Which of the AQUEOUS solutions is most probable to have resulted in the brightest glow of the bulb

Question

NH3(aq) because it is a week electrolyte
C12H22O11(aq) because it is a non-electrolyte
KCI(aq) because it is a strong electrolyte
CH3OH(aq) Because it is an Electrolyte

Answer 1

To determine which aqueous solution is most likely to produce the brightest glow in a bulb, we need to consider the nature of each solution in terms of its ability to conduct electricity, which is directly related to its status as an electrolyte.

1. NH3(aq) - Ammonia is a weak electrolyte. It partially ionizes in water to produce NH4+ and OH- ions, but it does not dissociate completely. Therefore, it would conduct electricity poorly.

2. C12H22O11(aq) - Sucrose (table sugar) is a non-electrolyte. It does not dissociate into ions in solution, meaning it cannot conduct electricity at all.

3. KCl(aq) - Potassium chloride is a strong electrolyte. It completely dissociates into K+ and Cl- ions in solution, allowing for efficient conduction of electricity. This would likely result in a bright glow in the bulb.

4. CH3OH(aq) - Methanol is also a non-electrolyte. While it can dissolve in water and participate in some reactions, it does not produce ions and therefore does not conduct electricity significantly.

Based on this analysis, KCl(aq) is the most probable solution to produce the brightest glow in the bulb, as it is a strong electrolyte that fully dissociates in solution and allows for maximum electrical conductivity.