Please look at the following below links which may help you in your analysis:

https://core.ac.uk/download/pdf/237468127.pdf

https://old.fssai.gov.in/Portals/0/Pdf/Draft_Manuals/OILS_AND_FAT.pdf

https://www.sciencedirect.com/topics/engineering/saponification-value

Thanks

Dear Basudev Biswal

The saponification value of an oil, which you already have calculated, equals the number of milligrams of KOH that you need to saponify one gram of it.

The only problem is that many lab analysts don't take into account that the usual purity of KOH pellets is usually as low as 88-92% (mean 90%). So, if KOH were 100% pure, one mole of KOH would be actually 39.10+16.00+1.01 = 56.11 g KOH, but it is 56.11/0.90 =62,34 g if, let's say, 90% pure. Anyway, it is much better to prepare a KOH solution and titrate it with HCl 0.5 mol/L.

So the amount of oil divided by 3 (because fats and oils are triglycerides and one of mol of them reacts with 3 mol KOH) that is completely saponified by 62,34 g KOH (90%) should represent the "average molecular mass" of the oil, which is in fact probably a complex mixture of many different triglycerides constituted by many different fatty acids.

I hope this info works for you.

Best regards from Costa Rica

Kindly check https://www.sciencedirect.com/topics/engineering/saponification-value