Find the Saponification value of given sample.

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written 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 years ago

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

Marks: 4M

Year: May 2015

engineering chemistry

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written 8.8 years ago by

teamques10 ★ 69k

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 \ \ …

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