JEE Advance - Chemistry (2010 - Paper 1 Offline - No. 14)

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

Explanation

To find the average titre value, first add up the three measurements provided and then divide by the number of measurements.

The sum of the measurements is:

$$ 25.2 \, \text{mL} + 25.25 \, \text{mL} + 25.0 \, \text{mL} = 75.45 \, \text{mL} $$

Since there are three measurements, divide this sum by 3 to calculate the average:

$$ \text{Average titre value} = \frac{75.45 \, \text{mL}}{3} = 25.15 \, \text{mL} $$

When reporting the average, we must consider the significant figures of the original measurements. The number of significant figures is determined by the least precise measurement, which in this case is 25.0 mL with three significant figures. Therefore, we should report the average value to three significant figures as well.

The average value of 25.15 mL has four significant figures, so we need to round it to three significant figures. However, this is slightly tricky since 25.15 already appears to be rounded to four significant figures. We should consult the original measurements to decide on the best course of action.

Looking at the individual measurements (25.2, 25.25, and 25.0), we should consider the lowest decimal place which they all have in common, which is the first decimal place. The third measurement has no second decimal place, indicating its level of precision. Thus, the number of significant figures for the average titre value should be in line with this level of precision. Since the average calculated is 25.15, when we adjust to the first decimal place for consistent significant figures, the average is 25.1 mL with three significant figures.

$$ \text{Corrected Average titre value} = 25.1 \, \text{mL} $$

Therefore, the number of significant figures in the average titre value is three: 25.1 mL.

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