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Sub-Atomic Particles

From 200 mg o...

Question

From 200 mg of $C O_{2}, 10^{21}$ molecules are removed. How many moles of $C O_{2}$ are left?

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Solution

One mole of atoms $= 6.023 \times 10^{23}$ atoms = Gram atomic mass of element
44 grams of $C O_{2} = 1 m o l e C O_{2} = 6 \times 10^{23}$ molecules of $C O_{2}$
$200 m g = \frac{200 \times 1}{44 \times 1000} m o l e s = \frac{200 \times 10^{23}}{44 \times 1000} \times 6.023$ molecular of $C O_{2}$
$= 27.37 \times 10^{20}$ molecular of $C O_{2}$
$= 2.737 \times 10^{21}$ molecules of $C O_{2}$
molecules left $= 1.737 \times 10^{21}$ molecular of $C O_{2}$
moles of $C O_{2}$ left $= \frac{1.737 \times 10^{4}}{6.023 \times 10^{23}}$
$= 0.29 \times 10^{- 2}$ moles of $C O_{2}$
$= 2.9$ mill moles

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