Draw a triacylglycerol (triglyceride) made from glycerol, myristic acid [CH3(CH2)12COOH], palmitic acid [CH3(CH2)14COOH], and ol
eic acid [CH3(CH2)7CH=CH(CH2)7COOH(cis)]. Place palmitic acid at the middle position. Cis/trans isometry is graded
1 answer:
Answer:
The triglyceride made from glycerol, myristic acid, palmitic acid and cis-oleic acid is shown in the attached pictures.
Explanation:
A triglyceride or triacylglyceride is an ester derived from glycerol and three fatty acids. The fatty acids of this exercise are myristic acid, palmitic acid and oleic acid.
A fatty acid is a biomolecule of lipid nature formed by a long linear hydrocarbon chain, of different length or number of carbon atoms, at the end of which is a carboxyl group.
The myristic acid has a carbon chain of 14 carbon atoms with a carboxyl group at one end.
Palmitic acid has a carbon chain of 16 carbon atoms with a carboxyl group at one end.
Oleic acid has a carbon chain of 18 carbon atoms with a carboxyl group at one end and a double bond at position 9.
Triglyceride is formed by the esterification of the glycerol oxidrile group and the carboxyl acid group.
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Answer:
i=1.62 .
Explanation:
Let, i be the Van't Hoff Factor.
Moles of benzamide,=
Molality of solution, m=
Now, we know
Depression in freezing point, .....1
It is given that,
Putting all these values we get,
Now, moles of ammonium chloride=
molality =
Putting all these values in eqn 1.
We get,
i=1.62 .
Hence, this is the required solution.