Question

How many milliliters of a 0.250 MNaOHMNaOH solution are needed to completely react with 500. gg of glyceryl tripalmitoleate (tripalmitolein)

Answer 1

Answer:

$7.48X10^3~mL$

Explanation:

For this question we have:

-) A solution NaOH 0.25 M

-) 500 g of glyceryl tripalmitoleate (tripalmitolein)

We can start with the reaction between NaOH and tripalmitolein. NaOH is a base and tripalmitolein is a triglyceride, therefore we will have a saponification reaction. The products of this reaction are glycerol and (E)-hexadec-9-enoate.

Now, with the reaction in mind, we can calculate the moles of NaOH that we need if we use the molar ratio between NaOH and tripalmitolein (3:1) and the molar mass of tripalmitolein (801.3 g/mol). So:

$500~g~tripalmitolein\frac{1~mol~tripalmitolein}{801.3~g~tripalmitolein}\frac{3~mol~NaOH}{1~mol~tripalmitolein}=1.87~mol~NaOH$

With the moles of NaOH we can calculate the volume (in litters) if we use the molarity equation and the Molarity value:

$M=\frac{mol}{L}$

$0.25~M=\frac{1.87~mol~NaOH}{L}$

$L=\frac{1.87~mol~NaOH}{0.25~M}$

$L=7.48$

Now we can do the conversion to mL:

$7.48~L~\frac{1000~mL}{1~L}=~7.48X10^3~mL$

I hope it helps!