Bihar Board Class 11 Science Chemistry · Ch 1: Some Basic Concepts of Chemistry

Topics in this test

Difficulty: 3 easy · 8 medium · 19 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. 2.0 g of a metal carbonate $MCO_3$ (molar mass $100\ g\ mol^{-1}$) is heated strongly so that it completely decomposes into the metal oxide $MO$ and carbon dioxide gas. What mass of $CO_2$ is released, and what is the mass of $MO$ left behind?

   • A. $0.88\ g$ of $CO_2$ and $1.12\ g$ of $MO$
   • B. $0.88\ g$ of $CO_2$ and $2.0\ g$ of $MO$
   • C. $1.12\ g$ of $CO_2$ and $0.88\ g$ of $MO$
   • D. $0.44\ g$ of $CO_2$ and $1.56\ g$ of $MO$
   Solution

   0.02 mol of $MCO_3$ gives 0.02 mol each of $MO$ and $CO_2$; mass of $CO_2 = 0.02\times44 = 0.88\ g$, so residue $MO = 2.0-0.88 = 1.12\ g$.

2. Q2. Carbon combines with oxygen to form two oxides. In the first oxide 12 g of carbon combines with 16 g of oxygen, and in the second oxide 12 g of carbon combines with 32 g of oxygen. The mass ratio of oxygen ($16:32 = 1:2$) illustrates the:

   • A. Law of conservation of mass
   • B. Law of multiple proportions
   • C. Law of definite proportions
   • D. Gay-Lussac's law of gaseous volumes
   Solution

   When one element combines with a fixed mass of another in different ratios that are small whole numbers, it illustrates the law of multiple proportions.

3. Q3. Consider the following statements: (i) Molarity changes with change in temperature. (ii) Molality changes with change in temperature. (iii) Mole fraction is independent of temperature. Which of the above statements are correct?

   • A. (ii) and (iii)
   • B. (i) and (ii)
   • C. Only (i)
   • D. (i) and (iii)
   Solution

   Molarity involves volume, which changes with temperature, while molality and mole fraction depend only on mass and moles, so they are temperature independent; hence (i) and (iii) are correct.

4. Q4. The total number of atoms present in 0.5 mol of carbon dioxide ($CO_2$) is:

   • A. $3.011 \times 10^{23}$
   • B. $1.806 \times 10^{24}$
   • C. $6.022 \times 10^{23}$
   • D. $9.033 \times 10^{23}$
   Solution

   0.5 mol $CO_2$ has $0.5 \times 6.022 \times 10^{23}$ molecules, and each molecule has 3 atoms, giving $9.033 \times 10^{23}$ atoms.

5. Q5. The molality of a solution made by dissolving 18 g of glucose (molar mass $= 180\ \text{g mol}^{-1}$) in 500 g of water is:

   • A. $0.2\ \text{m}$
   • B. $0.1\ \text{m}$
   • C. $0.4\ \text{m}$
   • D. $2\ \text{m}$
   Solution

   Moles of glucose $= 18/180 = 0.1$, and molality $= 0.1/0.5\ \text{kg} = 0.2\ \text{m}$.

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