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. hf, ch₃oh, ki, al(no₃)₃.

Al(NO₃)₃ >  KI > HF > CH₃OH

Explanation:

The electrical conductivities of the solutions will depend on the concentration of ions in solution.

Al(NO₃)₃ solution contains 0.1 M of Al³⁺ ions and 0.3 M of NO₃⁻ ions

KI solution contains 0.1 M of K⁺ ions and 0.1 M of I⁻ ions

HF solution contains less than 0.1 M of H⁺ ions and less then 0.1 M of F⁻ ions, because the HF acid will not dissociate completely

CH₃OH practically it does not dissociate, so in the solution will not be electrical conductive (comparative with the other solutions)

The solutions in order of decreasing intensity of the bulb are ranked as following:

Al(NO₃)₃ >  KI > HF > CH₃OH

the answer is 0.42 is that is an option. this question provides no info