Dear hosein

Thankyou so much for your answer.

ifmolecular weight of oil is 290 g/mol,and molecular weight of methanol is  34, and molar ratio of methanol to oil is 9:1, then for 1 litre of oil ,how much quantity of methanol is needed.if possible  please help me with the calculation.

A mixture is 290 g (290X1)  methanol  plus 9x34=  306 g of methanol has gives a total  mixture with a mass ration of 306/290 = 1.055 methanol to oil you need to know the SG of the oil to calculate volume ratio.  The SG methanol is .7924.  you haven't told us the SG of the oil but .85 is a reasonable estimate for a typical oil with an MW of 290.  1.055/.85/(1.055/.85+1/.7924) = 0.4984 since my SG of oil  was an estimate I  suspect your mixture is pretty close to 50/50 methanol and oil.

It depends on the type of oil and the temperature.

R

Thank you, Hosein, for your valuable help

Thank you, Rick sir for that wonderful explanation

Thank you Jerry sir, for that important suggestion.

Based on the density equation

Bineesh C Mathew

Apply to the values in the following equation.

d1 = Density of methanol

V1 = Volume of methanol

M1 = molecular weight of methanol

d2 = Density of oil

V2 = volume of oil

M2 = molecular weight of oil

9(d1*V1)/M1 = (d2*V2)/M2