Class 9 Atomic Foundations of Matter Mock Test (CBSE)

Topics in Atomic Foundations of Matter

• Law of conservation of mass
• Law of constant proportions
• Dalton's atomic theory
• Covalent bond
• Ionic bond
• Writing chemical formulae
• Properties of ionic and covalent compounds
• Molecular mass
• Formula unit mass

Atomic Foundations of Matter practice questions with answers

1. Q1Multiple choice·1 mark
   What is the molecular mass of nitric acid ($HNO_3$)? Given atomic masses: $H = 1\text{ u}$, $N = 14\text{ u}$, $O = 16\text{ u}$.

   1. (a)$63\text{ u}$
   2. (b)$31\text{ u}$
   3. (c)$49\text{ u}$
   4. (d)$65\text{ u}$
   Show answer and step marks

   1. Correct option: (a) $63\text{ u}$. Molecular mass of $HNO_3 = 1(\text{Mass of } H) + 1(\text{Mass of } N) + 3(\text{Mass of } O) = 1(1) + 1(14) + 3(16) = 1 + 14 + 48 = 63\text{ u}$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Which of the following represents the correct chemical formula for aluminum sulfate?

   1. (a)$AlSO_4$
   2. (b)$Al_2(SO_4)_3$
   3. (c)$Al_3(SO_4)_2$
   4. (d)$Al_2SO_4$
   Show answer and step marks

   1. Correct option: (b) $Al_2(SO_4)_3$. The valency of aluminum is $+3$ and the sulfate ion ($SO_4^{2-}$) has a valency of $-2$. By cross-multiplying the valencies, the formula is $Al_2(SO_4)_3$.+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   In a metallic oxide $M_2O_3$, the mass ratio of metal $M$ to oxygen is $7:3$. If the atomic mass of oxygen is $16\text{ u}$, the atomic mass of $M$ is:

   1. (a)$28\text{ u}$
   2. (b)$56\text{ u}$
   3. (c)$84\text{ u}$
   4. (d)$112\text{ u}$
   Show answer and step marks

   1. Correct option: (b) $56\text{ u}$. Setting up the mass ratio $\frac{2 A_M}{3 \times 16} = \frac{7}{3}$ from the formula $M_2O_3$ yields $A_M = 56\text{ u}$, corresponding to option (b).+1

   Answer: (b)
4. Q4Short answer·3 marks
   (a) Write the chemical formula of sodium carbonate. (b) Calculate the formula unit mass of sodium carbonate. (Given atomic masses: $\text{Na} = 23\text{ u}$, $\text{C} = 12\text{ u}$, $\text{O} = 16\text{ u}$)
   Show answer and step marks

   1. Writing correct symbols and valencies for sodium ($\text{Na}^+$) and carbonate ($\text{CO}_3^{2-}$), and crossing valencies to obtain $\text{Na}_2\text{CO}_3$.+1
   2. Setting up the formula unit mass expression: $2 \times \text{mass of Na} + \text{mass of C} + 3 \times \text{mass of O}$.+1
   3. Substituting values and calculating: $2(23) + 12 + 3(16) = 46 + 12 + 48 = 106\text{ u}$.+1

   Answer: (a) $\text{Na}_2\text{CO}_3$ (b) $106\text{ u}$

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