Back to Directory
NEET CHEMISTRYEasy

An element X has the following isotopic composition: 200^{200}X: 90% 199^{199}X: 8.0% 202^{202}X: 2.0% The weighted average atomic mass of the naturally-occurring element X is closest to:

A

200 amu

B

201 amu

C

202 amu

D

199 amu

Step-by-Step Solution

The weighted average atomic mass is calculated by summing the products of the atomic mass of each isotope and its fractional abundance.

Average Atomic Mass=(Massi×Fractional Abundancei)\text{Average Atomic Mass} = \sum (\text{Mass}_i \times \text{Fractional Abundance}_i)

  1. Convert Percentages to Decimals: 200^{200}X: 90%=0.9090\% = 0.90 199^{199}X: 8.0%=0.088.0\% = 0.08
  • 202^{202}X: 2.0%=0.022.0\% = 0.02
  1. Calculate Weighted Contributions: 200×0.90=180.0200 \times 0.90 = 180.0 199×0.08=15.92199 \times 0.08 = 15.92
  • 202×0.02=4.04202 \times 0.02 = 4.04
  1. Sum the Contributions: 180.0+15.92+4.04=199.96 amu180.0 + 15.92 + 4.04 = 199.96 \text{ amu}

  2. Round to Nearest Integer: The value 199.96199.96 is closest to 200200 amu.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: CHEMISTRY Question for NEET | Sushrut