Start learning →

ISC Class 12 Chemistry — practice by topic

Browse 437 questions across 7 Chemistry topics. Choose what you want to revise, then open a question to check its answer and worked solution.

Start learning ISC Class 12 with Quill →

Chemistry topics

Amines and Biomolecules (108)Chemical Kinetics (66)The d- and f-Block Elements & Coordination Compounds (57)Aldehydes, Ketones and Carboxylic Acids (56)Electrochemistry (50)Haloalkanes, Alcohols and Ethers (50)Solutions (50)

Try a Chemistry question

The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:Read the answer and worked solution →For a galvanic cell, the standard cell EMF is given by:Read the answer and worked solution →A solution is best defined as:Read the answer and worked solution →In a salt-in-water solution, the solvent is:Read the answer and worked solution →

Worked practice questions

The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:

  1. $1$
  2. $0.5$
  3. $2$
  4. $3$
Show answer and solution

Correct answer: C. $2$

NaCl dissociates in water: $\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$. One formula unit produces two particles in solution. Van't Hoff factor $i = \dfrac{\text{actual moles of particles}}{\text{moles of solute formula units}}$. For complete dissociation of NaCl, $i = 2$. In reality, ion pairing slightly reduces $i$ below the theoretical value, especially at higher concentrations. At dilute $0.01$ M the assumption of complete dissociation is reasonable, and the experimental $i$ is close to $1.9$. The question asks for the theoretical value assuming complete dissociation, so 2 is the cleanest answer. Compare with $\mathrm{CaCl_2}$: dissociates into 3 ions, so $i = 3$ (option D) — common trap.

Open this question →

For a galvanic cell, the standard cell EMF is given by:

  1. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} + E^{\circ}_{\text{anode}}$
  2. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{anode}} - E^{\circ}_{\text{cathode}}$
  3. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
  4. $E^{\circ}_{\text{cell}} = -2 E^{\circ}_{\text{anode}}$
Show answer and solution

Correct answer: C. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, where both potentials are listed as standard *reduction* potentials. For a spontaneous galvanic cell, $E^{\circ}_{\text{cell}} > 0$, which means the cathode has the higher reduction potential. This is just an algebraic restatement of "reduction happens at the cathode".

Open this question →

A solution is best defined as:

  1. Any mixture of a solid and a liquid
  2. A pure substance in a single phase
  3. A heterogeneous mixture of two or more substances
  4. A homogeneous mixture of two or more pure substances
Show answer and solution

Correct answer: D. A homogeneous mixture of two or more pure substances

A true solution is a *homogeneous* mixture of two or more pure substances. A binary solution has one solute and one solvent.

Open this question →

How to practise

  1. Choose a Chemistry question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.