You are given two fat-like solid substances and upon analysis you determine that Sample A has a higher melting point than sample
B. You thus come to the conclusion that sample A. a) is a triglyceride and sample B is a simple lipid. b) is a saturated fat and sample B is an unsaturated fat. c) has a lower number of carbon-carbon double bonds than sample B. d) has a higher number of carbon-carbon double bonds than sample B. e) is an unsaturated fat and sample B is a saturated fat.
c) has a lower number of carbon-carbon double bonds than sample B.
Explanation:
The two fat samples were compared on the basis of their melting point. Fats with no carbon-carbon double bonds in their hydrocarbon chains are called saturated ones. These fat molecules are packed tightly and have a higher melting point. On the other hand, the fats with one or more double bonds between carbon atoms are said to be unsaturated ones. These unsaturated fats have a lower melting point as the kink produced by double bonds does not allow their tight packing. Therefore, sample A with a higher melting point should have a less number of C-C double bonds than sample B.
