To calculate the osmolarity of 0.39M Na2CO3, we need to consider the number of particles that each molecule of Na2CO3 dissociates into when dissolved in water. Na2CO3 dissociates into 3 ions: 2 Na+ ions and 1 CO32- ion.

Therefore, the total number of particles contributed by 0.39M Na2CO3 is:

2 Na+ ions/molecule x 0.39M + 1 CO32- ion/molecule x 0.39M = 1.17M

Osmolarity is defined as the total concentration of solute particles per liter of solution, taking into account the dissociation of molecules into ions. So, the osmolarity of 0.39M Na2CO3 is:

1.17M

Therefore, the osmolarity of 0.39M Na2CO3 is 1.17 mOsm/L.