Answer:
[tex]n_{oil}=0.002mol[/tex]
Explanation:
Hello,
In this case, considering that the sodium hydroxide reacts with the cooking oil, the following relationship is applied:
[tex]n_{oil}=n_{NaOH}\\n_{oil}=M_{NaOH}*V_{NaOH}\\n_{oil}=0.4\frac{mol}{L} * 0.005L\\n_{oil}=0.002mol[/tex]
Best regards.