To determine the correct order of the relative strengths of the intermolecular forces in the three different states of a substance, let's analyze the physical states and their respective temperatures:
1. State A at [tex]\(342^\circ C\)[/tex]:
- This high temperature suggests that the substance is in a solid state.
- In solids, molecules are closely packed and have the strongest intermolecular forces among the states of matter.
2. State B at [tex]\(-18^\circ C\)[/tex]:
- This low temperature indicates that the substance is in a gaseous state.
- In gases, molecules are far apart and have the weakest intermolecular forces.
3. State C at [tex]\(-2^\circ C\)[/tex]:
- This medium temperature suggests that the substance is in a liquid state.
- In liquids, molecules are less packed than in solids but more packed than in gases, hence exhibiting moderate intermolecular forces.
Now, we can order the states based on the strength of the intermolecular forces:
- Solid (Strongest intermolecular forces)
- Liquid (Moderate intermolecular forces)
- Gas (Weakest intermolecular forces)
Therefore, the order of the relative strengths of the intermolecular forces from strongest to weakest is:
[tex]\[ \text{Solid} > \text{Liquid} > \text{Gas} \][/tex]
Given our states and their associated temperatures, we can now match them with this order:
- A (Solid at [tex]\(342^\circ C\)[/tex]) has the strongest intermolecular forces.
- C (Liquid at [tex]\(-2^\circ C\)[/tex]) has moderate intermolecular forces.
- B (Gas at [tex]\(-18^\circ C\)[/tex]) has the weakest intermolecular forces.
Thus, the correct order of the relative strengths of the intermolecular forces in the three states is:
[tex]\[ A > C > B \][/tex]
The correct choice is:
[tex]\[ b \quad A > C > B \][/tex]
