$\beta$-mercaptoethanol molarity
From the Aldrich data:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$

Converted to a more convenient unit:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$

Given the molar mass (also from Aldrich) of:
 $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$

We learn that one liter has:
 $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$

I can't speak to why your other source has a different number.