## Given density in g/L, how to calculate molar volume?

Answer:

\(\bar{V}=\frac{M}{D}\)

\(D \text { in } \mathrm{g} / \mathrm{L}\)

\(M \text { in } \mathrm{g} / \mathrm{mol}\)

You can do that with a unit conversion. In principle, if you can convert \(\text{from}\) the density \(\text{to}\) the molar volume, you can convert from the molar volume to the density, so I will show the former case.

Given a density in \(\frac{\mathrm{g}}{\mathrm{L}},\) you’re almost there already. A \(\text{molar volume}\) is known to have typical units of \(\frac{\mathrm{L}}{\mathrm{mol}} \text {. }\) So, take the reciprocal to get:

\(\Rightarrow \frac{\mathrm{L}}{\mathrm{g}}\)Now, since you want to cancel out \(\text{g}\) and have \(\mathrm{mol}^{-1}\) remaining, simply \(\text { multiply by the molar mass. }\)

\(\frac{\mathrm{L}}{g^{\prime}} \times \frac{g^{\prime}}{\mathrm{mol}}=\frac{\mathrm{L}}{\mathrm{mol}} \sqrt{ }\)What we’ve just done can be written in variables.

Given a density \(\text{D}\) ALREADY in \(\frac{g}{L}\) and a molar mass in \(\frac{\mathrm{g}}{\mathrm{mol}}\), the molar volume \(\nabla\) is:

\(\bar{V}=\frac{M}{D}\)