Question

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: 0.094MTrial 3: 10M

Answer 1

3.37M

Explanations:

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