Question: Q 4 : A 3 . 0 8 5 gram of fat sample was saponified by KOH solution ( M . wt . - 5

Q4: A 3.085 gram of fat sample was saponified by KOH solution (M.wt.-56gmol). The amount of the used base was neutralized by 21mL of 0.04MHCl. Calculate the saponification value of this sample. Titration of blank takes 1.5mL of the same solution of HCl.
(4 marks)
Qs: Choose the correct answer:
(6 marks)
The acid value of an oil sample with time.
a) Decreases
b) Increases
c) Remains constant
d) may increase or decrease
Which of the following term describe saponification?
a) Cleaving of ester molecules into carboxylic acid and alcohol
b) Dehydration synthesis by removing water
c) Hydrolysis of a salt by adding a weak acid
d) Synthesis of two alkyl groups to make an ether
Which of the following statement is true about saponification value of oil?
a) The shorter the chain of fatty acid, the higher is the saponification value
b) The shorter the chain of fatty acid, the higher is the saponification value
c) The higher the saturation in chain of fatty acid, the lower is the saponification value
d) The lower the saturation in chain of fatty acid, the higher is the saponification value
Good Luck
!!
 Q4: A 3.085 gram of fat sample was saponified by KOH

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