{SO}_{4}$ has valency $2$.

3. Cross-multiply the valencies: $\text{Na}_{2}\left( \text{SO}_{4} \right)_{1}$.

4. Formula: $\text{Na}_{2}\text{SO}_{4}$

1. Write symbols: $\text{K}$ and $\text{NO}_{3}$.

2. Write valencies: $\text{K}$ is $1$, $\text{NO}_{3}$ is $1$.

3. Cross-multiply: $\text{K}_{1}\left( \text{NO}_{3} \right)_{1}$.

4. Formula: $\text{KNO}_{3}$

1. Write symbols: $\text{Fe}$(ferric means valency 3) and $\text{PO}_{4}$(valency 3).

2. Simplify the ratio ($3:3$ becomes $1:1$) and cross-multiply: $\text{Fe}_{1}\left( \text{PO}_{4} \right)_{1}$.

3. Formula: $\text{FePO}_{4}$

1. Write symbols: $\text{Ca}$ and $\text{O}$.

2. Write valencies: $\text{Ca}$ is $2$, $\text{O}$ is $2$.

3. Simplify ratio ($2:2$ becomes $1:1$) and cross-multiply: $\text{Ca}_{1}\text{O}_{1}$.

4. Formula: $\text{CaO}$

1. Write symbols: $\text{Al}$ and $\text{OH}$.

2. Write valencies: $\text{Al}$ is $3$, $\text{OH}$ is $1$.

3. Cross-multiply: $\text{Al}_{1}\left( \text{OH} \right)_{3}$.

4. Formula: $\text{Al(OH)}_{3}$

• Answer Sodium has an atomic number of $11$, meaning its electron distribution is $2,8,1$. It has only $1$ electron in its outermost shell. To achieve a stable octet, it easily donates this single electron. Because it gives away $1$ electron, its combining capacity (valency) is $1$, making it monovalent.

• Answer

  • With Sulphate ($\text{SO}_{4}^{2 -}$): Metal M has valency $2$, and sulphate has valency $2$. Simplifying the ratio $2:2$ gives $1:1$. Cross-multiplying gives $\text{MSO}_{4}$.

  • With Phosphate ($\text{PO}_{4}^{3 -}$): Metal M has valency $2$, and phosphate has valency $3$. Cross-multiplying gives $\text{M}_{3}\left( \text{PO}_{4} \right)_{2}$.

• Answer Atoms are far too tiny to be weighed directly on a scale, so scientists needed a relative scale using a "reference atom" to compare how heavy different elements are.

  • First reference atom (Hydrogen): Initially chosen because it is the lightest element, with an assigned mass of $1$.

  • Second reference atom (Oxygen): Later used because oxygen readily reacts with many different elements.

  • Modern standard: Today, carbon-12 is universally used as the standard reference.

• Answer Unified atomic mass (symbol u, or dalton) is a standard unit used to express atomic and molecular masses. One unified atomic mass unit ($1\text{~u}$) is defined as exactly one-twelfth ($1/12^{\text{th}}$) of the mass of a neutral carbon-12 atom.

• Answer A mole is a standard counting unit in chemistry used to measure the amount of a substance. One mole of any substance always contains a fixed number of particles ($6.022 \times 10^{23}$ particles, known as Avogadro's number).

  • Example: One mole of water ($\text{H}_{2}\text{O}$) weighs $18\text{~g}$ and contains $6.022 \times 10^{23}$ water molecules.

• Name: Sodium sulphate

• Calculation: $(2 \times 23) + (1 \times 32) + (4 \times 16) = 46 + 32 + 64 = \mathbf{142}\text{~u}$

• Name: Potassium carbonate

• Calculation: $(2 \times 39) + (1 \times 12) + (3 \times 16) = 78 + 12 + 48 = \mathbf{138}\text{~u}$

• Name: Carbon dioxide

• Calculation: $(1 \times 12) + (2 \times 16) = 12 + 32 = \mathbf{44}\text{~u}$

• Name: Magnesium chloride

• Calculation: $(1 \times 24) + (2 \times 35.5) = 24 + 71 = \mathbf{95}\text{~u}$

