Tim conducted experiments with various aqueous solutions to identify the solution displaying the highest conductivity. The diagram illustrates the equipment he employed, while the table enumerates the aqueous solutions subjected to his testing. The diagram is a picture of a conductivity apparatus. Which of the aqueous solutions is most probable to have resulted in the brightest glow of the bulb?

Question

A. CH3OH (aq) because it is an electrolyte
B. C12H22O11 (aq) because it is a nonelectrolyte
C. KCI(aq) because it is a strong electrolyte
D. NH3(aq) because it is a weak electrolyte

Answer 1

To determine which aqueous solution is likely to produce the brightest glow of the bulb in Tim's conductivity apparatus, we can analyze each option based on their properties as electrolytes:

A. CH3OH (aq) - Methanol (CH3OH) is an alcohol and is not considered an effective electrolyte. It does not dissociate into ions in solution and would not conduct electricity well.

B. C12H22O11 (aq) - Sucrose (C12H22O11) is a sugar and is a nonelectrolyte. It does not ionize in aqueous solution, so it would not produce any conductivity and would not cause the bulb to glow.

C. KCl(aq) - Potassium chloride (KCl) is a strong electrolyte. It dissociates completely into K+ and Cl- ions in solution, resulting in a high level of electrical conductivity. This solution is likely to make the bulb glow brightly.

D. NH3(aq) - Ammonia (NH3), when dissolved in water, is classified as a weak electrolyte. It partially ionizes to form NH4+ and OH- ions, which means it has lower conductivity compared to strong electrolytes.

Considering the analysis, the solution most likely to produce the brightest glow of the bulb is:

C. KCl(aq) because it is a strong electrolyte.