The molar mass of the barium fluoride compound ($BaF_2 $) requires the relative atomic mass of each element. For barium ($Ba $), the relative atomic mass is about 137.33, and for fluorine ($F $), the relative atomic mass is about 19.00.

In $BaF_2 $, there is one barium atom and two fluorine atoms. The molar mass algorithm is the sum of the relative atomic mass of each atom multiplied by its atomic number.

Therefore, the molar mass of $BaF_2 $is: the relative atomic mass of barium and twice the relative atomic mass of fluorine are added, that is, 137.33 dollars + 2 × 19.00 = 137.33 + 38.00 = 175.33g/mol $.

From this, it can be seen that the molar mass of $BaF_2 $is about 175.33g/mol. This value is of great significance in the conversion of chemical calculations, such as the amount, mass, concentration, etc. It is also an indispensable basic data for chemical study and practice.