Do question "b" of the calculations part. Question "b" is in the image. The topic is about Titration.Part "a" answers:Trial 1; 0.102MTrial 2: 0.094…

Question

asked Feb 13, 2023 117k views

1 Answer

Lucsan · Answer 1 · 2023-02-18T10:54:57+0000

3.37M

Given the molarity of the base during each trial as show:

Trial 1; 0.102M

Trial 2: 0.094M

Trial 3: 10M

The average of the molarity is expressed as:

$\begin{gathered} average\text{ Mb}=(0.102+0.094+10)/(3) \\ average\text{ Mb}=(10.106)/(3) \\ average\text{ Mb}=3.37M \end{gathered}$

Hence the average molarity for the trials is 3.37M