A chemistry problem by Asif Anvar

Given/Known:

$P = 1 atm$

$T = 300 K$

$\rho = 5 \frac{mg}{cm^3}$

The first thing we need to do is make sure the numbers we'll be using are in the right units. Since P is in atm and T is in K, we'll need to use the gas constant R in units of L atm/mol K.

$R = 0.08206 \frac{L*atm}{mol*K}$

Also, we need to convert the density $\rho$ . Since we're using Liters, and since $\frac{mg}{cm^3} = \frac{g}{L}$ , it would make sense to convert $\rho$ from $\frac{mg}{cm^3}$ to $\frac{g}{L}$ . Since they're equivalent to each other, we still have:

$\rho = 5 \frac{g}{L}$

So now that we know those values, the next step is to arrange the ideal gas formula into an equation we can solve with the variables we have.

$PV=nRT$

Knowing that $n=\frac{m}{M}$ , and $\rho=\frac{m}{V}$ , we can rearrange the formula above like this:

$PV=\frac{m}{M}RT$ $\rightarrow$ $MP=\frac{m}{V}RT$ $\rightarrow$ $MP=\rho RT$

We can now plug our variables into $MP=\rho RT$ to solve for the molar mass, M:

$\text{ M} * (1 \text{atm} ) = (5 \frac{g}{L}) (0.08206 \frac{\text{ L*atm}}{\text{ mol*K}}) (300 \text{K}$ )

$\Rightarrow$ $M = 123.09 \frac{g}{mol}$

Now that we know the molar mass of the sample, we can divide it by the molar mass of 1 mole of acetic acid to find how many moles of acetic acid are in the sample.

Dividing the molar mass of the sample $M = 123.09 \frac{g}{mol}$ by the mass of 1 mole acetic acid is $60.052 \frac{g}{mol}$ gives us…

$\frac{123.09}{60.052} \approx 2 \text{ mols}$

Therefore, there must be 2 moles of acetic acid in the sample.