LUBRICANTS NUMERICALSSOLVED 1 1. PROBLEMSONSAPONIFICATIONVALUE: A. 2.0 gm of an oil is saponified with 30ml of N/2 alcoholic potassium hydroxide solution. After refluxing the mixture it requires25 ml of N/2 HCl solution. Find saponification value of the oil. Ans- Saponification value=Volume of KOH consumed*NKOH*56/Weight of the oil = (30-25)*0.5*56/2.0 =7.0 mgs of KOH B. 3 gms of lubricant wassaponified with KOH solution.After saponification,the mixture was titrated against 0.5N HCl solution. The burrete reading was found to be 12ml. if blank titration burrete reading was 36ml.calculate the saponification number of the lubricant. Ans - -Saponification value=Volume of KOH consumed*NKOH*56/Weight of the oil =(36-12)*28/3 =224 mgs of KOH (as NKOH not given,assume it to be 1) LUBRICANTS NUMERICALSSOLVED 2 2. 4 PROBLEMS ON ACID VALUE: A. 2.5 gms of an oil samplerequired 2.5ml of N/100 KOH to neutralize fatty acids in oil. Find the acid value of the sample. Ans Acid value= Volume of KOH consumed*NKOH*56/Weight of the oil =2.5*0.01*56/2.5 =0.56Mgs of KOH B. Find the acid value of a vegetable oil whose 5ml required 2ml of N/100 KOH during titration. Density of oil=0.92 Ans-i.Wt of oil=Density*volume=0.92*5=4.60gms ii.Acid value= Volume of KOH consumed*NKOH*56/Weight of the oil =2*0.01*56/4.45=0.236mg of KOH C. An oil blend was analysed for its acid value 5ml of oil required 2.5ml N/100 KOH. Find its acidicvalue. State whether the blend is useful for lubrication(density =0.89) Ans- i.Wt of oil=Density*volume=5*0.89=4.45gms ii.Acid value= Volume of KOH consumed*NKOH*56/Weight of the oil =2.5*0.01*56/4.45=0.315mg of KOH As this value is greater than 0.1,it cannot be used for lubrication. LUBRICANTS NUMERICALSSOLVED 3