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
Answer: 3
Solution
The three titre values are:
Average = (25.2 + 25.25 + 25.0) / 3 = 75.45 / 3 = 25.15 mL
The result should be reported to the least precise measurement. Since the least precise value (25.2 and 25.0) has 1 decimal place, the average should be rounded to 1 decimal place: 25.2 mL.
However, considering significant figures in the average:
25.15 mL has 4 significant figures, but should be rounded to match the precision of the least precise measurement.
The answer 25.2 mL has 3 significant figures.
- 25.2 mL (3 significant figures)
- 25.25 mL (4 significant figures)
- 25.0 mL (3 significant figures)
Average = (25.2 + 25.25 + 25.0) / 3 = 75.45 / 3 = 25.15 mL
The result should be reported to the least precise measurement. Since the least precise value (25.2 and 25.0) has 1 decimal place, the average should be rounded to 1 decimal place: 25.2 mL.
However, considering significant figures in the average:
25.15 mL has 4 significant figures, but should be rounded to match the precision of the least precise measurement.
The answer 25.2 mL has 3 significant figures.
JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer