A student performs a titration with different burettes and finds titre values of $25.2 \mathrm{~mL}, 25.25…
A student performs a titration with different burettes and finds titre values of $25.2 \mathrm{~mL}, 25.25 \mathrm{~mL}$, and $25.0 \mathrm{~mL}$.
The number of significant figures in the average titre value is
Solution
Average
$
\begin{aligned}
& =\frac{25.2+25.25+25.0}{3}=\frac{75.45}{3} \\
& =25.15=25.2 \mathrm{~mL}
\end{aligned}
$
Number of significant figure is 3 .
Basics in chemistry
Rule application
III (Difficulty level is high because of shear ignorance of such topics)