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) > KI > HF > CH₃OH

Explanation:

The electrical conductivity of an ionic solution depends upon (at constant temperature)

a) concentration of electrolyte (number of ions)

b) mobility of ions.

We are provided with four solutions of same concentrations.

Let us check about the number of ions produced by each of the solutions

a) HF : it is a weak acid and will dissociate less: so it will be conducting but will be less as compared to any salt like KI.

b) CH₃OH : this is non electrolytic (will hardly dissociate easily) so it will have least conductivity among the four solutions,

c) KI : it is a salt and will dissociate completely into two ions. It will have higher conductivity as compared to HF and methanol however will have low conductivity as compared to aluminium nitrate (which will furnish four ions on dissociation)

More the conductivity, more the intensity of the bulb

Hence the order will be

Al(NO₃) > KI > HF > CH₃OH

i believe the answer is 5,000