Question

Given one mole of each substance, which of the following will produce the FEWEST particles in aqueous solution? 1. sodium nitrate 2. CH2Cl2 3. K2SO4 4. sodium phosphate

Answer 1

Answer: The substance that produces fewest particles is $CH_2Cl_2$

Explanation:

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

Covalent compounds do not dissociate into ions when dissolved in aqueous solution.

For the given options:

• Option 1:  Sodium nitrate

The chemical formula of sodium nitrate is $NaNO_3$

The ionization reaction for the given compound follows:

$NaNO_3(aq.)\rightarrow Na^+(aq.)+NO_3^-(aq.)$

This produces in total of 2 ions.

• Option 2:  $CH_2Cl_2$

The given compound is a covalent compound and do not dissociate into its ions. It remains as such as a single unit.

• Option 3:  $K_2SO_4$

The chemical name for the given compound is potassium sulfate.

The ionization reaction for the given compound follows:

$K_2SO_4(aq.)\rightarrow 2K^+(aq.)+SO_4^{2-}(aq.)$

This produces in total of 3 ions.

• Option 4:  Sodium phosphate

The chemical formula of sodium phosphate is $Na_3PO_4$

The ionization reaction for the given compound follows:

$Na_3PO_4(aq.)\rightarrow 3Na^+(aq.)+PO_4^{3-}(aq.)$

This produces in total of 4 ions.

Hence, the substance that produces fewest particles is $CH_2Cl_2$