1.

0.436g of acetyl derivative of a polyhydric alcohol (molecular mass =92) required 0.336g KOH for hydrolysis. Calculate the number of hydroxy gropus in the alcohol.

Answer»

Solution :We known that , `(w)/(w) (M+42n)=56 N`
where w is the amount of alkali, W amount ofacetyl DERIVATIVE, M is the molecular mass of alcohol and n NUMBER of HYDROXYL groups.
`(0.336)/(0.436)(92+42n)=56n`
or `92+42n=(0.436)/(0.336)xx56n=(218)/(3)n`
or `276+126n=218n`
or `n=3 ` ( The alcohol is trihydric. )


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