Question

Source of power

Aqueous Solutions (aq)

C*H_{3}*OH

KCI

N*H_{3}

C_{12}*H_{22}*O_{11}

Which of the aqueous solutions is most likely to have resulted in the bulb glowing the brightest?

(1 point)

C_{12}*H_{22}*O_{11}(aq) because it is a nonelectrolyte

N*H_{3}(aq) because it is a weak electrolyte

C*H_{3}*OH(aq) because it is an electrolyte

OKCI(aq) because it is a strong electrolyte

Answer 1

To determine which aqueous solution is most likely to have resulted in the bulb glowing the brightest, we need to consider their properties in terms of electrical conductivity.

1. **C₁₂H₂₂O₁₁ (sucrose)**: This is a nonelectrolyte, meaning it does not dissociate into ions in solution and therefore will not conduct electricity effectively.

2. **NH₃ (ammonia)**: This is a weak electrolyte. It does partially ionize in water, but not to a significant extent, so it would conduct electricity poorly compared to strong electrolytes.

3. **C₂H₅OH (ethanol)**: Like sucrose, ethanol is also a nonelectrolyte and does not dissociate into ions in aqueous solution, which means it will not conduct electricity.

4. **KCl (potassium chloride)**: This is a strong electrolyte, as it dissociates completely into potassium ions (K⁺) and chloride ions (Cl⁻) when dissolved in water. This means it will conduct electricity very well.

Given this information, the aqueous solution most likely to result in the bulb glowing the brightest would be **KCl(aq) because it is a strong electrolyte**. Therefore, the correct answer is:

**OKCI(aq) because it is a strong electrolyte.**