As stated above me, degrees Plato are an expression of the percent mass of sugar in the wort.
1 ºPlato = 1% sugar by mass = (1 gram of sugar)/(100 grams of wort)
Likewise, if you have wort at 10 ºPlato (approximates to 1.040 OG...nice for a session ale) then you'll have 10 grams of sugar per 100 grams of wort.
So to calculate the molarity of this hypothetical 10 ºPlato beer, all we have to do is some dimensional analysis.
Beer is a complex solution of many organic and inorganic substances, including multiple types of sugars, so to simplify things, I'm going to focus on maltose.
The molar mass of maltose = 342.3 g/mol
10 grams of maltose * (1 mol of maltose / 342.3g) ≈ 0.02921 mols ≈ 0.029 mols of maltose per 100 grams of wort
Now, using our specific gravity, which is just the density of the wort compared to water, we can calculate the volume in mL per 100 grams of wort
(Specific Gravity) = (Density of solution)/(Density of water)
1.040 (SG) = (Density of solution)/(Density of water) ---> 1.040 * 1mg/mL = Density of solution = 1.040 grams /1mL
100 g of wort * (1 mL/1.040g) = 96.1538461538 mL ≈ 96.15 mL (ignoring significant figures here)
Now putting it all together:
(0.029 mols of maltose)/(100 grams of wort) * (100 grams of wort / 96.15 mL) * (1000 mL / L) = 0.3016 mols of maltose per L of wort, or as us science nerds would say, we have a 0.3016 molar solution of maltose
Using all of this information, you could convert to gallons and scale up to a 5 gallon batch... figure out the total number of mols per 5 gallon batch... convert to grams of sugar per batch...