The electrical conductivities of the following 0.100 M solutions were measured in an apparatus that contained a light bulb as the indicator of conductivity. Rank the solutions in order of decreasing intensity (brightest to dimmest) of the light bulb.

Rank from brightest to dimmest bulb. To rank items as equivalent, overlap them.

a. HF, CH3OH, KI, Al (NO3) 3

b. Brightest Dimmest

Question

Chemistry

Bentley

19 February, 04:59

The electrical conductivities of the following 0.100 M solutions were measured in an apparatus that contained a light bulb as the indicator of conductivity. Rank the solutions in order of decreasing intensity (brightest to dimmest) of the light bulb.

Rank from brightest to dimmest bulb. To rank items as equivalent, overlap them.

a. HF, CH3OH, KI, Al (NO3) 3

b. Brightest Dimmest

+1

Answers (1)

Know the Answer?

Marsh · Answer 1

Answer: KI> Al (NO3) 3> HF> CH3OH

Explanation:

Potassium iodide is a halide of group 1 which is very soluble in water and whose ions are very mobile in solution. It will readily dissociate and conduct electricity making the bulb brightest when KI is the electrolyte. Salts of group 13 are not as ionic as those of group 1 but dissociate appreciably and conduct electricity. HF is a halogen acid but self ionizes and shows some degree of conductivity. Methanol will have the lowest conductivity making the bulb dimmest when methanol is the electrolyte.