written 3.6 years ago by
teamques10
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Weight of oil = 0.5 g
Quantity of alcoholic KOH = 50 ml
Quantity of NHCl required ( Back titration ) = 22 ml
Saponification Value = $\dfrac{Volume\space of\space KOH\space consumed\times Normality\space of\space KOH\times56}{Weight \space of\space oil\space in\space gms.}$
$=\dfrac{[Volume\space of\space KOH\space added\space to \space oil\space - Back\space titration reading]\times N_{KOH}\times56 }{ Weight\space of\space oil\space in\space gms}$
$= \dfrac{[50-22]\times0.1\times56}{0.5}$
$={313.6 \space mgs \space KOH }$
Saponification of oil = 313.6 mgs KOH
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