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Some Basic Concepts of Chemistry question

2010 · Shift 1 · Q14
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Some Basic Concepts of Chemistry question

2010 · Shift 1 · Q14

JEE AdvancedChemistrySome Basic Concepts of ChemistryNumerical+2 / −1
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
Numerical answer
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Correct answer: 3

  1. Given titre values

    25.2 mL,25.25 mL,25.0 mL25.2\ \text{mL},\quad 25.25\ \text{mL},\quad 25.0\ \text{mL}25.2 mL,25.25 mL,25.0 mL

  2. Find the average titre value

    First add the three readings:

    25.2+25.25+25.0=75.4525.2 + 25.25 + 25.0 = 75.4525.2+25.25+25.0=75.45

    Now divide by 333:

    Average=75.453=25.15 mL\text{Average} = \frac{75.45}{3} = 25.15\ \text{mL}Average=375.45​=25.15 mL

  3. Apply significant figure rule

    The readings are obtained from different burettes, so the least precise reading limits the reported precision.

    • 25.225.225.2 has 3 significant figures
    • 25.2525.2525.25 has 4 significant figures
    • 25.025.025.0 has 3 significant figures

    Hence, the average should be reported with 3 significant figures.

  4. Round the average to 3 significant figures

    25.15≈25.225.15 \approx 25.225.15≈25.2

    Thus, the average titre value has 3 significant figures.

  5. Final answer

    3\boxed{3}3​

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