Final answer:
To saponify 100 g of glyceryl tripalmitoleate with a 0.250 M NaOH solution, 0.372 moles of NaOH are required, which equates to 1488 milliliters of the NaOH solution.
Step-by-step explanation:
To calculate how many milliliters of a 0.250 M NaOH solution are needed to saponify 100 g of glyceryl tripalmitoleate, we must first identify the saponification reaction. Saponification is a process that involves the hydrolysis of fats or oils by an alkali to form soap. In this case, glyceryl tripalmitoleate (also known as tripalmitolein), a type of fat, will be reacted with sodium hydroxide (NaOH).
Each triglyceride molecule requires three moles of NaOH to be completely saponified, yielding glycerol and sodium salts of the fatty acids. We'll need to calculate the moles of glyceryl tripalmitoleate and then use the molar ratio to determine the moles of NaOH needed.
The molar mass of tripalmitolein is approximately 807.0 g/mol. Therefore, 100 g of tripalmitolein is approximately:
100 g / 807.0 g/mol = 0.124 moles of tripalmitolein
Now, each mole of tripalmitolein requires three moles of NaOH:
0.124 moles of tripalmitolein x 3 moles of NaOH = 0.372 moles of NaOH needed
Finally, we can determine the volume of NaOH solution needed using its molarity (M):
Volume (L) = Moles of NaOH / Molarity of NaOH
Volume (L) = 0.372 moles / 0.250 M = 1.488 L
Volume (mL) = 1.488 L x 1000 mL/L = 1488 mL
Therefore, 1488 milliliters of a 0.250 M NaOH solution would be required to completely saponify 100 g of glyceryl tripalmitoleate.