Question

Methanol, CH₃OH, is a nonelectrolyte; ammonia, NH₃, is a weak electrolyte; and calcium hydroxide, Ca(OH)₂, is a strong electrolyte. What are the solute particles present in aqueous solutions of each compound? (Select all that apply.)

methanol	ammonia	calcium hydroxide
CH3, OH2+, H+, O2-, CH3O- ,OH- ,CH3OH	NH2+ , O2- , H+ , OH, NH3, NH4+	H+ ,Ca2+ , CaOH+ , Ca(OH)2, Ca(OH)3-,OH-

If 0.1 mol of each compound is dissolved in solution, which one contains about 0.3 mol of solute particles, which contains 0.1 mol of solute particles, and which contains somewhere between 0.1 and 0.2 mol of solute particles?

0.3 mol solute particles	---Select--- methanol ammonia  calcium hydroxide
0.1 mol solute particles	---Select--- methanol ammonia  calcium hydroxide
between 0.1 and 0.2 mol solute particles	---Select--- methanol ammonia  calcium hydroxide

Answer 1

Methanol: Since methanol is a non electrolyte, the only solute particles present is CH3OH.

Ammonia: Since ammonia is a weak electrolyte it will partially dissociate in to ions. Hence, solute particles present are NH3 and NH4+.

Calcium hydroxide: Since calcium hydroxide is a strong electrolyte, it will completely dissociate into ions. Hence solute particles present are Ca2+ and OH-.
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In the given compounds:

1 mole calcium hydroxide dissociates into 3 mole ions. Hence, 0.3 mol solute particles will be obtained from 0.1 mole calcium hydroxide.

As methanol does not dissociate, 0.1 mole solute ions will be obtained from 0.1 mole methanol.

Since, NH3 being weak electrolyte partially dissociates in its aqueous solution, so, between 0.1 and 0.2 mol solute particles will be obtained from 0.1 mole ammonia.