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Hint: Saponification is a process that involves conversion of fat, oil or lipid into soap and alcohol by the action of heat in the presence of aqueous alkali. To find the saponification number of peanut oil we need to find various quantities.

Complete step by step answer:

Saponification is defined as the hydrolysis of an ester with $NaOH$ or $KOH$ which gives alcohol and sodium or potassium salt of the acid. Saponification value is a measure of the content of ester linkages. It is determined by back titration of potassium oxide in the presence of phenolphthalein indicator with 0.5 N sulfuric or hydrochloric acid.

The given quantities in the question are : Weight of the oil = 1.5763 g, Molarity of KOH = 0.421 M, Volume of KOH = 25 ml, Volume of ${H}_{2}S{O}_{4}$ needed= 8.46 ml,

Molarity of ${H}_{2}S{O}_{4}$ = 0.2732 M.

To find the saponification number of peanut oil, we first need to find out the molar equivalents of KOH added and the molar equivalent of KOH left.

Molar equivalents of KOH added = Molarity $\times$ Volume

Molar equivalents of KOH added = 0.4210 $\times$ 25

Molar equivalents of KOH added = 10.525

Molar equivalents of KOH left = Molarity $\times$ Volume $\times$ Valence no. of ${ H }_{ 2 }S{O}_{ 4 }$

Molar equivalents of KOH left = 0.2732 $\times$ 8.46 $\times$ 2

Molar equivalents of KOH left = 4.623

Now, the molar equivalent of KOH used by the oil is given by:

Molar equivalents of KOH used by oil = M.eq. of KOH added - M.eq. of KOH

Molar equivalents of KOH used by oil = 10.525 - 4.623

Molar equivalents of KOH used by oil = 5.902

The molar equivalent of KOH is also given by:

$Molar\quad equivalents\quad of\quad KOH\quad used\quad by\quad oil\quad =\quad \dfrac { Weight\quad of\quad KOH }{ Equivalent\quad weight } \quad \times \quad 1000$

Or,$ Weight\quad of\quad KOH\quad =\quad \dfrac { M.eq.\quad of\quad KOH\quad used\quad \times \quad equivalent\quad weight }{ 1000 } \quad$

$\implies Weight\quad of\quad KOH\quad =\quad \dfrac { 5.902\quad \times \quad 56 }{ 1000 } \quad =\quad 0.3305$

Therefore, the saponification number can be calculated as:

$Saponification\quad number\quad =\quad \dfrac { Weight\quad of\quad KOH\quad in\quad mg }{ weight\quad of\quad oil\quad in\quad g }$

$\implies Saponification\quad number\quad =\quad \dfrac { 330.5 }{ 1.5763 } \quad =\quad 209.6$

Note: While calculating the molar equivalent of KOH left, it is actually calculated in reference with the ${H}_{2}S{O}_{4}$. Do make sure that you consider the valence no. of ${H}_{2}S{O}_{4}$.

Complete step by step answer:

Saponification is defined as the hydrolysis of an ester with $NaOH$ or $KOH$ which gives alcohol and sodium or potassium salt of the acid. Saponification value is a measure of the content of ester linkages. It is determined by back titration of potassium oxide in the presence of phenolphthalein indicator with 0.5 N sulfuric or hydrochloric acid.

The given quantities in the question are : Weight of the oil = 1.5763 g, Molarity of KOH = 0.421 M, Volume of KOH = 25 ml, Volume of ${H}_{2}S{O}_{4}$ needed= 8.46 ml,

Molarity of ${H}_{2}S{O}_{4}$ = 0.2732 M.

To find the saponification number of peanut oil, we first need to find out the molar equivalents of KOH added and the molar equivalent of KOH left.

Molar equivalents of KOH added = Molarity $\times$ Volume

Molar equivalents of KOH added = 0.4210 $\times$ 25

Molar equivalents of KOH added = 10.525

Molar equivalents of KOH left = Molarity $\times$ Volume $\times$ Valence no. of ${ H }_{ 2 }S{O}_{ 4 }$

Molar equivalents of KOH left = 0.2732 $\times$ 8.46 $\times$ 2

Molar equivalents of KOH left = 4.623

Now, the molar equivalent of KOH used by the oil is given by:

Molar equivalents of KOH used by oil = M.eq. of KOH added - M.eq. of KOH

Molar equivalents of KOH used by oil = 10.525 - 4.623

Molar equivalents of KOH used by oil = 5.902

The molar equivalent of KOH is also given by:

$Molar\quad equivalents\quad of\quad KOH\quad used\quad by\quad oil\quad =\quad \dfrac { Weight\quad of\quad KOH }{ Equivalent\quad weight } \quad \times \quad 1000$

Or,$ Weight\quad of\quad KOH\quad =\quad \dfrac { M.eq.\quad of\quad KOH\quad used\quad \times \quad equivalent\quad weight }{ 1000 } \quad$

$\implies Weight\quad of\quad KOH\quad =\quad \dfrac { 5.902\quad \times \quad 56 }{ 1000 } \quad =\quad 0.3305$

Therefore, the saponification number can be calculated as:

$Saponification\quad number\quad =\quad \dfrac { Weight\quad of\quad KOH\quad in\quad mg }{ weight\quad of\quad oil\quad in\quad g }$

$\implies Saponification\quad number\quad =\quad \dfrac { 330.5 }{ 1.5763 } \quad =\quad 209.6$

__Hence, the saponification number of peanut oil is 209.6.__Note: While calculating the molar equivalent of KOH left, it is actually calculated in reference with the ${H}_{2}S{O}_{4}$. Do make sure that you consider the valence no. of ${H}_{2}S{O}_{4}$.

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