A student performs a titration with different | vedclass

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(A) The given titre values are $25.2 \ mL$,$25.25 \ mL$,and $25.0 \ mL$. First,calculate the average: $	ext{Average} = \frac{25.2 + 25.25 + 25.0}{3}

Vedclass · Accepted Answer

$3$

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(A) The given titre values are $25.2 \ mL$,$25.25 \ mL$,and $25.0 \ mL$. First,calculate the average: $	ext{Average} = \frac{25.2 + 25.25 + 25.0}{3} = \frac{75.45}{3} = 25.15 \ mL$. According to the rules of significant figures in addition/subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. Here,$25.2$ and $25.0$ have one decimal place,so the sum $75.45$ should be rounded to one decimal place,which is $75.5$. Then,$75.5 / 3 = 25.166...$,which rounds to $25.2$. The value $25.2$ has $3$ significant figures.

$A$ student performs a titration with different burettes and finds titre values of $25.2 \ mL$,$25.25 \ mL$ and $25.0 \ mL$. The number of significant figures in the average titre value is

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