Question

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

Answer 1

`7.48X10^3~mL`

Step-by-step 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(1~mol~tripalmitolein)/(801.3~g~tripalmitolein)(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=(mol)/(L)`

`0.25~M=(1.87~mol~NaOH)/(L)`

`L=(1.87~mol~NaOH)/(0.25~M)`

`L=7.48`

Now we can do the conversion to mL:

`7.48~L~(1000~mL)/(1~L)=~7.48X10^3~mL`

I hope it helps!