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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Solve the problem: The natural isotopic abundance of 10B is 19.60% and 11B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.

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प्रश्न

Solve the problem:

The natural isotopic abundance of 10B is 19.60% and 11B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.

संख्यात्मक
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उत्तर

Isotope % Abundance Mass (u)
¹⁰B 19.60% 10.13 u
¹¹B 80.40% 11.009 u

Convert % to fractional abundance

For ¹⁰B: 19.60% = 0.1960

For ¹¹B: 80.40% = 0.8040

Apply formula for average atomic mass

Average atomic mass = (f1​ × m1​) + (f2​ × m2​)

= (0.1960 × 10.13) + (0.8040 × 11.009)

0.1960 × 10.13 = 1.98648

0.8040 × 11.009 = 8.84724

1.98648 + 8.84724 = 10.83372

Average atomic mass of boron = 10.83u (approx.)​

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पाठ 1: Some Basic Concepts of Chemistry - Exercises [पृष्ठ १२]

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बालभारती Chemistry [English] Standard 11 Maharashtra State Board
पाठ 1 Some Basic Concepts of Chemistry
Exercises | Q 4. (G) | पृष्ठ १२

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