2
$\begingroup$

Okay, so molarity is defined as moles of solute per liter of solution. It is used to describe the concentration. So in the lab, when I see a bottle labeled something like $\mathrm{3.5~M}~\ce{H2SO4}$, is it saying that the only chemical in the bottle is this $\ce{H2SO4}$?

Because the way I think of it is "there are 3.5 moles of $\ce{H2SO4}$ per liter of solution". The word "solution" is kind of throwing me off because it makes me feel like they are saying that there is more than just $\ce{H2SO4}$ in the bottle. Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

I don't know why I'm having such a hard time grasping this. But any help would be awesome!

$\endgroup$
7
$\begingroup$

Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

You are, unfortunately, incorrect. In wet chemistry, you need just a solvent$^\dagger$ and at least one solute to make a solution. While this does mean that you need at least two constituents to make up a solution, you do not need at least two solutes.

... is it saying that the ONLY chemical in the bottle is this $\ce{H2SO4}$?

It depends on your definition of a chemical:

  • If you consider water to be a chemical, then no: there are two chemicals in the bottle$^\ddagger$, $\ce{H2O}$ and $\ce{H2SO4}$.
  • If you are not considering water to be a chemical, then yes: the only chemical in the bottle$^\ddagger$ is $\ce{H2SO4}$.

$^\dagger$ If not specified, the solvent is almost always water.

$^\ddagger$ Except for small amounts of inevitable impurities.

$\endgroup$
6
$\begingroup$

Brian has the correct answer.

I'd like to add, however, that in the most general sense, we don't have to take a molar to mean a mole of solute per liter of solution. I've actually seen it used to mean "a mole of stuff per liter of space" for particular values of stuff.

Since the above sentence isn't exactly self-explanatory, let's take a look at a few examples.

Example 1: Water

At 4 °C, water has a density of 1000 g/L. Water has a molar mass of 18.02 g/mol. This means that in a liter of water at 4 °C, there are:

$$ 1~\mathrm{L} \cdot 1000 ~\frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{18.02 \mathrm{g}} = 55.5 ~\mathrm{mol} $$

Since there are 55.5 moles of water per liter of water, we can actually say that, at 4 °C, water is 55.5 molar. (See Example 3 on Wikipedia).

Note how this works: there is a certain amount of stuff (in this case, water) per amount of space (one liter, or one cubic decimeter). It doesn't matter that we don't have a solution--since there's stuff, and it takes up space, we can calculate a molarity for it.

Example 2: Ideal gas

An ideal gas takes up 22.4 L per mole, at STP. Taking the inverse of this, we find that it is 0.0446 moles/L. Therefore, the concentration of an idea gas at STP is 0.0446 molar.

Again, there's a certain amount of stuff (one mole of ideal gas) and a certain amount of space (22.4 liters), which allows us to calculate a molarity. The fact that the substance is pure doesn't stop us from writing a concentration for it.

A note of caution

This will NOT work for a measure of concentration like molality, because molality is defined by the amount of solvent used. It also won't usually work for things like normality, since you (usually) need to be in solution to generate ions.

$\endgroup$
  • 1
    $\begingroup$ Excellent points. I hate normality -- it doesn't even have to be ionic. For example, for EDTA, typically its normality is equal to its molarity, since it binds metals $1:1$. If you don't the 'customary' conversion from molarity to normality, its impossible to identify logically. Small wonder the term is obsolete! $\endgroup$ – hBy2Py Sep 3 '15 at 2:51
2
$\begingroup$

I think this is the best way to clear your doubt. Have you ever prepared tea? What are the main ingredients of tea?

$$\ce{water~(or~milk) + tea~leaves -> tea}$$

It's the addition of tea leaves which changes water to tea. Depending on how strong you want a tea you can add one teaspoon, two teaspoons or n number of teaspoons of tea leaves.

Similarly, in $\mathrm{3.5~M}~\ce{H2SO4}$ solution it contains:

$$\ce{H2SO4~(solute) + H2O~(solvent) -> H2SO4~solution}$$

To my knowledge there is no 100 percent pure $\ce{H2SO4}$ available commercially. The commercially available $\ce{H2SO4}$ is 96 percent $\ce{H2SO4}$ with a concentration approximately of 18 molar. See this link.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.