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

Topics in this test

Difficulty: 3 easy · 9 medium · 18 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. Assertion (A): Molality is preferred over molarity for expressing the concentration of a solution when experiments involve a change in temperature. Reason (R): Molality depends on the mass of the solvent, which does not change with temperature, whereas molarity depends on volume, which changes with temperature.

   • A. A is false but R is true
   • B. Both A and R are true, but R is not the correct explanation of A
   • C. Both A and R are true, and R is the correct explanation of A
   • D. A is true but R is false
   Solution

   Both statements are true, and R correctly explains A: because mass is temperature-independent while volume expands or contracts with temperature, molality stays constant but molarity varies.

2. Q2. A mass of $4.327\,\text{g}$ of a substance occupies a volume of $2.1\,\text{mL}$. Expressed to the correct number of significant figures, its density is

   • A. $2.1\,\text{g mL}^{-1}$
   • B. $2.06\,\text{g mL}^{-1}$
   • C. $2.0604\,\text{g mL}^{-1}$
   • D. $2\,\text{g mL}^{-1}$
   Solution

   $4.327/2.1 = 2.0604...$, but the answer can have only as many significant figures as the least precise value ($2.1$ has $2$), so it is rounded to $2.1\,\text{g mL}^{-1}$.

3. Q3. The total number of atoms present in $0.5$ mol of carbon dioxide ($\text{CO}_2$) gas is (Avogadro number $=6.022\times10^{23}$)

   • A. $3.011\times10^{23}$
   • B. $6.022\times10^{23}$
   • C. $1.806\times10^{24}$
   • D. $9.033\times10^{23}$
   Solution

   One $\text{CO}_2$ molecule has $3$ atoms, so atoms $=0.5\times 3\times 6.022\times10^{23}=9.033\times10^{23}$.

4. Q4. The mass percentage of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$ (molar mass $80\,\text{g mol}^{-1}$), is

   • A. $17.5\%$
   • B. $52.5\%$
   • C. $35\%$
   • D. $28\%$
   Solution

   There are $2$ nitrogen atoms contributing $2\times 14 = 28\,\text{g}$, so the percentage of nitrogen $=\frac{28}{80}\times 100 = 35\%$.

5. Q5. A solution is prepared by mixing $46\,\text{g}$ of ethanol (molar mass $46\,\text{g mol}^{-1}$) with $54\,\text{g}$ of water (molar mass $18\,\text{g mol}^{-1}$). The mole fraction of ethanol in the solution is

   • A. $0.25$
   • B. $0.75$
   • C. $0.46$
   • D. $0.33$
   Solution

   Moles of ethanol $=1$ and moles of water $=54/18=3$, so mole fraction of ethanol $=\frac{1}{1+3}=0.25$.

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