Metol MW is 344.38, MW of NaOH is 40, so 40g of Metol = 0.114M and we need 4*0.116*40 = 18.584g of NaOH for create Metol.base.Na2 salt ((C7H10NO)2.Na2). now we can rewrite Metolal formula as:

Sodium Sulfite 85g

Metol 40g + NaOH 18.584g = 34.18g of Metol.base.Na2 salt

NaOH 38-18.584=19.416g

Water 1l

i.e. we have 19.4g offreeNaOH in concentrate!

I think that this formula is very different from EZRodinal where concentrate haven't any free of NaOH. When we dilute Metotal 1:50 we have:

Sodium Sulfite 1.7g

Metol.base.Na2 salt 0.68g

NaOH 0.4g

Water 1l

I know only one receipt like this. Here is the link (russian) and here is translation to english. receipt is simple:

20ml of D23 stock

12ml of 20% solution of NaOH

Water to 300ml

recalculate to 1l of all components:

Sodium Sulfite 6g

Metol 0.45g

NaOH 7.2g

Water 1l

this is equivalent to

Sodium Sulfite 6g

Metol 0.45g + NaOH 0.2g = 0.385g of Metol.base.Na2 salt

NaOH 7g

Water 1l