• Name: Sodium hydroxide

• Calculation: $(1 \times 23) + (1 \times 16) + (1 \times 1) = \mathbf{40}\text{~u}$

• Name: Aluminium phosphate

• Calculation: $(1 \times 27) + (1 \times 31) + (4 \times 16) = 27 + 31 + 64 = \mathbf{122}\text{~u}$

• Name: Sodium bicarbonate (Sodium hydrogen carbonate)

• Calculation: $(1 \times 23) + (1 \times 1) + (1 \times 12) + (3 \times 16) = 23 + 1 + 12 + 48 = \mathbf{84}\text{~u}$

• Answer

  • Law Proved: This proves the Law of Constant Proportions.

  • Simple Explanation & Calculation:

    • In Sample m ($7\text{~g}$ total): Mass of calcium = $5\text{~g}$, Mass of oxygen = $2\text{~g}$. The ratio of calcium to oxygen is $5:2 = 2.5$.

    • In Sample n ($1.4\text{~g}$ total): Mass of calcium = $1.0\text{~g}$, Mass of oxygen = $0.4\text{~g}$. The ratio of calcium to oxygen is $1.0:0.4 = 2.5$.

    • Even though the samples differ in total weight and were made using different reactions, the elements are always combined in the exact same fixed proportion by mass, which is what the Law of Constant Proportions states.

(32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine)

• Beginner-Friendly Logic: To find molecules, find the moles first ($\text{Given~Mass} \div \text{Molecular~Mass}$) and multiply by Avogadro's number ($6.022 \times 10^{23}$).

• Calculations:

  • a. 32g oxygen ($\text{O}_{2}$):

    • Molecular mass = $32\text{~g/mol}$. Moles = $32 \div 32 = 1\text{~mol}$.

    • Molecules = $1 \times 6.022 \times 10^{23} = \mathbf{6.022 \times}\mathbf{10}^{\mathbf{23}}\text{~molecules}$.

  • b. 90g water ($\text{H}_{2}\text{O}$):

    • Molecular mass = $18\text{~g/mol}$. Moles = $90 \div 18 = 5\text{~moles}$.

    • Molecules = $5 \times 6.022 \times 10^{23} = \mathbf{3.011 \times}\mathbf{10}^{\mathbf{24}}\text{~molecules}$.

  • c. 8.8g carbon dioxide ($\text{CO}_{2}$):

    • Molecular mass = $44\text{~g/mol}$. Moles = $8.8 \div 44 = 0.2\text{~moles}$.

    • Molecules = $0.2 \times 6.022 \times 10^{23} = \mathbf{1.2044 \times}\mathbf{10}^{\mathbf{23}}\text{~molecules}$.

  • d. 7.1g chlorine ($\text{Cl}_{2}$):

    • Molecular mass = $71\text{~g/mol}$. Moles = $7.1 \div 71 = 0.1\text{~moles}$.

    • Molecules = $0.1 \times 6.022 \times 10^{23} = \mathbf{6.022 \times}\mathbf{10}^{\mathbf{22}}\text{~molecules}$.

(Sodium chloride, magnesium oxide, calcium carbonate)

• Beginner-Friendly Logic: To find grams, multiply moles by molecular mass ($\text{Grams} = \text{Moles} \times \text{Molecular~Mass}$).

• Calculations:

  • a. Sodium chloride ($\text{NaCl}$):

    • Molecular mass = $23 + 35.5 = 58.5\text{~g/mol}$.

    • Grams = $0.2 \times 58.5 = \mathbf{11.7}\text{~g}$.

  • b. Magnesium oxide ($\text{MgO}$):

    • Molecular mass = $24 + 16 = 40\text{~g/mol}$.

    • Grams = $0.2 \times 40 = \mathbf{8}\text{~g}$.

  • c. Calcium carbonate ($\text{CaCO}_{3}$):

    • Molecular mass = $40 + 12 + (3 \times 16) = 100\text{~g/mol}$.

    • Grams = $0.2 \times 100 = \mathbf{20}\text{~g}$.