Question

how much KOH is necessary to hydrolyse 1g . of trimyristin?

Answer 1

trimyristin is a triester containing 3 ester bondings. For complete hydrolysis all 3 ester linkages needed to be worked out. 1 KOH molecule is capable of hydrolyzing just 1 ester bond and hence 1 trimyristin molecule need 3 KOH molecules to achieve complete hydrolysis.Hence we write molar relationship as,

1 mole trimyristin 3 mole KOH

Molar mass of trimyristin = 723.177 g/mol

Molar mass of KOH = 56.106 g/mol

Mass of 3 moles of KOH = 3 * 56.106 = 168.318 g

Hence we write mass relationship as,

723.177 g of trimyristin 168.318 g KOH ........ (1)

We are given with 1.0 g trimyristin and asked to calculate mass of KOH would be required for complete neutralization. So we write,

If, 723.177 g of trimyristin 168.318 g KOH

then, 1.0 of trimyristin say "M" g KOH

On cross multiplication we have,

M * 723.177 = 1.0 * 168.318

M = 168.318 / 723.177

M = 0.233 g

0.233 g of KOH will be required for complete neutralization of 1.0 g of trimyristin.

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