What is the weight in g of 2.01 times 10^{23} | vedclass

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(A) The number of moles is calculated as: $n = \frac{	ext{Number of molecules}}{N_A} = \frac{2.01 	imes 10^{23}}{6.022 	imes 10^{23}} \approx \frac

Vedclass · Accepted Answer

$9.3$

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(A) The number of moles is calculated as: $n = \frac{	ext{Number of molecules}}{N_A} = \frac{2.01 	imes 10^{23}}{6.022 	imes 10^{23}} \approx \frac{1}{3} 	ext{ mol}$.The molar mass of $CO$ is $12 + 16 = 28 	ext{ g/mol}$.Weight = $	ext{moles} 	imes 	ext{molar mass} = \frac{1}{3} 	imes 28 = 9.33 	ext{ g}$.Rounding to the nearest option,the answer is $9.3 	ext{ g}$.

What is the weight in $g$ of $2.01 \times 10^{23}$ molecules of $CO$?

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