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$NEON has 2 ISOTOPES of differing mass and in different proportions shown below. Which option demonstrates how we calculate its relative atomic mass (A_(r)) in g/mol {}^20Ne:91% {}^22Ne:9% ((91+9)times (20+22))/(100)=42 ((91+20)times (9+22))/(100)=7.11 ((91times 9)+(20times 22))/(100)=12.6 ((91times 20)+(9times 22))/(100)=20.2$

NEON has 2 ISOTOPES of differing mass and in different proportions shown below. Which option demonstrates how we calculate its relative atomic mass (A_(r)) in g/mol {}^20Ne:91% {}^22Ne:9% ((91+9)times (20+22))/(100)=42 ((91+20)times (9+22))/(100)=7.11 ((91times 9)+(20times 22))/(100)=12.6 ((91times 20)+(9times 22))/(100)=20.2

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PiperExpert · Tutor for 3 years

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# Explanation: ## Step 1: Firstly, recognise we are given two isotopes of Neon (Ne), one with atomic mass 20 (Isotope 20Ne) being 91% abundant and another with atomic mass 22 (Isotope 22Ne) being 9% abundant. ## Step 2: The proper formula to calculate the relative atomic mass $ (A_r) $ is weighted average calculation based on the individual isotopes' relative abundance. This is done by multiplying the abundance by the atomic mass and summarize the results of all Isotopes. ### Formula: $ A_r = \sum \text{iso}_\text{i} (\text{Abundance}_\text{i} × \text{Mass}_\text{i}) $ * The above calculation should be performed for all the isotopes, i.e 'i'. ## Step 3: Replace values of the atomic mass and abundance of each considered Neon isotope into the formula. # Answer : Replacing the values into the formula we'll have: $ A_r = (0.91 × 20) + (0.09 × 22) $ So let's compute this: $ A_r = 18.2 + 1.98 = 20.18 g/mol $ Thus, the relative atomic mass (or 'molar mass') of Neon (Ne) is $ 20.18 g/mol $.

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