# If we have "24.9 L" of liquid ethanol, with a density of "789 g/L" at a certain temperature, how many mols do we have?

Apr 7, 2016

Let's consider the ideal gas law for a moment, for a transition point.

$P \setminus m a t h b f \left(V\right) = \setminus m a t h b f \left(n\right) R T$

Notice how we can divide by $n$ to achieve

$P \setminus m a t h b f \left(\overline{V}\right) = R T$,

where $\setminus m a t h b f \left(\overline{V} = \frac{V}{n}\right)$, the molar volume in $\text{L/mol}$. When we reciprocate that, guess what we get? The molar density, $\overline{\rho}$, in $\text{mol/L}$, more typically known as the molar concentration.

Obviously the ideal gas law is not supposed to work for liquids (liquid isn't even in the name), so what do we use instead of the molar density, which is really just the reciprocal molar volume?

The mass density, $\rho$, in $\text{g/L}$, which is incidentally a nice way to calculate the molar volumes of nonideal gases given their molar masses.

Given $\text{24.9 L}$ of liquid ethanol, whose density is $\text{789 g/L}$, it is fairly straightforward to convert to $\text{mol}$s. Generally, we have

$\text{mol" = cancel"L" xx cancel"g"/cancel"L" xx "mol"/cancel"g} = V \times \rho \times \frac{1}{M} _ r$,

so we see the calculation goes as follows:

=> 24.9 cancel"L EtOH" xx (789cancel"g EtOH")/cancel"L EtOH" xx "mol EtOH"/(46.07cancel"g EtOH"),

$\approx$ $\textcolor{b l u e}{\text{372 mol EtOH}}$

You can see at this point that the ideal gas law would clearly fail for a liquid. That's because of the significantly higher density than many gases. Generally liquids are approximately 2~4 orders of magnitude more dense.

CHALLENGE: An example of the density of a typical ideal gas is $1.98$ $g \text{/} L$, which is $C {O}_{2} \left(g\right)$. Can you figure out how to get the answer of $1.008$ $m o l$ of $C {O}_{2}$ in $22.4$ $L$ of $C {O}_{2}$?