# Could you help me with this? Justify the statement with an example: "Formality does not depend on what happens to a solute after it is dissolved in water, but depends on amount of solute dissolved in water."

Jul 5, 2015

Here's how you can think about molarity and formality.

#### Explanation:

Molarity and formality both express a solution's concentration as the ratio between the moles of solute and the liters of solution.

However, formality does not depend on what happens to the solute once it's dissolved in solution.

This means that the formality of a solution that contains an electrolyte, i.e. compounds that dissociate in aqueous solution to form cations and anions, will be the same as that of a solution that contains a non-electrolyte, i.e. a compound that does not dissociate in aqueous solution.

Molarity, on the other hand, does depend on what happens to the solute.

Both molarity and formality depend on how much solute you dissolve in water, since higher amounts of solute will imply higher concentrations.

A classic example involves salt, $N a C l$, and sugar, ${C}_{6} {H}_{12} {O}_{6}$.

Once salt is dissolved in aqueous solution, it dissociates completely to form $N {a}^{+}$ cations and $C {l}^{-}$ anions. This implies that $N a C l$ does not exist as an ionic compound in solution, it only exists as ions.

However, since formality does not take into account the form in which the solute exists in solution, you can say that if you dissolve 1 mole of $N a C l$ in 1 L of water you have a 1 F $N a C l$ solution.

When it comes to molarity, you can't actually say that you have a 1-M $N a C l$ solution, because sodium chloride does not exist in that form anymore.

When you dissolve 1 mole of glucose in 1 L of water, you once again have a 1 F solution, even though the glucose does not dissociate into ions.

This time, you can say that you also have a 1-M glucose solution, since glucose will continue to exist as molecules in aqueous solution.

As a conclusion, formality is indeed independent on what happens to the solute after it's dissolved in water, and depends on how much solute you actually dissolve.

Here's an example of a formality problem

http://socratic.org/questions/2-02g-of-kno3-are-dissolve-in-600-ml-of-solution-calculate-the-formality-of-the--2?source=search

Jul 5, 2015

To state a truism, "formality" is indeed a formalism; it depends on idealized concentrations that do not reflect solution phenomena.

#### Explanation:

Let's take an example, sulfuric acid. This is supplied as an 18 mol per litre solution. Given this concentration, and the density of the solution, you would be able to calculate the amount of sulfuric acid formally present in 1 L of conc. solution. But would this figure accurately reflect the number of actual sulfuric acid molecules? No, it would not, because in aqueous solution, a sulfuric acid molecule would speciate to 3 entities: the proton in water (and remember that this is also an abstraction); the bisulfate anion; and the sulfate dianion. So in summary, formality starting quantities, not species in solution.