Mumbai University > First Year Engineering > Sem1 > Applied Chemistry 1

Marks: 4M

Year: May 2015

Blank titration reading ($V_2$) = 50 ml

Back titration reading ($V_1$) = 22 ml

∴ 22ml of 0.1N NHCl = 50ml of xN KOH

Normality of KOH = 0.044 N

Saponification value = $\frac{(V2-V1)X Normality of KOH X 56}{weight \ \ of \ \ oil \ \ in \ \ grams} \\ = \frac{(50-22)X 0.044 X 56}{0.5} \\ = 137.98 mg$